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IB Maths Applications & Interpretation · SL · Topic 3

Geometry and trigonometry

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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 16 October 2026 · new questions added continuouslyFree access until 31 December 2026How every question is checked →

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+ Add to test✓ In your test1IBMI-032-00-01Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy5 marks

A community centre is installing a straight wheelchair ramp. The ramp rises a vertical height of 0.45 m over a horizontal distance of 5.2 m.

(a)

Find the angle that the ramp makes with the horizontal.

(2)

(b)

Find the length of the sloping surface of the ramp.

(2)

(c)

Local building guidance states that a ramp of this kind should make an angle of no more than 4.8° with the horizontal. State, with a reason, whether this ramp meets the guidance.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses tan θ = 0.45 / 5.2 M1	✓ 1
θ = 4.95° A1	✓ 1
Part (b)
Uses Pythagoras: √(5.2² + 0.45²) M1	✓ 1
5.22 m A1	✓ 1
Part (c)
No, because 4.95° is greater than 4.8° (follow through from part (a)) R1	✓ 1

Exam tip: Make sure your calculator is in degree mode before using inverse trigonometric functions; an answer of 0.0863 here would be the angle in radians.

+ Add to test✓ In your test2IBMI-032-00-02Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy6 marks

Amara stands on level ground 35 m from the foot of a vertical clock tower. Her eyes are 1.6 m above the ground. She measures the angle of elevation of the top of the tower as 38°.

(a)

Find the vertical distance from Amara's eye level to the top of the tower.

(2)

(b)

Hence find the height of the clock tower.

(1)

(c)

Amara walks directly towards the tower until the angle of elevation of the top is 55°. Find how far she walked.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses tan 38° = h / 35 M1	✓ 1
h = 27.3 m A1	✓ 1
Part (b)
27.345 + 1.6 = 28.9 m A1	✓ 1
Part (c)
New horizontal distance = 27.345 / tan 55° M1	✓ 1
= 19.1 m A1	✓ 1
Distance walked = 35 − 19.1 = 15.9 m A1	✓ 1

Common error: In part (c) the vertical distance in the triangle is still measured from eye level, so use 27.3 m, not the full tower height of 28.9 m.

+ Add to test✓ In your test3IBMI-032-00-03Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy5 marks

A delivery drone takes off from a depot P and flies in a straight line for 4.2 km on a bearing of 125° to a drop-off point D.

(a)

Find how far east of P the point D is.

(2)

(b)

Find how far south of P the point D is.

(2)

(c)

Write down the bearing on which the drone must fly to return directly from D to P.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
4.2 sin 125° (or 4.2 cos 35°) M1	✓ 1
3.44 km A1	✓ 1
Part (b)
4.2 cos 55° (or −4.2 cos 125°) M1	✓ 1
2.41 km A1	✓ 1
Part (c)
305° A1	✓ 1

Exam tip: A back bearing differs from the original bearing by exactly 180°; add 180° if the original is less than 180°, subtract if it is more.

+ Add to test✓ In your test4IBMI-032-00-04Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy6 marks

A triangular garden plot ABC has AB = 14 m, AC = 9.5 m and angle BAC = 72°.

(a)

Find the area of the plot.

(2)

(b)

Find the length BC.

(2)

(c)

The whole plot is to be covered with turf costing $6.40 per square metre. Turf can only be bought in whole square metres. Find the cost of the turf.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses ½ × 14 × 9.5 × sin 72° M1	✓ 1
63.2 m² A1	✓ 1
Part (b)
BC² = 14² + 9.5² − 2(14)(9.5)cos 72° M1	✓ 1
BC = 14.3 m A1	✓ 1
Part (c)
Rounds area up to 64 m² R1	✓ 1
64 × 6.40 = $409.60 A1	✓ 1

Common error: Rounding 63.2 m² to the nearest whole number gives 63 m², which would not cover the plot. In context the area must be rounded up.

+ Add to test✓ In your test5IBMI-032-06-01Volume and surface area in context Paper 1Easy5 marks

A farm stores rainwater in a closed cylindrical tank. The tank has internal radius 1.2 m and internal height 2.5 m.

(a)

Calculate the volume of the tank, in m³.

(2)

(b)

Write down the capacity of the tank in litres.

(1)

(c)

The empty tank is filled by a pump delivering 15 litres per minute. Find the time taken to fill the tank, giving your answer in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for substituting into $V=\pi r^2 h$: $\pi\times1.2^2\times2.5$	✓ 1
A1 for $V=11.3$ m³ (11.3097…)	✓ 1
Part (b)
A1 for 11 300 litres, using 1 m³ = 1000 litres; accept 11 310 or 11 309.7; follow through from their answer to part (a)	✓ 1
Part (c)
M1 for dividing their capacity by the rate, e.g. $11309.7\div15$	✓ 1
A1 for 754 minutes (753.98…), which is about 12.6 hours; accept 753 minutes as follow-through from 11 300 litres	✓ 1

Exam tip: Keep the unrounded volume (11.3097…) in your calculator and use it in later parts. A rounded value such as 11 300 litres gives 753 minutes rather than 754, and rounding early can cost accuracy marks when no follow-through is allowed.

+ Add to test✓ In your test6IBMI-032-06-02Volume and surface area in context Paper 1Easy4 marks

A traffic cone is modelled as a right circular cone with an open base. The base has diameter 36 cm and the slant height of the cone is 70 cm.

(a)

Find the vertical height of the cone.

(2)

(b)

The curved surface of the cone is painted with reflective paint. Find the area to be painted.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for Pythagoras with radius 18: $h=\sqrt{70^2-18^2}$	✓ 1
A1 for $h=67.6$ cm (67.646…)	✓ 1
Part (b)
M1 for using $\pi r l$: $\pi\times18\times70$	✓ 1
A1 for 3960 cm² (3958.4…)	✓ 1

Common error: Using 36 as the radius, or putting the vertical height into $\pi r l$ in place of the slant height, are both frequent slips. The formula needs the radius and the slant height.

+ Add to test✓ In your test7IBMI-032-06-03Volume and surface area in context Paper 1Easy5 marks

The roof of a planetarium is a hemispherical dome of radius 9 m. The outside curved surface of the dome is to be painted. One litre of paint covers 12 m². Paint is sold only in 5-litre tins costing $38.50 each.

(a)

Calculate the area of the curved surface of the dome.

(2)

(b)

Find the number of tins of paint that must be bought.

(2)

(c)

Find the total cost of the paint.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for half the surface area of a sphere: $\frac{1}{2}\times4\pi\times9^2$	✓ 1
A1 for 509 m² (508.938…)	✓ 1
Part (b)
M1 for $508.938\div12=42.4$ litres, then dividing by 5 to get 8.48 tins	✓ 1
A1 for 9 tins, rounded up in context	✓ 1
Part (c)
A1 for $9\times38.50=$346.50$	✓ 1

Common error: Rounding 8.48 to 8 tins leaves part of the dome unpainted. In a context like this, always round up to a whole number of tins.

+ Add to test✓ In your test8IBMI-032-06-04Volume and surface area in context Paper 1Easy6 marks

A glass aquarium is a cuboid with no lid. Its internal dimensions are length 80 cm, width 35 cm and height 45 cm. Ignore the thickness of the glass.

(a)

Calculate the total area of glass used.

(2)

(b)

The aquarium is filled with water to 5 cm below the top. Find the volume of water in litres.

(2)

(c)

Find the length of the longest thin straight rod that can fit inside the empty aquarium.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for adding the base and four sides only: $80\times35+2(80\times45)+2(35\times45)$	✓ 1
A1 for 13 150 cm² (accept 13 200 cm²)	✓ 1
Part (b)
M1 for $80\times35\times40=112\,000$ cm³	✓ 1
A1 for 112 litres	✓ 1
Part (c)
M1 for the space diagonal: $\sqrt{80^2+35^2+45^2}$	✓ 1
A1 for 98.2 cm (98.234…)	✓ 1

Exam tip: The longest rod lies along the space diagonal, from one bottom corner to the opposite top corner. All three dimensions go into one square root.

+ Add to test✓ In your test9IBMI-032-05-01Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy4 marks

A wheelchair ramp outside a library rises a vertical height of 0.45 m over a horizontal distance of 5.4 m.

(a)

Find the angle that the ramp makes with the horizontal.

(2)

(b)

Find the length of the sloping surface of the ramp.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
tan θ = 0.45 / 5.4 M1 ; θ = 4.76° (4.7636…) A1	✓ 2
Part (b)
√(5.4² + 0.45²) or equivalent trig method M1 ; 5.42 m (5.4187…) A1	✓ 2

Exam tip: Check that your calculator is in degree mode before using any trigonometric function. A ramp angle near 0.08 is a sign it is in radians.

+ Add to test✓ In your test10IBMI-032-05-02Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy6 marks

A surveyor stands on level ground 85 m from the base of a vertical radio tower. Her eye level is 1.6 m above the ground. The angle of elevation from her eye to the top of the tower is 32°.

(a)

Find the height of the tower.

(3)

(b)

The surveyor walks directly towards the tower until the angle of elevation to the top is 50°. Find the distance she walks.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
tan 32° = h / 85 M1 ; h = 53.11 m above eye level A1 ; height = 53.11 + 1.6 = 54.7 m A1	✓ 3
Part (b)
d = 53.11 / tan 50° M1 ; d = 44.57 m A1 ; distance walked = 85 − 44.57 = 40.4 m A1	✓ 3

Common error: The 1.6 m eye height affects only the total tower height. In part (b) the right-angled triangle still uses 53.11 m, measured from eye level, not 54.7 m.

+ Add to test✓ In your test11IBMI-032-05-04Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy4 marks

A triangular vegetable plot has two straight sides of length 14 m and 9 m, and the angle between these sides is 72°.

(a)

Find the length of the third side of the plot.

(2)

(b)

Find the area of the plot.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
c² = 14² + 9² − 2(14)(9)cos 72° M1 ; c = 14.1 m (14.111…) A1	✓ 2
Part (b)
Area = ½ × 14 × 9 × sin 72° M1 ; 59.9 m² A1	✓ 2

Common error: The area formula ½ab sin C needs the angle between the two given sides. Here that is 72°, so the third side is not needed for the area.

+ Add to test✓ In your test12IBMI-032-01-02Volume and surface area in context Paper 1Easy6 marks

A party hat is a right circular cone with no base. The base radius is 9 cm and the vertical height is 20 cm.

(a)

Find the slant height of the hat.

(2)

(b)

Find the volume of air inside the hat.

(2)

(c)

Find the area of card needed to make the hat, ignoring any overlap.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$l=\sqrt{9^2+20^2}$ M1 ; l = 21.9 cm (21.932…) A1	✓ 2
Part (b)
$V=\frac{1}{3}\pi(9^2)(20)$ M1 ; V = 1700 cm³ (1696.5…) A1	✓ 2
Part (c)
Curved surface area $=\pi r l=\pi(9)(21.932…)$ M1 ; 620 cm² (620.10…) A1	✓ 2

Exam tip: The curved surface area formula uses the slant height l, not the vertical height. Keep the unrounded value of l in your calculator for the next part.

+ Add to test✓ In your test13IBMI-032-01-03Volume and surface area in context Paper 1Easy6 marks

A gym orders spherical exercise balls, each with diameter 65 cm. Treat each ball as a perfect sphere and ignore the thickness of the rubber.

(a)

Find the volume of air in one ball, giving your answer in m³.

(2)

(b)

Find the surface area of one ball in m².

(2)

(c)

The rubber used costs 8.40 USD per m². Find the cost of the rubber needed to make 50 balls.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$V=\frac{4}{3}\pi(0.325)^3$ M1 ; 0.144 m³ (0.14379…) A1	✓ 2
Part (b)
$A=4\pi(0.325)^2$ M1 ; 1.33 m² (1.3273…) A1	✓ 2
Part (c)
1.3273… × 8.40 × 50 M1 ; 557.48 USD A1 (accept 557 or 557.5 USD; accept 558.60 USD from using 1.33)	✓ 2

Common error: Converting 65 cm to 0.65 m and then forgetting to halve it is a frequent slip. The radius is 0.325 m.

+ Add to test✓ In your test14IBMI-032-01-04Volume and surface area in context Paper 1Easy6 marks

A camping tent is a triangular prism. Each end is an isosceles triangle with base 2.4 m and vertical height 1.6 m. The tent is 3.0 m long. The tent has no fabric floor.

(a)

Find the volume of the tent.

(2)

(b)

Find the length of one sloping edge of a triangular end.

(2)

(c)

Find the total area of fabric used to make the two sloping sides and the two ends.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Area of end = ½ × 2.4 × 1.6 = 1.92 m² M1 ; V = 5.76 m³ A1	✓ 2
Part (b)
$\sqrt{1.2^2+1.6^2}$ M1 ; 2.0 m A1	✓ 2
Part (c)
2(2.0 × 3.0) + 2(1.92) M1 ; 15.8 m² (15.84) A1	✓ 2

Exam tip: Read the context carefully: the tent has no floor, so the rectangular base must not be included in the fabric area.

+ Add to test✓ In your test15IBMI-032-07-r2-01Voronoi diagrams Paper 1Easy6 marks

Two fire stations are located at A(2, 3) and B(8, 7) on a coordinate grid. Distances are in kilometres. The Voronoi diagram for the two stations has one edge, the perpendicular bisector of AB.

(a)

Find the coordinates of the midpoint of AB.

(1)

(b)

Find the equation of the perpendicular bisector of AB, giving your answer in the form $y = mx + c$.

(3)

(c)

A house is at H(4, 6). Find the distance from H to the nearest fire station.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(5, 5) A1	✓ 1
Part (b)
Gradient of AB $= \frac{7-3}{8-2} = \frac{2}{3}$ M1	✓ 1
Perpendicular gradient $= -\frac{3}{2}$ A1	✓ 1
$y - 5 = -1.5(x - 5)$ so $y = -1.5x + 12.5$ A1	✓ 1
Part (c)
Compares distances, e.g. HA $= \sqrt{13}$ and HB $= \sqrt{17}$, or uses the bisector to show H is in the cell of A M1	✓ 1
Nearest station is A, distance 3.61 km A1	✓ 1

Exam tip: A point lies in a site's Voronoi cell exactly when it is closer to that site than to any other, so comparing distances is a quick check on which side of an edge a point lies.

+ Add to test✓ In your test16IBMI-032-07-r2-02Voronoi diagrams Paper 1Easy6 marks

Three mobile phone masts are at P(0, 0), Q(6, 0) and R(0, 4), with coordinates in kilometres. A Voronoi diagram is drawn with the masts as sites.

(a)

Write down the equation of the perpendicular bisector of PQ and the equation of the perpendicular bisector of PR.

(2)

(b)

Hence write down the coordinates of the Voronoi vertex V.

(1)

(c)

Explain why the perpendicular bisector of QR must also pass through V.

(1)

(d)

Find the distance from V to each mast.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$x = 3$ A1	✓ 1
$y = 2$ A1	✓ 1
Part (b)
V(3, 2) A1	✓ 1
Part (c)
V is equidistant from P and Q and from P and R, so it is equidistant from Q and R and lies on the bisector of QR (or: the midpoint of QR is (3, 2)) R1	✓ 1
Part (d)
$\sqrt{3^2 + 2^2}$ M1	✓ 1
3.61 km A1	✓ 1

Common error: Writing $y = 3$ instead of $x = 3$ for the bisector of a horizontal segment. The bisector of a horizontal segment is a vertical line.

+ Add to test✓ In your test17IBMI-032-02-r2-01Voronoi diagrams Paper 1Easy6 marks

Two mobile phone masts, A and B, are located at A(2, 3) and B(8, 7) on a coordinate grid. Distances are in kilometres. A Voronoi diagram is used to model which mast each house connects to, assuming each house connects to its nearest mast.

(a)

Find the coordinates of the midpoint of AB.

(1)

(b)

Find the equation of the perpendicular bisector of AB. Give your answer in the form $y = mx + c$.

(3)

(c)

A house is located at H(4, 6). Determine which mast the house connects to, and find the distance from the house to this mast.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(5, 5) A1	✓ 1
Part (b)
Gradient of AB = $\frac{7-3}{8-2} = \frac{2}{3}$ A1	✓ 1
Perpendicular gradient = $-\frac{3}{2}$ M1	✓ 1
$y - 5 = -\frac{3}{2}(x - 5)$, so $y = -1.5x + 12.5$ A1	✓ 1
Part (c)
HA = $\sqrt{2^2+3^2} = \sqrt{13}$ and HB = $\sqrt{4^2+1^2} = \sqrt{17}$, compared M1	✓ 1
Mast A is nearest, distance 3.61 km A1	✓ 1

Exam tip: The perpendicular bisector always passes through the midpoint, and its gradient is the negative reciprocal of the gradient joining the two sites.

+ Add to test✓ In your test18IBMI-032-02-r2-02Voronoi diagrams Paper 1Easy5 marks

Three fire stations are located at P(0, 0), Q(10, 0) and R(0, 6) on a map grid, where units are kilometres. A Voronoi diagram is drawn with the fire stations as sites.

(a)

Write down the equation of the perpendicular bisector of (i) PQ; (ii) PR.

(2)

(b)

Hence write down the coordinates of the vertex of the Voronoi diagram.

(1)

(c)

A fire breaks out at this vertex. Find the distance from the fire to the nearest fire station.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(i) $x = 5$ A1	✓ 1
(ii) $y = 3$ A1	✓ 1
Part (b)
(5, 3) A1	✓ 1
Part (c)
Distance = $\sqrt{5^2 + 3^2}$ M1	✓ 1
= 5.83 km (equidistant from all three stations) A1	✓ 1

Exam tip: A Voronoi vertex is equidistant from the three sites whose cells meet there, so you only need one distance calculation.

+ Add to test✓ In your test19IBMI-032-07-r1-02Voronoi diagrams Paper 1Easy8 marks

Three fire stations are located at F₁(0, 0), F₂(10, 0) and F₃(4, 8), where distances are in kilometres. A Voronoi diagram is constructed with the three fire stations as sites.

(a)

Write down the equation of the perpendicular bisector of F₁F₂.

(1)

(b)

Show that the perpendicular bisector of F₁F₃ has equation y = −0.5x + 5.

(3)

(c)

Hence find the coordinates of the vertex V of the Voronoi diagram.

(2)

(d)

Find the distance from V to F₁, and interpret the significance of V in context.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
x = 5 A1	✓ 1
Part (b)
Midpoint of F₁F₃ = (2, 4) A1	✓ 1
Gradient of F₁F₃ = 8/4 = 2, so perpendicular gradient = −0.5 M1	✓ 1
y − 4 = −0.5(x − 2), leading to y = −0.5x + 5 A1	✓ 1
Part (c)
Substituting x = 5: y = −0.5(5) + 5 M1	✓ 1
V(5, 2.5) A1	✓ 1
Part (d)
VF₁ = √(5² + 2.5²) = 5.59 km A1	✓ 1
V is equidistant (5.59 km) from all three fire stations, so a fire at V is equally close to every station. Accept: V is the point in the region farthest from its nearest station. R1	✓ 1

Common error: A Voronoi vertex is equidistant from three sites, not just two. Check your vertex by confirming that its distance to the third site matches. A mismatch shows an algebra slip.

+ Add to test✓ In your test20IBMI-032-02-r1-04Voronoi diagrams Paper 1Easy7 marks

Three weather stations are located on a coordinate grid where 1 unit represents 1 km. The rainfall recorded at each station during one week is shown.
Station A (1, 5): 42 mm
Station B (7, 1): 35 mm
Station C (9, 8): 51 mm
A Voronoi diagram is drawn with the three stations as sites. The rainfall at a farm X(5, 4) is to be estimated using nearest-neighbour interpolation.

(a)

Calculate the distance from X to each of the three stations.

(3)

(b)

Hence state which Voronoi cell contains X and estimate the rainfall at X.

(2)

(c)

A rain gauge at the farm actually recorded 39 mm. Taking this as the exact value, calculate the percentage error in the estimate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Use of distance formula, e.g. XA = $\sqrt{4^2 + 1^2}$ M1	✓ 1
XA = $\sqrt{17}$ = 4.12 km, XB = $\sqrt{13}$ = 3.61 km A1	✓ 1
XC = $\sqrt{32}$ = 5.66 km A1	✓ 1
Part (b)
X is closest to B, so X lies in the cell of B R1	✓ 1
Estimated rainfall = 35 mm A1	✓ 1
Part (c)
$\frac{\|35 - 39\|}{39} \times 100$ M1	✓ 1
= 10.3% A1	✓ 1

Common error: Nearest-neighbour interpolation uses only the value at the site whose cell contains the point; do not average the values from several stations.

+ Add to test✓ In your test21IBMI-032-02-r1-01Voronoi diagrams Paper 1Easy7 marks

A town council models two public libraries as points on a coordinate grid, where 1 unit represents 1 km. Library A is at (2, 3) and library B is at (8, 7). The council draws a Voronoi diagram using A and B as sites.

(a)

Find the coordinates of the midpoint of AB.

(1)

(b)

Find the equation of the edge of the Voronoi diagram that separates the cells of A and B. Give your answer in the form $y = mx + c$.

(3)

(c)

A student lives at P(6, 4). Determine which library is nearer to P, and find the distance from P to that library.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(5, 5) A1	✓ 1
Part (b)
Gradient of AB = $\frac{7-3}{8-2} = \frac{2}{3}$ M1	✓ 1
Gradient of perpendicular bisector = $-\frac{3}{2}$ M1	✓ 1
$y - 5 = -1.5(x - 5)$ ⇒ $y = -1.5x + 12.5$ A1	✓ 1
Alternative: set $(x-2)^2+(y-3)^2=(x-8)^2+(y-7)^2$ (M1), simplify to $12x+8y=100$ (M1), giving $y=-1.5x+12.5$ (A1).	—
Part (c)
Valid method to decide which library is nearer M1 . Either find PA = $\sqrt{4^2+1^2} = \sqrt{17}$ ≈ 4.12 km and PB = $\sqrt{2^2+3^2} = \sqrt{13}$ and compare them. Or substitute x = 6 into the edge equation to get y = 3.5 < 4, so P lies on the same side of the edge as B.	✓ 1
P is nearer library B (P lies in B's cell) A1	✓ 1
Distance PB = $\sqrt{13}$ ≈ 3.61 km A1	✓ 1

Exam tip: A Voronoi edge between two sites is always the perpendicular bisector of the segment joining them: use the midpoint and the negative reciprocal of the gradient.

+ Add to test✓ In your test22IBMI-032-07-r1-01Voronoi diagrams Paper 1Easy6 marks

Two mobile phone masts, A and B, serve a rural area. On a map with coordinates in kilometres, A is at (2, 3) and B is at (8, 7). A Voronoi diagram is drawn with A and B as its sites.

(a)

Find the coordinates of the midpoint of AB.

(1)

(b)

Find the equation of the edge of the Voronoi diagram between the cells of A and B. Give your answer in the form y = mx + c.

(3)

(c)

A hiker stands at point P(6, 3). Determine which mast is closer to the hiker and find the distance from P to that mast.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint = ((2 + 8)/2, (3 + 7)/2) = (5, 5) A1	✓ 1
Part (b)
Gradient of AB = (7 − 3)/(8 − 2) = 2/3 M1	✓ 1
Gradient of perpendicular bisector = −3/2 M1	✓ 1
y − 5 = −1.5(x − 5), so y = −1.5x + 12.5 A1	✓ 1
Part (c)
Valid method to decide which mast is closer: either PA = √((6 − 2)² + (3 − 3)²) = 4 and PB = √((6 − 8)² + (3 − 7)²) = √20 ≈ 4.47, or at x = 6 the edge has y = 3.5, so P (with y = 3) lies on A's side of the edge M1	✓ 1
Mast A is closer, at a distance of 4 km A1	✓ 1

Exam tip: A Voronoi edge between two sites is always part of their perpendicular bisector. It passes through the midpoint and has gradient equal to the negative reciprocal of the gradient joining the sites.

+ Add to test✓ In your test23IBMI3S-3-05-01Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy4 marks

A wheelchair ramp rises 0.75 m above level ground. Its sloping surface makes an angle of 4.8° with the horizontal.

(a)

Calculate the length of the sloping surface of the ramp.

(2)

(b)

Calculate the horizontal distance covered by the ramp.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses $\sin 4.8^\circ=\dfrac{0.75}{L}$ or equivalent	✓ 1
A1 $L=8.96$ m	✓ 1
Part (b)
M1 uses $\tan 4.8^\circ=\dfrac{0.75}{x}$ (or Pythagoras with the answer to (a)) – ECF from (a) if Pythagoras is used	✓ 1
A1 $x=8.93$ m	✓ 1

Exam tip: Sketch the right-angled triangle and label the side opposite the angle before choosing sin, cos or tan.

+ Add to test✓ In your test24IBMI3S-3-05-02Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

A ferry leaves port A and sails 12 km on a bearing of 070° to a buoy B. It then sails 9 km on a bearing of 160° to an island C.

(a)

Show that angle ABC = 90°.

(2)

(b)

Calculate the direct distance AC.

(2)

(c)

Find the bearing of C from A.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of A from B is 070° + 180° = 250°	✓ 1
A1 angle ABC = 250° − 160° = 90° (AG)	✓ 1
Part (b)
M1 uses Pythagoras: $AC^2=12^2+9^2$	✓ 1
A1 $AC=15$ km	✓ 1
Part (c)
M1 $\tan(\angle BAC)=\dfrac{9}{12}$ giving angle $36.9^\circ$	✓ 1
A1 bearing = 070° + 36.9° = 107° (106.9°)	✓ 1

Common error: Measuring a bearing from the wrong point: the bearing of A from B is the reverse bearing, found by adding or subtracting 180°.

+ Add to test✓ In your test25IBMI3S-3-05-03Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy4 marks

A cylindrical rainwater tank has internal radius 1.4 m and height 2.2 m.

(a)

Calculate the volume of the tank, in m³.

(2)

(b)

The tank is filled from empty by a pump delivering 40 litres per minute. Find the time taken, to the nearest minute. (1 m³ = 1000 litres)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $V=\pi\times1.4^2\times2.2$	✓ 1
A1 $V=13.5$ m³	✓ 1
Part (b)
M1 converts to litres and divides: $\dfrac{13547}{40}$ (using unrounded volume)	✓ 1
A1 339 minutes	✓ 1

Exam tip: Convert units before dividing: a volume in m³ must become litres to match a rate in litres per minute.

+ Add to test✓ In your test26IBMI3S-3-05-04Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

Three air-quality sensors are placed on a map where 1 unit = 1 km: A(1, 1), B(9, 1) and C(3, 7). A Voronoi diagram is drawn so that each cell contains the points closest to one sensor.

(a)

Find the equation of the perpendicular bisector of AB.

(2)

(b)

Find the equation of the perpendicular bisector of AC.

(2)

(c)

Hence find the coordinates of the Voronoi vertex where the three cells meet.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AB is (5, 1) and AB is horizontal	✓ 1
A1 $x=5$	✓ 1
Part (b)
M1 gradient of AC = 3, so perpendicular gradient $-\frac13$ through (2, 4)	✓ 1
A1 $y=-\dfrac{x}{3}+\dfrac{14}{3}$ (accept equivalent)	✓ 1
Part (c)
M1 substitutes $x=5$ into the equation from (b)	✓ 1
A1 (5, 3)	✓ 1

Exam tip: A Voronoi edge lies on the perpendicular bisector of two sites, so you need the midpoint and the negative reciprocal gradient.

+ Add to test✓ In your test27IBMI3S-3-05b-01Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy5 marks

A wind turbine tower stands on level ground. From a point P, 120 m from the base of the tower, the angle of elevation of the hub (the top of the tower) is 33°.

(a)

Calculate the height of the hub above the ground.

(2)

(b)

A blade of length 45 m points vertically upwards from the hub. Find the angle of elevation of the tip of this blade from P.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use tan 33° = h/120	✓ 1
A1 h = 77.9 m	✓ 1
Part (b)
M1 height of tip = 77.9 + 45 = 122.9 m (or unrounded)	✓ 1
M1 attempt to use tan⁻¹ of (height of tip ÷ 120)	✓ 1
A1 45.7°	✓ 1

Common error: Using the hub height instead of the height of the blade tip in the second angle.

+ Add to test✓ In your test28IBMI3S-3-05b-02Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

A maintenance boat leaves harbour H and travels 18 km on a bearing of 072° to turbine T₁. It then travels 12 km on a bearing of 162° to turbine T₂.

(a)

Find the distance HT₂.

(3)

(b)

Find the bearing of T₂ from H, to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 back bearing of H from T₁ is 252°, so angle HT₁T₂ = 252° − 162°	✓ 1
A1 angle HT₁T₂ = 90°, so Pythagoras applies	✓ 1
A1 HT₂ = √(18² + 12²) = 21.6 km	✓ 1
Part (b)
M1 attempt to use tan⁻¹(12/18)	✓ 1
A1 angle T₁HT₂ = 33.7°	✓ 1
A1 bearing = 72° + 33.7° = 106° ≈ 106°	✓ 1

Exam tip: Draw a north line at each point and mark the angles from bearings before you use a triangle rule.

+ Add to test✓ In your test29IBMI3S-3-05b-03Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy4 marks

A solar water-heating system stores hot water in a cylindrical tank of radius 0.45 m and height 1.2 m.

(a)

Calculate the volume of the tank in m³.

(2)

(b)

The tank is filled at a rate of 15 litres per minute. Find the time taken to fill it completely, giving your answer in minutes. (1 m³ = 1000 litres)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use V = πr²h	✓ 1
A1 V = 0.763 m³	✓ 1
Part (b)
M1 volume in litres = 763 and divide by 15	✓ 1
A1 50.9 minutes	✓ 1

Common error: Forgetting to convert m³ to litres before dividing by the flow rate.

+ Add to test✓ In your test30IBMI3S-3-05b-04Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy5 marks

Two vaccination centres, V₁ and V₂, are located on a map grid with units in km at V₁(1, 2) and V₂(7, 6). People are assumed to attend the nearest centre.

(a)

Find the equation of the boundary between the regions served by the two centres, giving your answer in the form y = mx + c.

(3)

(b)

A care home is at (3, 5). Determine which centre it should use and calculate its distance from that centre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 midpoint (4, 4)	✓ 1
M1 gradient of V₁V₂ = 4/6 and perpendicular gradient −3/2	✓ 1
A1 y = −1.5x + 10	✓ 1
Part (b)
M1 attempt to compare distances (or the side of the boundary) for (3, 5)	✓ 1
A1 V₁ is nearer, distance √13 = 3.61 km	✓ 1

Exam tip: The Voronoi boundary between two sites is the perpendicular bisector of the line joining them: use the midpoint and the negative reciprocal gradient.

+ Add to test✓ In your test31IBMI3S-3-01r6-01Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy4 marks

A wind turbine stands on level ground. An observer stands at a point on the ground 85 m from the base of the tower. The angle of elevation of the centre of the hub from the observer is 38°. Ignore the height of the observer.

(a)

Calculate the height of the hub above the ground. Give your answer to 3 significant figures.

(2)

(b)

Calculate the straight-line distance from the observer to the centre of the hub. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use $\tan 38° = \frac{h}{85}$	✓ 1
A1 $h = 66.4$ m	✓ 1
Part (b)
M1 attempt to use $\cos 38° = \frac{85}{d}$ or Pythagoras with the unrounded height	✓ 1
A1 $108$ m	✓ 1

Exam tip: Check your calculator is in degree mode before using any trigonometric function; a quick sense check is that the hypotenuse must be longer than both other sides.

+ Add to test✓ In your test32IBMI3S-3-01r6-02Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy7 marks

A service boat leaves a harbour H and sails 6.4 km on a bearing of 070° to a turbine T. It then sails 4.8 km on a bearing of 160° to a second turbine U.

(a)

Show that angle HTU = 90°.

(2)

(b)

Find the direct distance HU.

(2)

(c)

Find the bearing of U from H, correct to 0.1°.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of H from T is $070° + 180° = 250°$ (or equivalent angle reasoning)	✓ 1
A1 $250° - 160° = 90°$	✓ 1
Part (b)
M1 attempt to use Pythagoras, $\sqrt{6.4^2 + 4.8^2}$	✓ 1
A1 HU = 8.0 km	✓ 1
Part (c)
M1 attempt to find angle UHT using $\tan(\text{UHT}) = \frac{4.8}{6.4}$	✓ 1
A1 angle UHT = 36.9°	✓ 1
A1 bearing = $70° + 36.9° = 106.9°$	✓ 1

Common error: Bearings are measured clockwise from north; to find the bearing of H from T, add or subtract 180° before using any angle in the triangle.

+ Add to test✓ In your test33IBMI3S-3-01r6-03Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy7 marks

Two libraries are at the points L₁(2, 3) and L₂(10, 7) on a map, where each unit is 1 km. A Voronoi diagram is drawn so that each region contains the points nearer to one library than to the other.

(a)

Find the coordinates of the midpoint of L₁L₂.

(1)

(b)

Find the gradient of the edge separating the two regions.

(2)

(c)

Find the equation of the edge, giving your answer in the form $y = mx + c$.

(2)

(d)

A house is at H(5, 9). Find the distance from H to each library and state which library is nearer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 (6, 5)	✓ 1
Part (b)
M1 gradient of L₁L₂ $= \frac{7-3}{10-2} = 0.5$	✓ 1
A1 perpendicular gradient $= -2$	✓ 1
Part (c)
M1 substitute (6, 5) into $y - 5 = -2(x - 6)$	✓ 1
A1 $y = -2x + 17$	✓ 1
Part (d)
M1 attempt to use the distance formula for at least one library	✓ 1
A1 H to L₁ $= 6.708$ km, H to L₂ $= 5.385$ km; L₂ is nearer	✓ 1

Exam tip: The edge of a Voronoi region between two sites is the perpendicular bisector of the line joining them, so it always passes through their midpoint.

+ Add to test✓ In your test34IBMI3S-3-01r6-04Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy4 marks

The base of a wind turbine sits on a solid concrete foundation in the shape of a cylinder of radius 8.5 m and depth 2.5 m.

(a)

Calculate the volume of concrete in the foundation, to 3 significant figures.

(2)

(b)

Concrete costs £116 per cubic metre. Using the unrounded volume, find the cost of the concrete, to the nearest £.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use $V = \pi r^2 h$	✓ 1
A1 $567$ m³	✓ 1
Part (b)
M1 multiply volume by 116	✓ 1
A1 £65,824	✓ 1

Common error: Round only at the end: using a rounded volume in part (b) can change the final pound value.

+ Add to test✓ In your test35IBMI3S-3-02r8-01Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Easy6 marks

Three libraries are modelled as points on a map with coordinates in km: A(2, 1), B(8, 1) and C(5, 7). A Voronoi diagram is drawn so that each cell contains the points closest to one library.

(a)

Find the equation of the perpendicular bisector of AB.

(2)

(b)

Find the coordinates of the vertex where the three edges of the diagram meet.

(3)

(c)

A new house is built at the vertex found in (b). Find its distance from the nearest library.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AB is (5, 1) and AB is horizontal	✓ 1
A1 x = 5	✓ 1
Part (b)
M1 midpoint of AC = (3.5, 4) and gradient of AC = 2	✓ 1
M1 perpendicular bisector y − 4 = −½(x − 3.5), with x = 5 substituted	✓ 1
A1 (5, 3.25)	✓ 1
Part (c)
A1 3.75 km (equal distance to all three libraries)	✓ 1

Common error: The perpendicular bisector passes through the midpoint and has gradient equal to the negative reciprocal of the segment's gradient.

+ Add to test✓ In your test36IBMI3S-3-02r8-05Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Easy6 marks

Maria stands 85 m from the base of a wind turbine tower on level ground. Her eye is 1.6 m above the ground. The angle of elevation of the hub (the top of the tower) from her eye is 24.5°. The blades are 22 m long, and one blade points vertically upwards from the hub.

(a)

Find the height of the hub above the ground, to 3 significant figures.

(3)

(b)

Find the angle of elevation of the tip of the upward blade from Maria's eye, to 3 significant figures. Use the unrounded height from (a).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 vertical rise = 85 tan 24.5°	✓ 1
A1 38.7 m	✓ 1
A1 40.3 m (add eye height 1.6 m)	✓ 1
Part (b)
M1 height of tip above her eye = 40.3367… + 22 − 1.6 = 60.7367…	✓ 1
M1 tan⁻¹(60.7367… ÷ 85)	✓ 1
A1 35.5°	✓ 1

Common error: The angle of elevation is measured from the observer's eye, so subtract the eye height when forming the right-angled triangle for the tip.

+ Add to test✓ In your test37IBMI3S-3-02r8-09Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Easy6 marks

A swimmer starts at S on one shore of a lake and swims 400 m on a bearing of 130°. She then turns and swims 250 m on a bearing of 220° to reach F.

(a)

Explain why the two legs of the swim are at right angles.

(1)

(b)

Find the direct distance SF, to 3 significant figures.

(2)

(c)

Find the bearing of S from F, to the nearest tenth of a degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R1 the change of bearing is 220° − 130° = 90°	✓ 1
Part (b)
M1 √(400² + 250²)	✓ 1
A1 472 m	✓ 1
Part (c)
M1 angle at S between first leg and SF = tan⁻¹(250 ÷ 400) = 32.0°	✓ 1
M1 bearing of F from S = 130° + 32.0° = 162.0°	✓ 1
A1 bearing of S from F = 342.0°	✓ 1

Common error: A bearing is measured clockwise from north; to reverse a bearing add or subtract 180° so the answer lies between 0° and 360°.

+ Add to test✓ In your test38IBMI3S-3-02r8-10Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Easy7 marks

A pool is 20 m long and 8 m wide. The floor slopes uniformly, so the depth increases from 1.0 m at the shallow end to 2.8 m at the deep end. The pool is completely filled.

(a)

Calculate the volume of water in the pool.

(3)

(b)

The pool is filled at 0.8 m³ per minute. Find the time to fill it, in hours and minutes.

(2)

(c)

Chemical treatment costs €0.12 per m³ of water. Find the cost of treating a full pool.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area of side cross-section = ½ × (1.0 + 2.8) × 20 = 38 m²	✓ 1
M1 multiply by the width 8 m	✓ 1
A1 304 m³	✓ 1
Part (b)
M1 304 ÷ 0.8 = 380 minutes	✓ 1
A1 6 hours 20 minutes	✓ 1
Part (c)
M1 304 × 0.12	✓ 1
A1 €36.48	✓ 1

Exam tip: A prism's volume is cross-section area × length; here the cross-section is a trapezium, not a rectangle.

+ Add to test✓ In your test39IBMI3S-3-02r9-01Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy4 marks

A robotic arm used to sort fish crates has a rigid section of length 45 cm. The section is fixed at its lower end to a base and is tilted at 32° above the horizontal.

(a)

Calculate the vertical height of the free end of the section above its fixed end. Give your answer to 3 significant figures.

(2)

(b)

Calculate the horizontal distance between the two ends of the section. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 45 sin 32°	✓ 1
A1 23.8 cm	✓ 1
Part (b)
M1 45 cos 32°	✓ 1
A1 38.2 cm	✓ 1

Common error: Check that the calculator is in degree mode before using sin, cos or tan with an angle in degrees.

+ Add to test✓ In your test40IBMI3S-3-02r9-02Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

A fishing boat leaves harbour H and sails 12 km on a bearing of 070° to a point X. It then sails 9 km on a bearing of 160° to a point Y.

(a)

Show that angle HXY is 90°.

(2)

(b)

Find the distance HY.

(2)

(c)

Find the bearing of Y from H. Give your answer to the nearest 0.1°.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the bearing of H from X is 070° + 180° = 250° (back bearing)	✓ 1
A1 angle HXY = 250° − 160° = 90° (AG)	✓ 1
Part (b)
M1 use Pythagoras: HY² = 12² + 9²	✓ 1
A1 HY = 15 km	✓ 1
Part (c)
M1 angle XHY = tan⁻¹(9 ÷ 12) = 36.87…°	✓ 1
A1 bearing = 70° + 36.87…° = 106.9°	✓ 1

Common error: A bearing is measured clockwise from north; sketch the north lines at both points before finding any angles.

+ Add to test✓ In your test41IBMI3S-3-02r9-03Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

A fashion designer makes conical party hats from a single piece of fabric. Each hat has a base radius of 14 cm and a slant height of 30 cm. The fabric costs €0.0042 per cm². Use unrounded values in your calculations and ignore any overlap or waste.

(a)

Calculate the curved surface area of one hat. Give your answer to 3 significant figures.

(2)

(b)

Calculate the perpendicular height of the hat. Give your answer to 3 significant figures.

(2)

(c)

Find the cost of the fabric for one hat, to the nearest cent.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 14 × 30	✓ 1
A1 1320 cm²	✓ 1
Part (b)
M1 √(30² − 14²)	✓ 1
A1 26.5 cm	✓ 1
Part (c)
M1 multiply the curved surface area by 0.0042	✓ 1
A1 €5.54	✓ 1

Exam tip: The slant height is the length along the surface, so it is the hypotenuse of the right-angled triangle formed with the radius and the height.

+ Add to test✓ In your test42IBMI3S-3-02r9-04Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

Three fish farms are located at A(2, 2), B(10, 2) and C(6, 8), where the coordinates are in kilometres from a reference point. A Voronoi diagram is drawn for the three farms.

(a)

Find the equation of the perpendicular bisector of AB.

(2)

(b)

Find the equation of the perpendicular bisector of AC. Give your answer in the form y = mx + c.

(2)

(c)

Find the coordinates of the Voronoi vertex where the cells of A, B and C meet. Give the y-coordinate to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AB is (6, 2) and AB is horizontal	✓ 1
A1 x = 6	✓ 1
Part (b)
M1 gradient of AC = 6 ÷ 4 = 1.5, so the perpendicular gradient is −2/3, through the midpoint (4, 5)	✓ 1
A1 y = −(2/3)x + 23/3	✓ 1
Part (c)
M1 substitute x = 6 into the equation from part (b)	✓ 1
A1 (6, 3.67)	✓ 1

Exam tip: A Voronoi vertex is the point where the perpendicular bisectors meet, and it is equidistant from the three nearest sites.

+ Add to test✓ In your test43IBMI3S-3-01r11-01Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy4 marks

A wind farm has a cylindrical oil storage tank with internal radius 1.8 m and internal height 4.5 m.

(a)

Calculate the volume of the tank, in m³, correct to 3 significant figures.

(2)

(b)

The tank is empty. Oil is pumped in at a constant rate of 0.35 m³ per minute. Find the time taken to fill the tank completely, in minutes, correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
V = π × 1.8² × 4.5	✓ 1
V = 45.8 m³	✓ 1
Part (b)
time = volume ÷ rate = 45.8 ÷ 0.35	✓ 1
= 130.9 minutes	✓ 1

Exam tip: Keep the unrounded volume on your calculator for the next part so that rounding does not shift the final digit.

+ Add to test✓ In your test44IBMI3S-3-01r11-05Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy6 marks

Two swimming pools are located at A(1, 2) and B(7, 6) on a map where each unit represents 1 km. The region closer to A than to B is separated from the region closer to B than to A by a straight boundary.

(a)

Find the coordinates of the midpoint of AB.

(1)

(b)

Find the equation of the boundary, giving your answer in the form y = mx + c.

(3)

(c)

A house is at H(5, 3). Calculate the distance from H to each pool, correct to 2 decimal places, and state which pool is nearer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
midpoint = (4, 4)	✓ 1
Part (b)
gradient of AB = 4/6 = 2/3, so the boundary has gradient −3/2	✓ 1
passes through (4, 4): y − 4 = −1.5(x − 4)	✓ 1
y = −1.5x + 10	✓ 1
Part (c)
HA = √(4² + 1²) = 4.12 km, HB = √(2² + 3²) = 3.61 km	✓ 1
B is nearer	✓ 1

Common error: The boundary passes through the midpoint but is perpendicular to AB: use the negative reciprocal of the gradient of AB, not the gradient itself.

+ Add to test✓ In your test45IBMI3S-3-01r11-09Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy4 marks

From a point on level ground 85 m from the base of a wind turbine tower, the angle of elevation of the hub is 32°. When a blade points vertically upwards from the hub, the angle of elevation of the blade tip from the same point is 44°.

(a)

Calculate the height of the hub above the ground, correct to 3 significant figures.

(2)

(b)

Find the length of a blade, correct to 3 significant figures. Use unrounded values in your working.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
tan 32° = height ÷ 85	✓ 1
height = 53.1 m	✓ 1
Part (b)
height of tip = 85 tan 44° = 82.08 m	✓ 1
blade length = 82.08 − 53.11 = 29.0 m	✓ 1

Exam tip: Make sure your calculator is in degree mode, and keep unrounded values from earlier parts when a later part uses them.

+ Add to test✓ In your test46IBMI3S-3-02r13-01Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Easy4 marks

An air ambulance flies 40 km in a straight line from Northfield Hospital to a village on a bearing of 065°. Assume the ground is flat and that bearings are measured clockwise from north.

(a)

Calculate how far east of the hospital the village is. Give your answer to 3 significant figures.

(2)

(b)

Calculate how far north of the hospital the village is. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 east distance = 40 sin 65°	✓ 1
A1 36.3 km	✓ 1
Part (b)
M1 north distance = 40 cos 65°	✓ 1
A1 16.9 km	✓ 1

Common error: With bearings measured from north, the sine of the bearing gives the eastward component and the cosine gives the northward component; swapping them is a frequent slip.

+ Add to test✓ In your test47IBMI3S-3-02r13-02Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Easy5 marks

A beekeeper's triangular meadow has two straight boundaries of lengths 120 m and 95 m, and the angle between these boundaries is 68°.

(a)

Calculate the area of the meadow. Give your answer to 3 significant figures.

(3)

(b)

The beekeeper places 32 hives in the meadow, spaced so that each hive has the same area of the meadow. Using the unrounded area from part (a), find the area available per hive, to the nearest square metre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area = ½ × 120 × 95 × sin 68°	✓ 1
M1 correct substitution evaluated	✓ 1
A1 5280 m² (5284.9… unrounded)	✓ 1
Part (b)
M1 divide the unrounded area by 32	✓ 1
A1 165 m²	✓ 1

Exam tip: Use the formula ½ab sin C only when the angle lies between the two known sides.

+ Add to test✓ In your test48IBMI3S-3-02r13-03Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Easy5 marks

A beekeeper stores honey in a cylindrical tank of internal radius 0.35 m and internal height 0.90 m.

(a)

Calculate the volume of the tank in litres. (1 m³ = 1000 litres.) Give your answer to 3 significant figures.

(3)

(b)

The tank is filled to 80% of its capacity. Using the unrounded volume from part (a), find the volume of honey in the tank, to the nearest litre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume = π × 0.35² × 0.90	✓ 1
A1 0.346 m³	✓ 1
M1 convert to litres: 346 litres	✓ 1
Part (b)
M1 0.80 × the unrounded volume	✓ 1
A1 277 litres	✓ 1

Common error: Convert units at the end: 1 m³ is 1000 litres, not 100 litres.

+ Add to test✓ In your test49IBMI3S-3-02r13-04Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Easy6 marks

On a map of a region, the positions of two hospitals are H₁(1, 2) and H₂(7, 6), where one unit represents 1 km. A Voronoi diagram is drawn for these two sites.

(a)

Find the equation of the Voronoi edge (the perpendicular bisector of H₁H₂). Give your answer in the form y = mx + c.

(4)

(b)

An incident occurs at the point (5, 8). Determine which hospital is nearer to the incident and find the distance to it. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of H₁H₂	✓ 1
A1 midpoint (4, 4)	✓ 1
M1 gradient of H₁H₂ = 2/3, so the perpendicular gradient is −3/2	✓ 1
A1 y = −1.5x + 10	✓ 1
Part (b)
M1 distance to each hospital (√52 ≈ 7.21 km to H₁, and √8 to H₂)	✓ 1
A1 H₂ is nearer, at 2.83 km	✓ 1

Exam tip: A Voronoi edge between two sites is their perpendicular bisector: it passes through the midpoint and has gradient equal to the negative reciprocal.

+ Add to test✓ In your test50IBMI3S-3-02r15-01Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

Two libraries, A and B, are located at A(2, 1) and B(8, 5) on a map where each unit represents 1 km. A town planner uses a Voronoi diagram with these two sites.

(a)

Find the equation of the perpendicular bisector of AB, giving your answer in the form ax + by = c, where a, b and c are integers.

(4)

(b)

A new housing estate is at the point P(5, 6). Find the distance from P to each library, and hence state which library is nearer to P.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 midpoint of AB is (5, 3)	✓ 1
M1 gradient of AB = (5 − 1)/(8 − 2) = 2/3, so perpendicular gradient = −3/2	✓ 1
A1 y − 3 = −(3/2)(x − 5)	✓ 1
A1 3x + 2y = 21	✓ 1
Part (b)
M1 uses the distance formula for both libraries: PA = √34 = 5.83 km	✓ 1
A1 PB = √10 = 3.16 km, so library B is nearer	✓ 1

Exam tip: As a check, substitute the midpoint of the two sites into your equation of the bisector; it must satisfy the equation.

+ Add to test✓ In your test51IBMI3S-3-02r15-05Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy5 marks

A swimming pool is a prism of length 25 m and width 10 m. Its depth increases uniformly along the length, from 1.0 m at the shallow end to 2.5 m at the deep end, so each end-on side view of the length is a trapezium.

(a)

Calculate the area of the trapezium formed by the side of the pool (length 25 m, depths 1.0 m and 2.5 m).

(2)

(b)

Hence find the volume of water in the pool when it is completely full.

(1)

(c)

The pool is filled from empty by a pump that delivers 2.5 m³ of water per minute. Find the time taken to fill the pool completely, in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses 1/2 (1.0 + 2.5) × 25	✓ 1
A1 area = 43.75 m²	✓ 1
Part (b)
A1 volume = 43.75 × 10 = 437.5 m³	✓ 1
Part (c)
M1 time = volume ÷ rate = 437.5 ÷ 2.5	✓ 1
A1 175 minutes	✓ 1

Exam tip: Identify the prism's cross-section first (here, the trapezium side view); volume = cross-sectional area × length across the width.

+ Add to test✓ In your test52IBMI3S-3-02r15-09Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy5 marks

A student stands 74 m from the base of a wind turbine tower on level ground. Her eyes are 1.7 m above the ground. She measures the angle of elevation from her eyes to the hub at the top of the tower as 38°. Give answers to 3 significant figures.

(a)

Calculate the height of the hub above the level of the student's eyes.

(2)

(b)

Hence find the height of the hub above the ground.

(1)

(c)

The student walks directly towards the tower until the angle of elevation of the hub is 55°. Find her new horizontal distance from the base of the tower. Use the unrounded value of h from part (a).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses tan 38° = h/74	✓ 1
A1 h = 57.8 m	✓ 1
Part (b)
A1 57.82 + 1.7 = 59.5 m	✓ 1
Part (c)
M1 uses tan 55° = h/x	✓ 1
A1 x = 40.5 m	✓ 1

Common error: The angle of elevation is measured from the observer's eye level, so add the eye height at the end to obtain the height above ground.

+ Add to test✓ In your test53IBMI3S-3-02r15-14Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

A lifeguard sits on a chair with her eyes 2.4 m above the level of the water surface. She looks down at a swimmer at an angle of depression of 18°. Assume the swimmer and the base of the chair are at the same level.

(a)

Calculate the horizontal distance from the swimmer to the base of the chair, giving your answer to 3 significant figures.

(2)

(b)

A second swimmer is 12 m from the base of the chair, in the same direction as the first swimmer. Find the angle of depression from the lifeguard's eyes to this swimmer, giving your answer to 3 significant figures.

(2)

(c)

Find the straight-line distance from the lifeguard's eyes to the second swimmer, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 18° = 2.4/x	✓ 1
A1 x = 7.39 m	✓ 1
Part (b)
M1 tan θ = 2.4/12	✓ 1
A1 θ = 11.3°	✓ 1
Part (c)
M1 Pythagoras √(12² + 2.4²)	✓ 1
A1 distance = 12.2 m	✓ 1

Common error: The angle of depression is measured from the horizontal at the observer, so it equals the angle of elevation from the swimmer, not the angle between the line of sight and the vertical.

+ Add to test✓ In your test54IBMI3S-3-01r17-01Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy4 marks

A sports scientist uses a cylindrical drinks bottle with internal radius 3.5 cm and internal height 22 cm.

(a)

Calculate the volume of the bottle, in cm³, correct to 3 significant figures.

(2)

(b)

An athlete pours 600 cm³ of water into the empty bottle, which stands upright. Find the depth of the water, in cm, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of V = πr²h with r = 3.5 and h = 22	✓ 1
A1 V = 846.7 = 847 cm³	✓ 1
Part (b)
M1 600 = π(3.5)²h (or h = 600 ÷ (π × 3.5²))	✓ 1
A1 h = 15.6 cm	✓ 1

Exam tip: Keep the unrounded value on your calculator between steps and round only the final answer.

+ Add to test✓ In your test55IBMI3S-3-01r17-02Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy4 marks

In an astronomy exercise, the Moon is modelled as a sphere of radius 1737 km and the Earth as a sphere of radius 6371 km.

(a)

Calculate the surface area of the Moon. Give your answer in the form a × 10^k, where 1 ≤ a < 10 and k is an integer, with a correct to 3 significant figures, in km².

(2)

(b)

Find how many times larger the volume of the Earth is than the volume of the Moon. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of A = 4πr² with r = 1737	✓ 1
A1 A = 3.79 × 10⁷ km²	✓ 1
Part (b)
M1 (6371 ÷ 1737)³ or ratio of (4/3)π(6371)³ to (4/3)π(1737)³	✓ 1
A1 49.3	✓ 1

Exam tip: The ratio of volumes of similar solids is the cube of the ratio of their lengths, so the common factor 4π/3 cancels.

+ Add to test✓ In your test56IBMI3S-3-01r17-03Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy7 marks

Two sports-science testing centres are at A(1, 2) and B(7, 6), where the coordinates are in km. Each athlete is assigned to the nearer centre. The boundary between the two catchment areas is the perpendicular bisector of AB.

(a)

Find the coordinates of the midpoint of AB.

(2)

(b)

Find the equation of the boundary between the two catchment areas. Give your answer in the form y = mx + c.

(3)

(c)

An athlete lives at R(6, 4). Find the distance from R to the nearer centre, in km, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to average the coordinates, ((1 + 7)/2, (2 + 6)/2)	✓ 1
A1 (4, 4)	✓ 1
Part (b)
M1 gradient of AB = (6 − 2)/(7 − 1) = 2/3	✓ 1
M1 perpendicular gradient = −3/2 and use of the midpoint (4, 4) (ECF from (a))	✓ 1
A1 y = −1.5x + 10	✓ 1
Part (c)
M1 distance to A = √(5² + 2²) = 5.39 and distance to B = √(1² + 2²) (both attempted)	✓ 1
A1 B is nearer, at 2.24 km	✓ 1

Exam tip: A point is in the cell of a site if no other site is closer, so comparing two distances is always a valid check.

+ Add to test✓ In your test57IBMI3S-3-01r17-04Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy4 marks

A railway line climbs a hillside. The first section of track is 4200 m long and rises at a constant angle of 2.5° to the horizontal.

(a)

Calculate the vertical height gained along the first section, in metres, correct to 3 significant figures.

(2)

(b)

The second section of track rises 95 m over a horizontal distance of 2100 m. Find the angle this section makes with the horizontal, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height = 4200 sin 2.5°	✓ 1
A1 183 m	✓ 1
Part (b)
M1 tan θ = 95/2100 (or θ = tan⁻¹(95/2100))	✓ 1
A1 θ = 2.59°	✓ 1

Common error: Check that your calculator is in degree mode before using sin, cos or tan on an angle given in degrees.

+ Add to test✓ In your test58IBMI3S-3-02r19-01Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Easy6 marks

A water pipeline is laid up a straight hillside that rises at a constant angle of 12° to the horizontal. The pipeline covers a horizontal distance of 850 m.

(a)

Calculate the vertical height gained by the pipeline, to the nearest metre.

(2)

(b)

Calculate the length of the pipeline, to the nearest metre.

(2)

(c)

A second section of pipe continues up the same slope and gains a further 40 m in height. Find the extra horizontal distance it covers, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height = 850 × tan 12°	✓ 1
A1 181 m	✓ 1
Part (b)
M1 length = 850 ÷ cos 12°	✓ 1
A1 869 m	✓ 1
Part (c)
M1 40 ÷ tan 12°	✓ 1
A1 188 m	✓ 1

Exam tip: Label the right-angled triangle with opposite, adjacent and hypotenuse before choosing sin, cos or tan.

+ Add to test✓ In your test59IBMI3S-3-02r19-08Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Easy5 marks

A syringe barrel is a cylinder of internal radius 0.75 cm and length 10 cm. Note that 1 cm³ = 1 mL.

(a)

Calculate the capacity of the syringe in mL, to 3 significant figures.

(2)

(b)

A medicine has a concentration of 40 mg per mL. A patient needs a dose of 85 mg. Find the length of the barrel that this dose occupies, in cm, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 0.75² × 10	✓ 1
A1 17.7 mL	✓ 1
Part (b)
M1 volume = 85 ÷ 40 = 2.125 mL	✓ 1
M1 2.125 = π × 0.75² × L	✓ 1
A1 1.20 cm	✓ 1

Common error: Convert the dose to a volume first; the cylinder formula gives volume, not mass.

+ Add to test✓ In your test60IBMI3S-3-02r19-11Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Easy8 marks

A section of concrete drainage pipe is a hollow cylinder of length 2.5 m, outer radius 0.60 m and inner radius 0.45 m.

(a)

Calculate the volume of concrete in one section, to 3 significant figures.

(2)

(b)

The density of the concrete is 2400 kg m⁻³. Using the unrounded volume, find the mass of one section, to 3 significant figures.

(2)

(c)

A lorry can carry a maximum load of 24 000 kg. Find the greatest number of whole sections it can carry.

(2)

(d)

The inside curved surface of one section is lined with a waterproof coating. Calculate the area to be coated, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 0.60² × 2.5 − π × 0.45² × 2.5 (or π(0.60² − 0.45²) × 2.5)	✓ 1
A1 1.24 m³	✓ 1
Part (b)
M1 mass = density × volume = 2400 × 1.2370…	✓ 1
A1 2970 kg	✓ 1
Part (c)
M1 24 000 ÷ 2968.8…	✓ 1
A1 8 sections (8.08… rounded down)	✓ 1
Part (d)
M1 2π × 0.45 × 2.5	✓ 1
A1 7.07 m²	✓ 1

Common error: Subtract the volumes (or radii squared) of the inner and outer cylinders; subtracting the radii first gives the wrong volume.

+ Add to test✓ In your test61IBMI3S-3-02r19-13Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Easy7 marks

Two pharmacies A(1, 2) and B(7, 6) are shown on a map with coordinates in km. People use their nearest pharmacy, so the boundary between their regions is the perpendicular bisector of AB.

(a)

Find the equation of the boundary, in the form ax + by = c where a, b and c are integers.

(3)

(b)

Determine which pharmacy is nearer to a clinic at (3, 5).

(2)

(c)

Find the distance from the clinic to this nearer pharmacy, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AB is (4, 4)	✓ 1
M1 gradient of AB is 4/6 = 2/3, so perpendicular gradient is −3/2	✓ 1
A1 3x + 2y = 20 (or equivalent)	✓ 1
Part (b)
M1 squared distances: to A 2² + 3² = 13, to B 4² + 1² = 17	✓ 1
A1 pharmacy A	✓ 1
Part (c)
M1 √13	✓ 1
A1 3.61 km	✓ 1

Exam tip: The perpendicular bisector passes through the midpoint and has gradient equal to the negative reciprocal of the gradient of the line joining the two sites.

+ Add to test✓ In your test62IBMI3S-3-03r20-01Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy4 marks

A fish farm keeps trout in a cylindrical tank of internal diameter 2.4 m. The water in the tank is 1.5 m deep.

(a)

Calculate the volume of water in the tank, in m³.

(2)

(b)

The farm stocks the tank with 18 kg of trout per m³ of water. Using your unrounded answer to part (a), find the total mass of trout, in kg, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of $V=\pi r^2h$ with $r=1.2$	✓ 1
A1 6.786 (accept 6.79), so 6.79 m³ to 3 s.f.	✓ 1
Part (b)
M1 multiply volume by 18, i.e. 6.7858 × 18 = 122.15	✓ 1
A1 122 kg (122 kg)	✓ 1

Exam tip: Convert the diameter to a radius before substituting into the volume formula; using 2.4 as the radius gives four times the correct volume.

+ Add to test✓ In your test63IBMI3S-3-03r20-02Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy7 marks

Three charging stations for warehouse robots are at A(2, 1), B(8, 1) and C(5, 7), where coordinates are in metres on a floor plan. The regions of the floor closest to each station are shown on a Voronoi diagram.

(a)

Find the equation of the perpendicular bisector of AB.

(2)

(b)

Find the equation of the perpendicular bisector of AC. Give your answer in the form $y=mx+c$.

(3)

(c)

Hence find the coordinates of the vertex where the regions of A, B and C meet.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AB is (5, 1)	✓ 1
A1 $x=5$ (AB is horizontal, so the bisector is vertical)	✓ 1
Part (b)
M1 midpoint of AC is (3.5, 4) and gradient of AC is $\frac{6}{3}=2$	✓ 1
M1 perpendicular gradient $-\frac{1}{2}$ used with the midpoint	✓ 1
A1 $y=-0.5x+5.75$	✓ 1
Part (c)
M1 substitute $x=5$ into the equation from (b), i.e. $y=-0.5(5)+5.75$	✓ 1
A1 (5, 3.25) (ECF from (b))	✓ 1

Common error: A perpendicular bisector must pass through the midpoint of the two sites; using one of the sites as the point on the line is a frequent slip.

+ Add to test✓ In your test64IBMI3S-3-03r20-03Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy7 marks

A delivery robot starts at P and travels 12 m on a bearing of 060° to reach Q. It then travels 9 m on a bearing of 150° to reach R.

(a)

Show that angle PQR = 90°.

(2)

(b)

Find the distance PR.

(2)

(c)

Find the bearing of R from P, giving your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of P from Q is 240° (60° + 180°)	✓ 1
A1 240° − 150° = 90°, so angle PQR = 90° (AG)	✓ 1
Part (b)
M1 use of Pythagoras, $PR^2=12^2+9^2$	✓ 1
A1 PR = 15 m	✓ 1
Part (c)
M1 angle QPR = $\tan^{-1}\left(\frac{9}{12}\right)=36.87°$	✓ 1
M1 add to the bearing of Q from P, i.e. 60° + 36.87°	✓ 1
A1 96.9° (096.9°)	✓ 1

Exam tip: Bearings are measured clockwise from north and quoted with three figures; sketch a north line at each point before choosing which angles to add.

+ Add to test✓ In your test65IBMI3S-3-03r20-04Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy4 marks

A fashion student makes a conical party hat from fabric. The hat has a circular base of radius 9 cm and a slant height of 24 cm.

(a)

Calculate the curved surface area of the hat, in cm², to the nearest cm².

(2)

(b)

The fabric cut for each hat is 15% larger than the curved surface area, to allow for overlaps. Using your unrounded value from (a), find the area of fabric cut, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of $A=\pi rl$ with $r=9$ and $l=24$	✓ 1
A1 678.6 so 679 cm²	✓ 1
Part (b)
M1 multiply by 1.15, 678.58 × 1.15 = 780.37	✓ 1
A1 780 cm²	✓ 1

Common error: The slant height, not the vertical height, is used in the curved surface area formula for a cone.

+ Add to test✓ In your test66IBMI3S-3-02r22-01Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy6 marks

A fish farm uses a cylindrical net pen of radius 6.5 m and depth 8.0 m.

(a)

Calculate the volume of the pen, to the nearest cubic metre.

(2)

(b)

The pen is stocked with 15 kg of fish per cubic metre. Find the total mass of fish, in tonnes, to 3 significant figures.

(2)

(c)

The net covers the curved side and the circular base of the pen (the top is open). Find the area of net, to the nearest square metre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 6.5² × 8.0	✓ 1
A1 1062 m³	✓ 1
Part (b)
M1 volume × 15 (1062 × 15 = 15 930 kg)	✓ 1
A1 15.9 tonnes	✓ 1
Part (c)
M1 2π × 6.5 × 8.0 + π × 6.5²	✓ 1
A1 459 m²	✓ 1

Common error: A net with an open top has no top circle: add the curved surface 2πrh and one circular end πr² only.

+ Add to test✓ In your test67IBMI3S-3-02r22-02Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy7 marks

A robot arm has its pivot fixed 0.40 m above the floor. The arm is a rigid rod of length 1.2 m, and it is raised to an angle of 35° above the horizontal.

(a)

Find the height of the tip of the arm above the floor, in metres, to 3 significant figures.

(2)

(b)

Find the horizontal distance from the pivot to the tip of the arm, to 3 significant figures.

(2)

(c)

The robot must place an item on a shelf 1.50 m above the floor using the tip of the arm. Find the angle above the horizontal at which the arm must be raised, to the nearest 0.1°.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.40 + 1.2 sin 35°	✓ 1
A1 1.09 m	✓ 1
Part (b)
M1 1.2 cos 35°	✓ 1
A1 0.983 m	✓ 1
Part (c)
M1 sin θ = (1.50 − 0.40) / 1.2	✓ 1
M1 use of inverse sine, θ = sin⁻¹(1.1/1.2)	✓ 1
A1 66.4°	✓ 1

Exam tip: Draw a right-angled triangle for the arm alone: its vertical side is the height above the pivot, so remember to add the pivot height afterwards.

+ Add to test✓ In your test68IBMI3S-3-02r22-03Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy6 marks

A fishing boat leaves port P and sails 24 km on a bearing of 070° to a fishing ground A. It then sails 18 km on a bearing of 160° to a second ground B.

(a)

Show that angle PAB is 90°.

(2)

(b)

Find the distance PB.

(2)

(c)

Find the bearing of B from P, to the nearest degree.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the bearing of P from A is 070° + 180° = 250°	✓ 1
A1 250° − 160° = 90°	✓ 1
Part (b)
M1 Pythagoras, PB² = 24² + 18²	✓ 1
A1 30 km	✓ 1
Part (c)
M1 angle APB = tan⁻¹(18/24) = 36.9° and add to 070°	✓ 1
A1 107°	✓ 1

Common error: For a return (back) bearing add or subtract 180°; mixing up the angle at the turning point is the most common source of an incorrect triangle.

+ Add to test✓ In your test69IBMI3S-3-02r22-08Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy7 marks

Two robot charging stations are located at A(1, 2) and B(7, 6) on a grid, where units are metres. A Voronoi diagram is drawn for the two sites.

(a)

Find the midpoint of AB.

(1)

(b)

Find the gradient of the edge that separates the region nearest A from the region nearest B.

(2)

(c)

Find the equation of this edge, giving your answer in the form y = mx + c.

(2)

(d)

A robot is at R(5, 5). Determine which charging station is nearer and find its distance from R, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 (4, 4)	✓ 1
Part (b)
M1 gradient of AB = (6 − 2)/(7 − 1) = 2/3	✓ 1
A1 perpendicular gradient −3/2	✓ 1
Part (c)
M1 y − 4 = −(3/2)(x − 4) (ECF)	✓ 1
A1 y = −1.5x + 10	✓ 1
Part (d)
M1 calculates AR = 5 and BR = √5 and compares them	✓ 1
A1 B is nearer, at 2.24 m	✓ 1

Exam tip: The edge between two sites is the perpendicular bisector: it passes through the midpoint with gradient equal to the negative reciprocal of the gradient joining the sites.

+ Add to test✓ In your test70IBMI3S-3-01r23-01Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy5 marks

Two mobile-phone masts, A and B, are built on a flat plain. Taking 1 unit as 1 km, A is at (2, 1) and B is at (8, 5). A Voronoi diagram is drawn for the two masts as sites.

(a)

Find the coordinates of the midpoint of AB and the gradient of AB.

(2)

(b)

Find the equation of the perpendicular bisector of AB, giving your answer in the form ax + by = c where a, b and c are integers.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 midpoint (5, 3)	✓ 1
A1 gradient of AB = (5−1)/(8−2) = 2/3	✓ 1
Part (b)
A1 perpendicular gradient = −3/2 (ECF from (a))	✓ 1
M1 uses midpoint and perpendicular gradient: y − 3 = −(3/2)(x − 5)	✓ 1
A1 3x + 2y = 21 (any equivalent integer form)	✓ 1

Common error: The edge of a Voronoi cell is the perpendicular bisector, so use the negative reciprocal of the gradient of AB, not the gradient itself.

+ Add to test✓ In your test71IBMI3S-3-01r23-02Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy5 marks

Three mountain huts, H₁(1, 2), H₂(7, 3) and H₃(4, 9), are marked on a map where 1 unit represents 1 km. The map is divided into Voronoi cells, one for each hut, so that every point lies in the cell of its nearest hut. A climber is at the point P(5, 5).

(a)

Calculate the distance from P to each hut.

(2)

(b)

State, with a reason, which hut’s cell contains P.

(1)

(c)

Find the equation of the Voronoi edge between the cells of H₁ and H₂, in the form y = mx + c.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses distance formula for at least one hut	✓ 1
A1 PH₁ = 5.00 km, PH₂ = √8 = 2.83 km, PH₃ = √17 = 4.12 km (all three correct)	✓ 1
Part (b)
R1 H₂, because it is the nearest hut (smallest distance)	✓ 1
Part (c)
M1 midpoint (4, 2.5) and gradient of H₁H₂ = 1/6 so perpendicular gradient −6	✓ 1
A1 y = −6x + 26.5	✓ 1

Exam tip: Compare squared distances if you only need to decide which site is nearest; it avoids unnecessary square roots.

+ Add to test✓ In your test72IBMI3S-3-01r23-06Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

A soup can is a cylinder with internal radius 3.4 cm and height 11.2 cm.

(a)

Calculate the volume of the can, in cm³, correct to 3 significant figures.

(2)

(b)

The can is filled with 0.38 litres of soup. Find the percentage of the can’s volume that is filled. (1 litre = 1000 cm³.)

(2)

(c)

Calculate the total outer surface area of the can, in cm², to 3 significant figures. Treat the outside dimensions as equal to the inside ones.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 3.4² × 11.2	✓ 1
A1 406.7 cm³ = 407 cm³	✓ 1
Part (b)
M1 380 ÷ volume (use the unrounded volume)	✓ 1
A1 93.4%	✓ 1
Part (c)
M1 2π × 3.4² + 2π × 3.4 × 11.2	✓ 1
A1 312 cm²	✓ 1

Exam tip: Convert litres to cm³ before comparing with a volume in cm³.

+ Add to test✓ In your test73IBMI3S-3-01r23-10Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy4 marks

A spherical chocolate truffle has diameter 3.0 cm.

(a)

Calculate the volume of the truffle, correct to 3 significant figures.

(2)

(b)

The chocolate has density 1.2 g cm⁻³. Find the mass of one truffle, to 3 significant figures. Use the unrounded volume.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (4/3)π(1.5)³	✓ 1
A1 14.1 cm³	✓ 1
Part (b)
M1 1.2 × volume	✓ 1
A1 17.0 g	✓ 1

Common error: Check whether you are given the diameter or the radius before using the sphere formula.

+ Add to test✓ In your test74IBMI3S-3-03r29-01Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams Paper 1Easy4 marks

A cargo ship leaves port P and sails 18.0 km on a bearing of 065° to a buoy Q. Both P and Q are at sea level on a flat chart, with north pointing up the page.

(a)

Find how far east of P the buoy Q is. Give your answer to the nearest 0.1 km.

(2)

(b)

Find how far north of P the buoy Q is. Give your answer to the nearest 0.1 km.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 east distance = 18 sin 65°	✓ 1
A1 16.3 km	✓ 1
Part (b)
M1 north distance = 18 cos 65°	✓ 1
A1 7.6 km	✓ 1

Common error: A bearing is measured clockwise from north, so the angle in the triangle is from the north line: use sine for the east component and cosine for the north component.

+ Add to test✓ In your test75IBMI3S-3-03r29-02Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams Paper 1Easy7 marks

A spotlight is fixed 6.4 m above the stage floor. From the spotlight, the angle of depression of an actor standing on the stage is 38°. Treat the spotlight as a point and the stage as horizontal.

(a)

Calculate the horizontal distance from the point on the floor directly below the spotlight to the actor. Give your answer to 3 significant figures.

(2)

(b)

Calculate the straight-line distance from the spotlight to the actor. Give your answer to 3 significant figures.

(2)

(c)

The actor walks directly towards the point below the spotlight until the angle of depression is 52°. Calculate how far the actor has walked. Use unrounded values in your working and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 38° = 6.4 / d	✓ 1
A1 d = 8.19 m	✓ 1
Part (b)
M1 sin 38° = 6.4 / s (or Pythagoras with their d)	✓ 1
A1 s = 10.4 m	✓ 1
Part (c)
M1 new horizontal distance = 6.4 / tan 52° (= 5.0002...)	✓ 1
M1 subtracts new distance from the answer to (a) (unrounded 8.1916...)	✓ 1
A1 3.19 m	✓ 1

Exam tip: The angle of depression is measured from the horizontal at the spotlight, so it equals the angle of elevation from the actor to the spotlight. Mark it on your sketch before choosing sine, cosine or tangent.

+ Add to test✓ In your test76IBMI3S-3-03r29-03Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams Paper 1Easy5 marks

A small cinder cone formed by a volcano is modelled as a solid cone of base radius 45 m and perpendicular height 12 m.

(a)

Calculate the volume of the cone. Give your answer to the nearest 100 m³.

(2)

(b)

Calculate the curved surface area of the cone. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 V = (1/3)π(45)²(12)	✓ 1
A1 25 400 m³ (25 446.9...)	✓ 1
Part (b)
M1 slant height = √(45² + 12²) = 46.57...	✓ 1
M1 curved surface area = π × 45 × 46.57...	✓ 1
A1 6580 m² (6584.0...)	✓ 1

Common error: The slant height, not the perpendicular height, is used in the curved surface area formula. Find it with Pythagoras first.

+ Add to test✓ In your test77IBMI3S-3-03r29-13Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

A theatre prop is a solid pyramid with a square base of side 1.8 m and a perpendicular height of 2.4 m. The apex is directly above the centre of the base.

(a)

Calculate the volume of the pyramid.

(2)

(b)

Show that the slant height of each triangular face, measured from the apex to the midpoint of a base edge, is 2.56 m to 3 significant figures.

(2)

(c)

The four triangular faces are covered with fabric. Calculate the area of fabric needed. Use the unrounded slant height and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 V = (1/3) × 1.8² × 2.4	✓ 1
A1 2.592 m³ (accept 2.59)	✓ 1
Part (b)
M1 slant height = √(2.4² + 0.9²)	✓ 1
A1 √6.57 = 2.563... = 2.56 m (shown)	✓ 1
Part (c)
M1 4 × ½ × 1.8 × 2.563...	✓ 1
A1 9.23 m² (9.2275...)	✓ 1

Common error: Use the perpendicular height of the pyramid for the volume, but the slant height of a face for its area. They are different lengths.

+ Add to test✓ In your test78IBMI3S-3-01r34-01Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams Paper 1Easy4 marks

A cylindrical grain jar was excavated at a Bronze Age settlement. Its inside radius is 18 cm and its inside height is 45 cm. Give answers to 3 significant figures.

(a)

Calculate the volume of the jar, in cm³.

(2)

(b)

Archaeologists estimate that stored grain had a density of 0.78 g cm⁻³. Estimate the mass of grain in a completely full jar, in kg. Use your unrounded answer from (a).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of $V=\pi r^2h$ with $r=18$, $h=45$	✓ 1
A1 $V=45\,800$ cm³ (45 804.4…)	✓ 1
Part (b)
M1 multiply volume by density (ECF from (a)) and convert from g to kg (÷1000)	✓ 1
A1 35.7 kg (35.73…)	✓ 1

Common error: Forgetting to convert grams to kilograms, or using the diameter in place of the radius in the cylinder formula.

+ Add to test✓ In your test79IBMI3S-3-01r34-05Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams Paper 1Easy4 marks

A cycling route climbs a hill. Over a horizontal distance of 2600 m the road rises vertically by 185 m. Assume that the road is straight.

(a)

Find the angle of elevation of the road above the horizontal, in degrees, to 3 significant figures.

(2)

(b)

Find the length of the road, to the nearest metre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\tan\theta=\dfrac{185}{2600}$	✓ 1
A1 $\theta=4.07^\circ$ (4.0696…)	✓ 1
Part (b)
M1 Pythagoras $\sqrt{2600^2+185^2}$, or $\dfrac{185}{\sin\theta}$ with the unrounded angle	✓ 1
A1 2607 m (2606.6…)	✓ 1

Exam tip: Set your calculator to degree mode before using inverse trigonometric functions, and sketch the right-angled triangle first.

+ Add to test✓ In your test80IBMI3S-3-01r34-09Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams Paper 1Easy4 marks

A drone hovers 48 m vertically above flat ground, directly above point D. It records an excavation trench at point T on the ground. The angle of depression of T from the drone is 32°.

(a)

Find the horizontal distance from D to T, to 3 significant figures.

(2)

(b)

Find the straight-line distance from the drone to T, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\tan32^\circ=\dfrac{48}{DT}$ (angle of depression equals the angle of elevation of the drone from T)	✓ 1
A1 $DT=76.8$ m (76.82…)	✓ 1
Part (b)
M1 $\sin32^\circ=\dfrac{48}{\text{distance}}$, or Pythagoras with the unrounded value from (a)	✓ 1
A1 90.6 m (90.58…; accept ECF from (a))	✓ 1

Common error: Using 32° as the angle at the drone. The angle of depression is measured from the horizontal, so the angle at the drone is 58°.

+ Add to test✓ In your test81IBMI3S-3-01r34-11Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams Paper 1Easy4 marks

A bike-share company has repair stations at A(2, 8), B(9, 6) and C(5, 1), with coordinates in kilometres. A cyclist with a puncture is at P(6, 5). The cyclist walks to the nearest station, measured in a straight line.

(a)

Find the distance from P to each of the three stations. Give each answer to 3 significant figures.

(3)

(b)

State which station the cyclist walks to and name the part of a Voronoi diagram that contains P.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of the distance formula $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$	✓ 1
A1 two distances correct	✓ 1
A1 PA = 5.00 km, PB = 3.16 km, PC = 4.12 km	✓ 1
Part (b)
A1 station B; P lies in the Voronoi cell of site B	✓ 1

Common error: Stopping after finding one distance. To decide which site is nearest, every distance must be compared.

+ Add to test✓ In your test82IBMI3S-3-03r37-01Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy5 marks

A volcano is modelled as a cone of height 1.8 km and base radius 4.5 km.

(a)

Calculate the volume of the volcano. Give your answer in km³, correct to 3 significant figures.

(3)

(b)

Find the angle that the side of the volcano makes with the horizontal ground. Give your answer to the nearest tenth of a degree.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses $V=\frac{1}{3}\pi r^2h$	✓ 1
A1 correct substitution $\frac{1}{3}\pi(4.5)^2(1.8)$	✓ 1
A1 38.2 km³	✓ 1
Part (b)
M1 $\tan\theta=\frac{1.8}{4.5}$	✓ 1
A1 21.8°	✓ 1

Exam tip: The volume formula for a cone is in the formula booklet, but you must choose the right measurement for the height: use the vertical height, not the sloping side.

+ Add to test✓ In your test83IBMI3S-3-03r37-02Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy7 marks

A theatre prop is a solid cylinder of radius 0.40 m and height 1.2 m with a solid hemisphere of radius 0.40 m fixed on top of it, so that the prop is a single solid. The prop is to be painted on its curved surfaces only (the circular base on the floor is not painted).

(a)

Calculate the total volume of the prop. Give your answer in m³ to 3 significant figures.

(3)

(b)

Calculate the area to be painted. Give your answer in m² to 3 significant figures.

(2)

(c)

Each tin of paint covers 6 m² of surface and costs €14.50. The prop needs two coats of paint. Find the least cost of the paint needed. Use the unrounded area from part (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses $\pi r^2h$ for the cylinder (0.603 m³) and $\frac{2}{3}\pi r^3$ for the hemisphere (0.134 m³)	✓ 1
M1 adds the two volumes	✓ 1
A1 0.737 m³	✓ 1
Part (b)
M1 curved area of cylinder $2\pi rh$ (3.02) plus hemisphere $2\pi r^2$ (1.01)	✓ 1
A1 4.02 m²	✓ 1
Part (c)
M1 area for two coats $2\times4.0212=8.04$ m², so $8.04\div6=1.34$ tins, rounded up to 2 tins	✓ 1
A1 cost = 2 × 14.50 = €29.00	✓ 1

Exam tip: Paint and material are bought in whole units. Always round a quantity like this up to the next whole tin, then multiply by the price.

+ Add to test✓ In your test84IBMI3S-3-03r37-06Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy5 marks

The lamp of a lighthouse is 32 m above sea level. A cargo ship is sailing directly towards the base of the lighthouse. From the lamp, the angle of depression of the ship is first measured as 3.5°. A short time later the angle of depression is 6.0°. Treat the ship as a point at sea level.

(a)

Calculate the horizontal distance from the ship to the lighthouse when the angle of depression is 3.5°. Give your answer to the nearest metre.

(2)

(b)

Calculate the distance that the ship sails between the two measurements. Give your answer to the nearest metre.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\tan3.5^{\circ}=\frac{32}{d}$	✓ 1
A1 $d=523$ m	✓ 1
Part (b)
M1 second distance $\frac{32}{\tan6^{\circ}}$	✓ 1
A1 304 m	✓ 1
A1 distance sailed = 219 m	✓ 1

Common error: Mixing up the angle of depression with the angle at the ship's mast or at the lighthouse base leads to the wrong triangle; the angle at the ship equals the angle of depression.

+ Add to test✓ In your test85IBMI3S-3-03r37-11Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

Two volcano monitoring stations are at A(2, 1) and B(8, 5) on a map where the units are kilometres. These stations are sites in a Voronoi diagram.

(a)

Find the coordinates of the midpoint of AB.

(1)

(b)

Find the equation of the edge of the Voronoi diagram between the cells of A and B. Give your answer in the form $y=mx+c$.

(3)

(c)

A new vent opens at P(6, 6). Find the distance from P to the nearer station, and state which station it is. Give the distance in km to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 (5, 3)	✓ 1
Part (b)
M1 gradient of AB $=\frac{5-1}{8-2}=\frac{2}{3}$	✓ 1
M1 perpendicular gradient $-\frac{3}{2}$ and uses the midpoint (5, 3)	✓ 1
A1 $y=-1.5x+10.5$	✓ 1
Part (c)
M1 distances $\sqrt{(6-2)^2+(6-1)^2}=6.40$ and $\sqrt{(6-8)^2+(6-5)^2}$	✓ 1
A1 B is nearer, at 2.24 km	✓ 1

Exam tip: The edge between two Voronoi cells is the perpendicular bisector of the line joining the two sites, so it passes through the midpoint at right angles.

+ Add to test✓ In your test86IBMI3S-3-01r40-01Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy6 marks

A rail company uses a Voronoi diagram to decide which of two signal-control stations monitors each point of its network. Station A is at (3, 4) and station B is at (11, 4), where coordinates are in kilometres from a reference point. A trackside mast M is at (6, 9).

(a)

Find the coordinates of the midpoint of AB.

(1)

(b)

Find the equation of the edge of the Voronoi diagram between the cells of A and B.

(2)

(c)

By calculating the distance from M to each station, determine which station monitors M, and state this distance.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 midpoint is (7, 4)	✓ 1
Part (b)
R1 AB is horizontal (gradient 0), so the perpendicular bisector is vertical	✓ 1
A1 $x = 7$	✓ 1
Part (c)
M1 attempt to use the distance formula for at least one station	✓ 1
A1 MA = $\sqrt{34}$ = 5.83 km and MB = $\sqrt{50}$ = 7.07 km	✓ 1
A1 station A monitors M, at a distance of 5.83 km	✓ 1

Exam tip: A point lies in the cell of the site it is closest to, so compare the distances rather than guessing from the sketch.

+ Add to test✓ In your test87IBMI3S-3-01r40-02Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy4 marks

A sports scientist tests runners on a treadmill whose belt is 2.40 m long. The belt is raised to make an angle of 7.0° with the horizontal.

(a)

Calculate the vertical height through which the front of the belt is raised.

(2)

(b)

A gradient of 15% means that vertical rise divided by horizontal distance is 0.15. Find the angle the belt must make with the horizontal to give a gradient of 15%, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height = 2.40 sin 7.0°	✓ 1
A1 0.292 m	✓ 1
Part (b)
M1 $\tan\theta = 0.15$	✓ 1
A1 $\theta = 8.53°$	✓ 1

Exam tip: Check that your calculator is in degree mode before using sin, cos or tan with an angle in degrees.

+ Add to test✓ In your test88IBMI3S-3-01r40-03Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy5 marks

A hydrotherapy pool for athletes is a cylinder of radius 0.80 m. When full, the water is 1.20 m deep.

(a)

Calculate the volume of water in the full pool, in m³, correct to 3 significant figures.

(2)

(b)

The pool is filled at a constant rate of 15 litres per minute. Given that 1 m³ = 1000 litres, find the time taken to fill the pool, correct to the nearest minute. (The answer is the same whether you use the rounded or the unrounded volume.)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $V = \pi \times 0.80^2 \times 1.20$	✓ 1
A1 2.41 m³	✓ 1
Part (b)
A1 volume is about 2410 litres (2413 litres unrounded)	✓ 1
M1 time = volume (in litres) ÷ 15	✓ 1
A1 161 minutes	✓ 1

Exam tip: Convert units before dividing: cubic metres and litres cannot be mixed in the same calculation.

+ Add to test✓ In your test89IBMI3S-3-01r40-04Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Easy5 marks

A train leaves a junction J and travels 14.0 km in a straight line on a bearing of 060°.

(a)

Find how far north of J the train is at the end of this journey.

(2)

(b)

Find how far east of J the train is at the end of this journey, correct to 3 significant figures.

(2)

(c)

Write down the bearing on which the train must travel to return directly to J.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 14.0 cos 60°	✓ 1
A1 7.00 km	✓ 1
Part (b)
M1 14.0 sin 60°	✓ 1
A1 12.1 km	✓ 1
Part (c)
A1 240°	✓ 1

Exam tip: A back bearing differs from the original bearing by 180°; add 180° if the bearing is below 180° and subtract it otherwise.

+ Add to test✓ In your test90IBMI3S-3-03r42-01Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; 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Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy5 marks

Two branches of a county library are at A(1, 2) and B(7, 6) on a map, where each unit represents 1 km. The council draws a Voronoi diagram with the two branches as sites.

(a)

Find the coordinates of the midpoint of AB.

(1)

(b)

Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.

(3)

(c)

State what is true about a household located on this line.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 midpoint (4, 4)	✓ 1
Part (b)
M1 gradient of AB = (6 − 2)/(7 − 1) = 2/3	✓ 1
A1 gradient of perpendicular bisector = −3/2 (or −1.5)	✓ 1
A1 y = −1.5x + 10	✓ 1
Part (c)
R1 it is the same distance from branch A as from branch B (equidistant from both branches)	✓ 1

Common error: A Voronoi boundary uses the gradient perpendicular to the line joining the sites, not the gradient of that line itself, and it passes through the midpoint.

+ Add to test✓ In your test91IBMI3S-3-03r42-09Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; 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Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; 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Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

A ramp leads up to the entrance of a library. The ramp is a straight slope that makes an angle of 5.2° with the horizontal and rises 0.72 m.

(a)

Calculate the length of the ramp surface, giving your answer to 3 significant figures.

(2)

(b)

Calculate the horizontal distance covered by the ramp, giving your answer to 3 significant figures.

(2)

(c)

A building guideline says that the gradient of a ramp must not exceed 1/12. Calculate the gradient of this ramp and state whether it meets the guideline.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.72/sin 5.2°	✓ 1
A1 7.94 m	✓ 1
Part (b)
M1 0.72/tan 5.2°	✓ 1
A1 7.91 m	✓ 1
Part (c)
A1 gradient = tan 5.2° = 0.0910 (3 s.f.: 0.091)	✓ 1
R1 0.091 > 1/12 = 0.0833, so the ramp does not meet the guideline	✓ 1

Exam tip: Gradient = vertical rise ÷ horizontal run = tan of the angle. Compare the same kind of quantity (decimal with decimal) when checking a limit.

+ Add to test✓ In your test92IBMI3S-3-03r42-14Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; 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volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy4 marks

A circular paddling pool at a swimming club has radius 1.8 m. It is filled with water to a depth of 0.45 m. The pool is a cylinder.

(a)

Calculate the volume of water in the pool, in m³, to 3 significant figures.

(2)

(b)

Water flows into the pool from a hose at 35 litres per minute. Given that 1 m³ = 1000 litres, find the time to fill the pool to this depth, to the nearest minute.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 1.8² × 0.45	✓ 1
A1 4.58 m³	✓ 1
Part (b)
M1 4580 ÷ 35 (using 4.58 m³ = 4580 litres is acceptable)	✓ 1
A1 131 minutes	✓ 1

Exam tip: Convert units before dividing: 1 m³ is 1000 litres, so a volume in cubic metres must be multiplied by 1000 to compare with a flow rate in litres.

+ Add to test✓ In your test93IBMI3S-3-03r44-01Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy5 marks

A vent on a volcano has built up a pile of ash in the shape of a cone. The pile has base radius 18 m and perpendicular height 7.5 m.

(a)

Calculate the volume of the pile, giving your answer to the nearest m³.

(2)

(b)

Find the area of the curved surface of the pile, giving your answer to the nearest m².

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of V = ⅓πr²h with r = 18 and h = 7.5	✓ 1
A1 2545 m³ (2544.69…)	✓ 1
Part (b)
M1 slant height = √(18² + 7.5²) = 19.5 m	✓ 1
M1 use of curved area = πrl = π × 18 × 19.5	✓ 1
A1 1103 m² (1102.70…)	✓ 1

Common error: Use the slant height, not the perpendicular height, in the curved surface area formula πrl.

+ Add to test✓ In your test94IBMI3S-3-03r44-02Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy5 marks

A cargo ship leaves harbour H and sails 28 km on a bearing of 063°. Its position is then S.

(a)

Calculate how far east of H the ship is at S. Give your answer to 3 significant figures.

(2)

(b)

Calculate how far north of H the ship is at S. Give your answer to 3 significant figures.

(2)

(c)

Write down the bearing of H from S.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 28 sin 63°	✓ 1
A1 24.9 km	✓ 1
Part (b)
M1 28 cos 63°	✓ 1
A1 12.7 km	✓ 1
Part (c)
A1 243°	✓ 1

Exam tip: Draw a north line at the start point and mark the bearing clockwise from it before choosing sine or cosine.

+ Add to test✓ In your test95IBMI3S-3-03r44-03Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy7 marks

Two seismic stations are located at A(2, 3) and B(10, 7) on a map in which one unit represents 1 km. The boundary between the region nearer to A and the region nearer to B is part of a Voronoi diagram.

(a)

Find the coordinates of the midpoint of AB.

(1)

(b)

Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.

(3)

(c)

A new sensor is placed at P(7, 5). Determine which station is nearer to P and find the distance from P to that station, giving your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 (6, 5)	✓ 1
Part (b)
M1 gradient of AB = (7 − 3)/(10 − 2) = 1/2	✓ 1
M1 perpendicular gradient −2 and line through (6, 5)	✓ 1
A1 y = −2x + 17	✓ 1
Part (c)
M1 PA² = 5² + 2² = 29 and PB² = 3² + 2² = 13 (or equivalent comparison)	✓ 1
A1 B is nearer	✓ 1
A1 3.61 km	✓ 1

Common error: The perpendicular bisector passes through the midpoint of AB, not through either station.

+ Add to test✓ In your test96IBMI3S-3-03r44-04Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Easy6 marks

In a theatre, a spotlight is fixed 9.2 m above the stage floor. An actor stands on the floor at a horizontal distance of 6.5 m from the point directly below the spotlight. The beam points straight at the actor's feet.

(a)

Calculate the angle of depression of the beam. Give your answer to the nearest 0.1°.

(2)

(b)

Calculate the length of the beam from the spotlight to the actor's feet. Give your answer to 3 significant figures.

(2)

(c)

The actor walks a further 2.0 m directly away from the point below the spotlight. Find the new angle of depression of the beam to the nearest 0.1°.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan θ = 9.2/6.5	✓ 1
A1 54.8°	✓ 1
Part (b)
M1 √(9.2² + 6.5²) or 9.2/sin 54.8°	✓ 1
A1 11.3 m	✓ 1
Part (c)
M1 tan θ = 9.2/8.5	✓ 1
A1 47.3°	✓ 1

Exam tip: The angle of depression is measured from the horizontal at the spotlight, and equals the angle of elevation from the actor.

+ Add to test✓ In your test97IBMI3S-3-01r46-01Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

A honeycomb cell built by bees is a prism whose cross-section is a regular hexagon of side 3.00 mm. The cell is 11.0 mm deep.

(a)

Show that the area of the hexagonal cross-section is 23.4 mm², correct to 3 significant figures.

(2)

(b)

Calculate the volume of one cell, giving your answer in mm³ to 3 significant figures.

(2)

(c)

A beekeeper fills 800 cells completely with honey. Calculate the volume of honey, in cm³, to 3 significant figures. Use the unrounded volume of one cell from part (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 splits the hexagon into six congruent triangles with two sides of 3 mm and an included angle of 60°, or equivalent method	✓ 1
A1 area = 6 × ½ × 3 × 3 × sin 60° = 23.38… which rounds to 23.4 mm² (AG)	✓ 1
Part (b)
M1 volume = area × depth = 23.38… × 11	✓ 1
A1 volume = 257 mm³	✓ 1
Part (c)
M1 800 × 257.2… = 205768 mm³ and conversion using 1 cm³ = 1000 mm³	✓ 1
A1 volume = 206 cm³	✓ 1

Common error: Forgetting that 1 cm³ is 1000 mm³ (not 10 mm³) when converting volumes.

+ Add to test✓ In your test98IBMI3S-3-01r46-02Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

A camera is mounted on a gantry 6.50 m above a straight section of a Formula 1 track. A car passes the camera's line of sight at a horizontal distance d₁ when the angle of depression from the camera to the car is 14°. Exactly 0.60 seconds later the angle of depression to the car is 5°. The car travels directly away from the gantry along a straight line, in the same vertical plane as the camera.

(a)

Calculate the horizontal distance d₁, in metres, to 3 significant figures.

(2)

(b)

Calculate the horizontal distance d₂ at the later time, in metres, to 3 significant figures.

(2)

(c)

Hence find the average speed of the car, in m s⁻¹, over this 0.60 s. Use the unrounded values of d₁ and d₂. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 14° = 6.5 / d₁	✓ 1
A1 d₁ = 26.1 m	✓ 1
Part (b)
M1 tan 5° = 6.5 / d₂	✓ 1
A1 d₂ = 74.3 m	✓ 1
Part (c)
M1 distance travelled = d₂ − d₁ = 48.23 m, then divided by 0.6	✓ 1
A1 speed = 80.4 m s⁻¹	✓ 1

Exam tip: The angle of depression is measured from the horizontal, so it equals the angle of elevation of the camera as seen from the car.

+ Add to test✓ In your test99IBMI3S-3-01r46-07Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy5 marks

Two apiaries are at A(2, 6) and B(10, 2) on a coordinate grid where each unit is 1 km. Bees from a hive are assumed to forage at the nearest apiary's side of the Voronoi boundary. A flower field is at F(5, 5).

(a)

Find the midpoint of AB.

(1)

(b)

Find the equation of the Voronoi edge (the perpendicular bisector of AB).

(2)

(c)

By calculating two distances, determine which apiary is closer to the flower field F. Give the distances to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 midpoint = (6, 4)	✓ 1
Part (b)
M1 gradient of AB = −4/8 = −1/2, so the perpendicular gradient is 2	✓ 1
A1 y − 4 = 2(x − 6), so y = 2x − 8	✓ 1
Part (c)
M1 FA = √(3² + 1²) = 3.16 km and FB = √(5² + 3²) = 5.83 km	✓ 1
A1 apiary A is closer, since 3.16 < 5.83	✓ 1

Common error: Using the midpoint's gradient instead of the negative reciprocal; the boundary is perpendicular to AB, not parallel to it.

+ Add to test✓ In your test100IBMI3S-3-01r46-10Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy4 marks

A wheelchair ramp at a hospital entrance is a straight slope 5.20 m long. The top of the ramp is 0.45 m above the ground.

(a)

Calculate the angle the ramp makes with the horizontal, to 3 significant figures.

(2)

(b)

A new design rule requires the angle with the horizontal to be at most 4.8° for the same height of 0.45 m. Calculate the shortest possible length of the ramp, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin θ = 0.45 / 5.2	✓ 1
A1 θ = 4.96°	✓ 1
Part (b)
M1 sin 4.8° = 0.45 / L	✓ 1
A1 L = 5.38 m	✓ 1

Common error: Using tan instead of sin: the ramp length is the hypotenuse, so the 0.45 m height is opposite the angle and 5.2 m is the hypotenuse.

+ Add to test✓ In your test101IBMI3S-3-03r46-01Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy5 marks

A cable car on a mountain railway rises at a constant angle of elevation of 24° along a straight cable of length 1800 m. The lower station is 1320 m above sea level.

(a)

Find the vertical height gained between the two stations. Give your answer to 3 significant figures.

(2)

(b)

Find the horizontal distance between the two stations. Give your answer to 3 significant figures.

(2)

(c)

Using the unrounded height from part (a), find the altitude of the upper station above sea level, to the nearest metre.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height = 1800 sin 24° (or equivalent right-angled trigonometry)	✓ 1
A1 732 m	✓ 1
Part (b)
M1 horizontal distance = 1800 cos 24°	✓ 1
A1 1640 m	✓ 1
Part (c)
A1 1320 + 732.13… = 2052 m (ECF from (a) if rounded value used and consistent)	✓ 1

Common error: Calculators must be in degree mode. Check that sin 24° is about 0.407 before you multiply.

+ Add to test✓ In your test102IBMI3S-3-03r46-02Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy5 marks

A mobile-phone mast B stands 8.5 km from a relay tower A, on a bearing of 072° from A.

(a)

Find how far east of A the mast B is, to 3 significant figures.

(2)

(b)

Find how far north of A the mast B is, to 3 significant figures.

(2)

(c)

Write down the bearing of A from B.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 east distance = 8.5 sin 72°	✓ 1
A1 8.08 km	✓ 1
Part (b)
M1 north distance = 8.5 cos 72°	✓ 1
A1 2.63 km	✓ 1
Part (c)
A1 072° + 180° = 252°	✓ 1

Exam tip: Bearings are measured clockwise from north, so the east component uses sine and the north component uses cosine.

+ Add to test✓ In your test103IBMI3S-3-03r46-03Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

Two Wi-Fi hotspots are at A(2, 1) and B(8, 5) on a campus map, where one unit represents 100 m. Laptops always connect to the nearest hotspot. The edge of the Voronoi diagram between A and B lies on the perpendicular bisector of AB.

(a)

Find the coordinates of the midpoint of AB.

(1)

(b)

Find the equation of the perpendicular bisector of AB, giving your answer in the form $y = mx + c$.

(3)

(c)

A laptop is at P(4, 6). Find the distance, in units, from P to its nearest hotspot. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 (5, 3)	✓ 1
Part (b)
M1 gradient of AB = 4/6 = 2/3	✓ 1
M1 perpendicular gradient = −3/2 and substitution of (5, 3)	✓ 1
A1 $y = -1.5x + 10.5$	✓ 1
Part (c)
M1 uses distance formula to find PA = √29 ≈ 5.39 and PB	✓ 1
A1 nearest is B, distance √17 = 4.12 units	✓ 1

Exam tip: To decide which site is nearest to a point, compare distances; you do not need the equation of the edge.

+ Add to test✓ In your test104IBMI3S-3-03r46-04Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Easy6 marks

A cylindrical jar of honey has internal radius 4.2 cm and internal height 11.5 cm. The jar is completely full. The density of honey is 1.42 g cm⁻³.

(a)

Calculate the volume of honey in the jar, to 3 significant figures.

(2)

(b)

Calculate the mass of honey in the jar, in grams, to 3 significant figures. Use the unrounded volume.

(2)

(c)

A paper label covers the whole curved surface of the jar. Calculate the area of the label, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $V = \pi \times 4.2^2 \times 11.5$	✓ 1
A1 637 cm³	✓ 1
Part (b)
M1 mass = 637.30… × 1.42	✓ 1
A1 905 g	✓ 1
Part (c)
M1 $A = 2\pi \times 4.2 \times 11.5$	✓ 1
A1 303 cm²	✓ 1

Common error: The curved surface of a cylinder has area 2πrh; do not include the circular ends when a label wraps only the side.

+ Add to test✓ In your test105IBMI-032-00-05Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium8 marks

A ferry leaves harbour H and sails 18 km on a bearing of 040° to an island A. It then sails 25 km on a bearing of 115° to a second island B.

(a)

Show that angle HAB = 105°.

(2)

(b)

Find the distance HB.

(3)

(c)

Find the bearing of B from H.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Bearing of H from A is 220° (or uses alternate angles with the north line at A) M1	✓ 1
220° − 115° = 105° A1	✓ 1
Part (b)
HB² = 18² + 25² − 2(18)(25)cos 105° M1	✓ 1
HB² = 1182 (1181.9…) A1	✓ 1
HB = 34.4 km A1	✓ 1
Part (c)
sin AHB / 25 = sin 105° / 34.38 M1	✓ 1
Angle AHB = 44.6° A1	✓ 1
Bearing = 40° + 44.6° = 084.6° A1	✓ 1

Exam tip: Keep the unrounded value of HB in your calculator for part (c); using 34.4 still gives 084.6°, but early rounding can shift answers in longer chains.

+ Add to test✓ In your test106IBMI-032-00-06Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium7 marks

A coastguard stands at the top of a vertical cliff 62 m above sea level. Two buoys, X and Y, lie due east of the foot of the cliff, in a straight line with it, with X nearer the cliff. The angles of depression from the coastguard to X and to Y are 28° and 17° respectively.

(a)

Find the horizontal distance from the foot of the cliff to X.

(2)

(b)

Find the distance between the two buoys.

(3)

(c)

A kayak travels in a straight line from X to Y in 40 seconds. Find its average speed in km h⁻¹.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
tan 28° = 62 / d, or equivalent M1	✓ 1
d = 117 m (116.6…) A1	✓ 1
Part (b)
Distance to Y = 62 / tan 17° (= 202.8… m) M1	✓ 1
Subtracts the distance to X from the distance to Y: 202.8 − 116.6 M1	✓ 1
XY = 86.2 m. Accept 86 m from rounded values (203 − 117) A1	✓ 1
Part (c)
Speed = their XY / 40 (= 2.155 m s⁻¹) and converts by × 3.6, or equivalent M1	✓ 1
7.76 km h⁻¹. Accept 7.74 from XY = 86 m; follow through from their XY A1	✓ 1

Exam tip: Keep full calculator values for the distances to X and Y, and round only at the end. Early rounding changes the final speed in the third significant figure. The angle of depression equals the angle of elevation from the buoy, so use it directly with tan.

+ Add to test✓ In your test107IBMI-032-00-07Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium6 marks

A rectangular storeroom is a cuboid measuring 6.0 m long, 4.5 m wide and 2.8 m high. A straight cable is fixed from one top corner of the room to the bottom corner diagonally opposite it.

(a)

Find the length of the diagonal of the floor.

(2)

(b)

Find the length of the cable.

(2)

(c)

Find the angle between the cable and the floor.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
√(6.0² + 4.5²) M1	✓ 1
7.50 m A1	✓ 1
Part (b)
√(7.5² + 2.8²) M1	✓ 1
8.01 m A1	✓ 1
Part (c)
tan θ = 2.8 / 7.5 (or equivalent with sin or cos) M1	✓ 1
θ = 20.5° A1	✓ 1

Exam tip: For the angle between a line and a plane, identify the right-angled triangle formed by the line, its projection onto the plane, and the vertical drop.

+ Add to test✓ In your test108IBMI-032-00-09Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium7 marks

A triangular sail for a small dinghy has sides of length 4.8 m, 3.6 m and 5.5 m.

(a)

Find the size of the largest angle of the sail.

(3)

(b)

Find the area of the sail.

(2)

(c)

Sailcloth costs $34.75 per square metre. The sailmaker buys 15% more cloth than the area of the sail to allow for waste. Find the cost of the cloth.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Identifies the angle opposite the 5.5 m side R1	✓ 1
cos θ = (4.8² + 3.6² − 5.5²) / (2 × 4.8 × 3.6) M1	✓ 1
θ = 80.4° A1	✓ 1
Part (b)
½ × 4.8 × 3.6 × sin 80.4° M1	✓ 1
8.52 m² A1	✓ 1
Part (c)
8.52 × 1.15 × 34.75 M1	✓ 1
$340.46 (accept $340.48 from 8.52) A1	✓ 1

Common error: The largest angle is always opposite the longest side; choosing the wrong side in the cosine rule gives one of the smaller angles.

+ Add to test✓ In your test109IBMI-032-00-10Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium7 marks

Two hikers leave a campsite C at the same time. Hiker A walks 3.5 km on a bearing of 030°. Hiker B walks 5.2 km on a bearing of 120°.

(a)

Explain why angle ACB is a right angle.

(1)

(b)

Find the distance between the two hikers.

(2)

(c)

Find angle CAB.

(2)

(d)

Find the bearing of hiker B from hiker A.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
120° − 30° = 90° R1	✓ 1
Part (b)
√(3.5² + 5.2²) M1	✓ 1
6.27 km A1	✓ 1
Part (c)
tan CAB = 5.2 / 3.5 M1	✓ 1
CAB = 56.1° A1	✓ 1
Part (d)
Bearing of C from A is 210° M1	✓ 1
210° − 56.1° = 154° A1	✓ 1

Exam tip: Sketch the north line at A and mark the back bearing to C first; then decide whether to add or subtract the triangle angle.

+ Add to test✓ In your test110IBMI-032-06-05Volume and surface area in context Paper 1Medium8 marks

A ridge tent is modelled as a triangular prism of length 3 m. Its cross-section is an isosceles triangle. The two sloping sides are each 2.1 m long, and the angle between them at the ridge is 70°. The tent is made entirely of fabric, including the two triangular ends and a rectangular groundsheet.

(a)

Find the area of the triangular cross-section.

(2)

(b)

Hence find the volume of the tent.

(1)

(c)

Find the width of the base of the tent.

(2)

(d)

Find the total area of fabric needed.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for using $\frac{1}{2}ab\sin C$: $\frac{1}{2}\times2.1^2\times\sin70°$	✓ 1
A1 for 2.07 m² (2.07202…)	✓ 1
Part (b)
A1 for $2.07202\times3=6.22$ m³	✓ 1
Part (c)
M1 for the cosine rule, $\sqrt{2.1^2+2.1^2-2(2.1)(2.1)\cos70°}$, or for $2\times2.1\sin35°$	✓ 1
A1 for 2.41 m (2.40902…)	✓ 1
Part (d)
M1 for finding the three rectangular face areas: two sloping rectangles $2\times2.1\times3=12.6$ and the groundsheet $2.40902\times3=7.227$	✓ 1
M1 for adding the two triangular ends, $2\times2.07202=4.144$, to the rectangular areas, so that all five faces are summed	✓ 1
A1 for 24.0 m² (23.971…)	✓ 1

Common error: The usual way to lose marks here is to leave out the groundsheet or one of the triangular ends. List all five faces of the prism before adding them.

+ Add to test✓ In your test111IBMI-032-06-06Volume and surface area in context Paper 1Medium7 marks

A grain silo is modelled as a cylinder of radius 3.5 m and height 12 m, with a hemispherical roof of the same radius on top. The silo stands on a concrete base.

(a)

Find the total volume of the silo.

(3)

(b)

The silo is filled to 80% of its volume with grain of density 750 kg m⁻³. Find the mass of grain in the silo.

(2)

(c)

The outer wall and roof are to be given a weatherproof coating. The base is not coated. Find the area to be coated.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for the cylinder volume, $\pi\times3.5^2\times12=461.8$	✓ 1
M1 for the hemisphere volume, $\frac{2}{3}\pi\times3.5^3=89.80$	✓ 1
A1 for 552 m³ (551.61…)	✓ 1
Part (b)
M1 for $0.8\times551.61\times750$	✓ 1
A1 for 331 000 kg (331 tonnes)	✓ 1
Part (c)
M1 for adding the curved cylinder area and half a sphere: $2\pi(3.5)(12)+2\pi(3.5)^2$	✓ 1
A1 for 341 m² (340.86…)	✓ 1

Exam tip: Only the exposed surfaces are coated. The circle where the roof meets the cylinder is not a surface, and the base is excluded, so do not add any $\pi r^2$ terms.

+ Add to test✓ In your test112IBMI-032-06-08Volume and surface area in context Paper 1Medium5 marks

Three tennis balls, each a sphere of diameter 6.7 cm, are sold in a closed cylindrical tube. The balls fit exactly: they touch the curved wall, and the stack touches both ends of the tube.

(a)

Find the volume of the tube.

(2)

(b)

Find the percentage of the tube's volume that is empty space.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for radius 3.35 and height 3 × 6.7 = 20.1 substituted in $\pi r^2h$	✓ 1
A1 for 709 cm³ (708.656…)	✓ 1
Part (b)
M1 for the volume of the balls: $3\times\frac{4}{3}\pi(3.35)^3=472.44$ cm³	✓ 1
M1 for $\frac{708.66-472.44}{708.66}\times100$ (or equivalent, including use of 709)	✓ 1
A1 for 33.3% (exactly one third); accept 33.4% from using 709	✓ 1

Exam tip: The height of the tube is three diameters, 20.1 cm. Keep unrounded values in your calculator until the final step so your percentage does not drift.

+ Add to test✓ In your test113IBMI-032-06-10Volume and surface area in context Paper 1Medium6 marks

A water trough for livestock is a prism of length 4 m. Its cross-section is an isosceles trapezium, with an open top of width 0.9 m, a base of width 0.5 m and a vertical depth of 0.4 m.

(a)

Find the area of the cross-section.

(2)

(b)

Find the capacity of the trough in litres.

(2)

(c)

Find the angle that each sloping side makes with the horizontal.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{1}{2}(0.9+0.5)\times0.4$	✓ 1
A1 for 0.28 m²	✓ 1
Part (b)
M1 for $0.28\times4=1.12$ m³	✓ 1
A1 for 1120 litres	✓ 1
Part (c)
M1 for a horizontal offset of 0.2 m on each side, with $\tan\theta=\frac{0.4}{0.2}$	✓ 1
A1 for 63.4°	✓ 1

Exam tip: Each sloping side covers half of the 0.4 m difference in widths, so the horizontal offset is 0.2 m, not 0.4 m.

+ Add to test✓ In your test114IBMI-032-05-06Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium6 marks

From the top of a vertical cliff 72 m above sea level, a coastguard sees two kayaks. Both kayaks are directly out to sea, in a straight line with the foot of the cliff. The angles of depression of the kayaks are 41° and 28°.

(a)

Find the horizontal distance from the foot of the cliff to the nearer kayak.

(2)

(b)

Find the distance between the two kayaks.

(3)

(c)

State one assumption you have made in your calculations.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
tan 41° = 72 / d M1 ; d = 82.8 m A1	✓ 2
Part (b)
Distance to the further kayak = 72 / tan 28° M1 ; = 135.4 m A1 ; separation = 135.41 − 82.83 = 52.6 m A1	✓ 3
Part (c)
For example: the cliff is vertical / the sea is level / the kayaks and the foot of the cliff lie on one straight line / the kayaks are treated as points R1	✓ 1

Common error: The angle of depression is measured down from the horizontal at the observer. By alternate angles it equals the angle of elevation at the kayak, not the angle at the top of the cliff inside the triangle.

+ Add to test✓ In your test115IBMI-032-05-10Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium6 marks

The inside of a shipping container is a cuboid 12.0 m long, 2.35 m wide and 2.39 m high. A company wants to transport long steel rods inside it.

(a)

Find the length of the diagonal of the floor of the container.

(2)

(b)

Find the length of the longest straight rod that fits inside the container.

(2)

(c)

Find the angle between this rod and the floor of the container.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
√(12.0² + 2.35²) M1 ; 12.2 m (12.228…) A1	✓ 2
Part (b)
√(12.228² + 2.39²) or √(12.0² + 2.35² + 2.39²) M1 ; 12.5 m (12.459…) A1	✓ 2
Part (c)
tan θ = 2.39 / 12.228 (or equivalent) M1 ; θ = 11.1° A1	✓ 2

Exam tip: In 3D problems, find a right-angled triangle that contains the line you need. The angle between a line and a plane uses the line's projection onto the plane, which here is the floor diagonal.

+ Add to test✓ In your test116IBMI-032-05-11Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium7 marks

A vertical wind-turbine tower TF has height 60 m, where F is its foot on horizontal ground. Two walkers stand at points A and B on the ground. The angle of elevation of T from A is 35° and from B is 22°. The angle AFB is 70°.

(a)

Find the distance AF.

(2)

(b)

Find the distance BF.

(2)

(c)

Find the distance between the two walkers.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
AF = 60 / tan 35° M1 ; 85.7 m A1	✓ 2
Part (b)
BF = 60 / tan 22° M1 ; 149 m (148.50…) A1	✓ 2
Part (c)
Use of the cosine rule in the horizontal triangle AFB M1 ; AB² = 85.69² + 148.50² − 2(85.69)(148.50)cos 70° A1 ; AB = 144 m A1	✓ 3

Common error: The 70° angle lies in the horizontal plane at F. Use it with the horizontal distances AF and BF, not with the sloping lines of sight TA and TB.

+ Add to test✓ In your test117IBMI-032-01-05Volume and surface area in context Paper 1Medium6 marks

A grain silo consists of a cylinder of radius 3.5 m and height 12 m, topped by a hemisphere of the same radius.

(a)

Find the total volume of the silo.

(3)

(b)

The silo is filled to 85% of its capacity with wheat of density 760 kg m⁻³. Find the mass of the wheat in tonnes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Cylinder: π(3.5)²(12) = 461.81… m³ M1 ; Hemisphere: $\frac{2}{3}\pi(3.5)^3$ = 89.80… m³ M1 ; Total = 552 m³ (551.61…) A1	✓ 3
Part (b)
Volume of wheat: 0.85 × 551.61… = 468.87… m³ M1 ; Mass: 468.87… × 760 = 356 341 kg (approx.) M1 ; 356 tonnes A1 (accept 357 tonnes by follow-through from 552 m³)	✓ 3

Common error: Using the full sphere formula for the dome doubles its contribution. A hemisphere is half of $\frac{4}{3}\pi r^3$. Keep full calculator values between parts to avoid rounding drift.

+ Add to test✓ In your test118IBMI-032-01-07Volume and surface area in context Paper 1Medium7 marks

A drinks company makes closed cylindrical cans that hold 330 cm³. The current design has radius 3.2 cm. Assume the metal used for every can has the same uniform thickness, so the amount of metal is proportional to the total surface area.

(a)

Find the height of the current can.

(2)

(b)

Find the total surface area of the current can.

(2)

(c)

A new design with radius 3.8 cm also holds 330 cm³. Determine whether the new design uses less metal, justifying your answer.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
π(3.2)²h = 330 M1 ; h = 10.3 cm (10.258…) A1	✓ 2
Part (b)
2π(3.2)² + 2π(3.2)(10.258…) M1 ; 271 cm² (270.59…) A1	✓ 2
Part (c)
New height = 330/(π × 3.8²) = 7.274… cm M1 ; surface area = 2π(3.8)² + 2π(3.8)(7.274…) = 264 cm² (264.41…) A1 ; 264 < 271, so the new design uses less metal R1	✓ 3

Exam tip: A closed can has two circular ends. In a comparison question, state a clear conclusion that refers to both values.

+ Add to test✓ In your test119IBMI-032-01-08Volume and surface area in context Paper 1Medium6 marks

A water trough for cattle is a prism 2.5 m long. Its cross-section is an isosceles trapezium with a horizontal base of width 50 cm, an open top of width 80 cm and a depth of 30 cm.

(a)

Find the capacity of the trough in litres.

(3)

(b)

The water in the trough is 20 cm deep. Find the volume of water as a percentage of the capacity of the trough.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Area = ½(0.8 + 0.5)(0.3) = 0.195 m² M1 ; × 2.5 = 0.4875 m³ M1 ; 488 litres (487.5) A1	✓ 3
Part (b)
Width of water surface = 50 + 20 = 70 cm M1 ; volume = ½(0.5 + 0.7)(0.2)(2.5) = 0.3 m³ M1 ; 61.5% A1	✓ 3

Common error: Assuming the trough is two-thirds full because the water is two-thirds of the depth ignores the narrowing cross-section. The width at the water surface is less than 80 cm.

+ Add to test✓ In your test120IBMI-032-07-r1-03Voronoi diagrams Paper 1Medium6 marks

Four villages are located at P(0, 0), Q(8, 0), R(8, 6) and S(0, 10), with coordinates in kilometres. The Voronoi diagram for the four villages has exactly two vertices, V₁(4, 3) and V₂(2.5, 5). A new landfill site must be built inside the quadrilateral PQRS, and the council will only consider the two Voronoi vertices as possible locations.

(a)

Verify that V₁ is equidistant from P, Q and R.

(2)

(b)

Find the distance from V₂ to the nearest village.

(2)

(c)

The landfill should be as far as possible from the nearest village. State which vertex should be chosen, justifying your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses distance formula, e.g. V₁P = √(4² + 3²) M1	✓ 1
V₁P = V₁Q = V₁R = 5 km A1	✓ 1
Part (b)
V₂P = √(2.5² + 5²) (= V₂R = V₂S), V₂Q = √(5.5² + 5²) = 7.43 M1	✓ 1
Nearest distance = 5.59 km A1	✓ 1
Part (c)
V₂ A1	✓ 1
Its nearest village is 5.59 km away, which is greater than 5 km for V₁ R1	✓ 1

Exam tip: For a 'furthest from all sites' problem, compare the distance from each candidate point to its NEAREST site, then choose the largest of these minimum distances.

+ Add to test✓ In your test121IBMI-032-07-r2-03Voronoi diagrams Paper 1Medium8 marks

A village has three wells at A(1, 2), B(5, 2) and C(3, 6), with coordinates in kilometres. Each household collects water from its nearest well. The perpendicular bisector of AB has equation $x = 3$.

(a)

Show that the perpendicular bisector of AC has equation $y = -0.5x + 5$.

(3)

(b)

Find the coordinates of the Voronoi vertex V where the three cells meet.

(2)

(c)

Explain why V is the same distance from all three wells.

(1)

(d)

A fourth well is to be dug inside triangle ABC as far as possible from the existing wells. It is placed at V. Find its distance from the nearest existing well.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint of AC is (2, 4) A1	✓ 1
Gradient of AC $= 2$, so perpendicular gradient $= -0.5$ M1	✓ 1
$y - 4 = -0.5(x - 2)$ leading to $y = -0.5x + 5$ A1	✓ 1
Part (b)
Substitutes $x = 3$: $y = -1.5 + 5$ M1	✓ 1
V(3, 3.5) A1	✓ 1
Part (c)
V lies on the bisectors of AB and AC, so VA = VB and VA = VC R1	✓ 1
Part (d)
$\sqrt{(3-1)^2 + (3.5-2)^2}$ M1	✓ 1
2.5 km A1	✓ 1

Exam tip: A 'show that' answer must show every step, from the midpoint through the perpendicular gradient to the rearranged equation. Simply checking that the given equation works earns fewer marks.

+ Add to test✓ In your test122IBMI-032-02-r1-02Voronoi diagrams Paper 1Medium7 marks

Two electric-vehicle charging stations, A and B, are modelled as sites of a Voronoi diagram on a grid where 1 unit represents 1 km. Station A is at (2, 1). The location of B has been lost from the records, but the edge separating the two cells is known to lie on the line $x + 2y = 9$.

(a)

Find the gradient of the edge, and hence write down the gradient of the line segment AB.

(2)

(b)

Write down the equation of the line through A and B.

(1)

(c)

Find the coordinates of M, the point where line AB meets the edge.

(2)

(d)

Hence find the coordinates of B.

(1)

(e)

Find the distance between the two charging stations.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$y = -\frac{1}{2}x + 4.5$, so gradient of edge = $-\frac{1}{2}$ A1	✓ 1
The edge is the perpendicular bisector of AB, so gradient of AB = 2 R1	✓ 1
Part (b)
$y = 2x - 3$ A1	✓ 1
Part (c)
Substituting: $x + 2(2x - 3) = 9$ M1	✓ 1
$x = 3$, $y = 3$, so M(3, 3) A1	✓ 1
Part (d)
M is the midpoint of AB, so B = (2×3 − 2, 2×3 − 1) = (4, 5) A1	✓ 1
Part (e)
AB = $\sqrt{2^2 + 4^2} = \sqrt{20}$ = 4.47 km A1	✓ 1

Common error: M is the midpoint of AB, not the position of B. B is found by going the same displacement again from M as from A to M.

+ Add to test✓ In your test123IBMI-032-02-r2-04Voronoi diagrams Paper 1Medium8 marks

A district is modelled by the square region $0 \le x \le 10$, $0 \le y \le 10$, with units in kilometres. Four libraries are located at A(2, 2), B(8, 2), C(2, 8) and D(8, 8). The Voronoi diagram for these four sites consists of the lines $x = 5$ and $y = 5$. A fifth library, E, is built at (5, 5) and the Voronoi diagram is updated.

(a)

Show that the perpendicular bisector of AE has equation $x + y = 7$.

(2)

(b)

Write down the equations of the perpendicular bisectors of BE, CE and DE.

(2)

(c)

Find the coordinates of the vertices of the Voronoi cell of library E.

(2)

(d)

Find the area of the region served by library E.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint (3.5, 3.5) and gradient of AE = 1, so perpendicular gradient = −1 M1	✓ 1
$y - 3.5 = -(x - 3.5)$ giving $x + y = 7$ A1	✓ 1
Part (b)
BE: $y = x - 3$; CE: $y = x + 3$ A1	✓ 1
DE: $x + y = 13$ A1	✓ 1
Part (c)
Solving pairs of bisector equations simultaneously M1	✓ 1
(5, 2), (8, 5), (5, 8), (2, 5) A1	✓ 1
Part (d)
The cell is a square (rhombus) with diagonals of length 6 km, area = $\frac{6 \times 6}{2}$ M1	✓ 1
= 18 km² A1	✓ 1

Exam tip: When a new site is added, only the bisectors between the new site and its neighbours are needed; symmetry can save a lot of working.

+ Add to test✓ In your test124IBMI-032-07-r1-04Voronoi diagrams Paper 1Medium8 marks

Three weather stations record monthly rainfall. Coordinates are in kilometres.
Station W₁ is at (2, 2) and recorded 34 mm.
Station W₂ is at (10, 4) and recorded 41 mm.
Station W₃ is at (6, 10) and recorded 28 mm.
Rainfall at other locations is estimated using nearest-neighbour interpolation, based on the Voronoi diagram of the three stations.

(a)

A farm is located at F(7, 5). Estimate the monthly rainfall at F, showing clear working.

(3)

(b)

Find the equation of the perpendicular bisector of W₂W₃.

(3)

(c)

A second farm G lies on the edge between the cells of W₂ and W₃ and has x-coordinate 11. Find the y-coordinate of G. Suggest, with a reason, a suitable rainfall estimate for G.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Calculates the distances from F to all three stations: FW₁ = √34 ≈ 5.83 km, FW₂ = √10 ≈ 3.16 km, FW₃ = √26 ≈ 5.10 km M1	✓ 1
F lies in the cell of W₂, the nearest station A1	✓ 1
Estimated rainfall = 41 mm A1	✓ 1
Part (b)
Midpoint = (8, 7) A1	✓ 1
Gradient of W₂W₃ = 6/(−4) = −1.5, so the perpendicular gradient = 2/3 M1	✓ 1
y − 7 = (2/3)(x − 8), i.e. y = (2/3)x + 5/3 (or 2x − 3y + 5 = 0) A1	✓ 1
Part (c)
y = (2/3)(11) + 5/3 = 9 A1	✓ 1
G is equidistant (√26 km) from W₂ and W₃, so nearest-neighbour interpolation does not give a single value. Award the mark for any sensible estimate with this reason, for example the mean (41 + 28)/2 = 34.5 mm, or either 41 mm or 28 mm with a stated tie-break rule R1	✓ 1

Exam tip: Nearest-neighbour interpolation assigns the value of the site whose cell contains the point. Justify the cell by comparing distances, not by judging position by eye. A point on a cell edge is equidistant from two sites, so state how you resolve the tie.

+ Add to test✓ In your test125IBMI-032-02-r1-05Voronoi diagrams Paper 1Medium7 marks

Three villages are located at A(0, 0), B(8, 0) and C(2, 6), where 1 unit represents 1 km. A recycling plant is to be built inside triangle ABC at the point that is as far as possible from the nearest village. This point is the vertex of the Voronoi diagram of the three villages.

(a)

Write down the equation of the perpendicular bisector of AB.

(1)

(b)

Find the equation of the perpendicular bisector of AC.

(2)

(c)

Find the coordinates of the Voronoi vertex V.

(1)

(d)

Find the distance from V to each village.

(2)

(e)

Interpret your answer to the previous part in context.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$x = 4$ A1	✓ 1
Part (b)
Midpoint (1, 3) and gradient of AC = 3, so perpendicular gradient $-\frac{1}{3}$ M1	✓ 1
$y = -\frac{1}{3}x + \frac{10}{3}$ (or equivalent, e.g. $x + 3y = 10$) A1	✓ 1
Part (c)
At $x = 4$: $y = -\frac{4}{3} + \frac{10}{3} = 2$, so V(4, 2) A1	✓ 1
Part (d)
VA = $\sqrt{4^2 + 2^2}$, VB = $\sqrt{4^2 + 2^2}$, VC = $\sqrt{2^2 + 4^2}$ M1	✓ 1
Each distance = $\sqrt{20}$ ≈ 4.47 km A1	✓ 1
Part (e)
The plant is 4.47 km from all three villages. Moving it anywhere else in the triangle brings it closer to at least one village, so 4.47 km is the greatest possible distance to the nearest village R1	✓ 1

Exam tip: The Voronoi vertex is equidistant from all three sites, so checking that VA = VB = VC is a quick way to verify your coordinates.

+ Add to test✓ In your test126IBMI-032-02-r2-03Voronoi diagrams Paper 1Medium8 marks

A supermarket chain has three stores in a city, at A(1, 1), B(9, 1) and C(5, 7), where units are kilometres. A rival company wants to open a new store inside triangle ABC. It chooses the point whose distance to the nearest existing store is as large as possible. For this triangle, that point is the vertex of the Voronoi diagram for A, B and C.

(a)

Show that the perpendicular bisector of AC has equation $2x + 3y = 18$.

(3)

(b)

Write down the equation of the perpendicular bisector of AB.

(1)

(c)

Find the coordinates of the vertex of the Voronoi diagram for the three stores.

(2)

(d)

The rival opens its store at this vertex. Find the distance from the new store to the nearest existing store.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint of AC = (3, 4) A1	✓ 1
Gradient of AC = $\frac{6}{4} = \frac{3}{2}$, so the perpendicular gradient = $-\frac{2}{3}$ M1	✓ 1
$y - 4 = -\frac{2}{3}(x-3)$ leading to $3y - 12 = -2x + 6$, i.e. $2x + 3y = 18$ AG A1	✓ 1
Part (b)
$x = 5$ A1	✓ 1
Part (c)
Substituting $x = 5$ into $2x + 3y = 18$: $10 + 3y = 18$ M1	✓ 1
$\left(5, \frac{8}{3}\right)$ or (5, 2.67) A1	✓ 1
Part (d)
For example, the distance to A: $\sqrt{(5-1)^2 + \left(\frac{8}{3}-1\right)^2}$ M1	✓ 1
$= \frac{13}{3}$ km, or 4.33 km (3 s.f.) A1	✓ 1
The vertex is equidistant from A, B and C, so a correct distance to any one store earns full marks.	—

Common error: A common error is substituting the rounded value y = 2.67 into the distance calculation, which can give a slightly wrong final answer. Use the exact value 8/3 or the full calculator value.

+ Add to test✓ In your test127IBMI3S-3-05-06Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium6 marks

A surveyor wants the width of a river. She marks two points A and B, 60 m apart, on her bank. A tree T stands on the opposite bank. She measures angle TAB = 72° and angle TBA = 55°.

(a)

Find angle ATB.

(1)

(b)

Use the sine rule to calculate the distance AT.

(3)

(c)

Hence find the shortest distance from T to the line AB, which is the width of the river.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $180^\circ-72^\circ-55^\circ=53^\circ$	✓ 1
Part (b)
M1 $\dfrac{AT}{\sin55^\circ}=\dfrac{60}{\sin53^\circ}$	✓ 1
A1 correct substitution	✓ 1
A1 $AT=61.5$ m	✓ 1
Part (c)
M1 $AT\times\sin72^\circ$ (ECF from (b))	✓ 1
A1 $58.5$ m	✓ 1

Common error: The sine rule pairs each side with the angle opposite it; here 60 m is opposite angle T, not angle A.

+ Add to test✓ In your test128IBMI3S-3-05-07Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium6 marks

A paper cup for a water dispenser is a cone with base radius 6.0 cm and slant height 15 cm.

(a)

Calculate the vertical height of the cone.

(2)

(b)

Calculate the volume of the cup, in cm³.

(2)

(c)

Calculate the area of paper in the curved surface of the cup.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $h=\sqrt{15^2-6^2}$	✓ 1
A1 $h=13.7$ cm	✓ 1
Part (b)
M1 $V=\frac13\pi\times6^2\times13.75$ (ECF from (a))	✓ 1
A1 $V=518$ cm³	✓ 1
Part (c)
M1 $\pi rl=\pi\times6\times15$	✓ 1
A1 $283$ cm²	✓ 1

Common error: Do not use the slant height in the volume formula; the volume needs the perpendicular height.

+ Add to test✓ In your test129IBMI3S-3-05-08Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium6 marks

A grain silo is a cylinder of radius 2.5 m and height 6.0 m, with a hemispherical roof of the same radius on top.

(a)

Calculate the total volume of the silo.

(3)

(b)

The outside curved surfaces of the silo (not the base) are to be painted. Calculate the area to be painted.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cylinder $\pi\times2.5^2\times6=117.8$	✓ 1
M1 hemisphere $\frac23\pi\times2.5^3=32.7$	✓ 1
A1 $151$ m³	✓ 1
Part (b)
M1 curved cylinder $2\pi\times2.5\times6=94.2$	✓ 1
M1 hemisphere $2\pi\times2.5^2=39.3$	✓ 1
A1 $134$ m²	✓ 1

Exam tip: For a composite solid, list each piece separately and decide which faces are actually exposed before adding areas.

+ Add to test✓ In your test130IBMI3S-3-05b-05Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium6 marks

Three clinics A, B and C serve a region. AC = 9.5 km, BC = 12.8 km and AB = 15.4 km.

(a)

Calculate the size of angle ACB.

(3)

(b)

Find the area of the triangle ABC.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the cosine rule with AB opposite angle C	✓ 1
M1 correct substitution: cos C = (9.5² + 12.8² − 15.4²)/(2 × 9.5 × 12.8)	✓ 1
A1 C = 86.0°	✓ 1
Part (b)
M1 attempt to use area = ½ab sin C	✓ 1
A1 correct substitution using the value of C from part (a) (ECF)	✓ 1
A1 60.7 km²	✓ 1

Common error: Putting the side opposite the required angle in the wrong place in the cosine rule.

+ Add to test✓ In your test131IBMI3S-3-05b-06Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium6 marks

Points A and B are on level ground 60 m apart. A wind turbine is built at point P on the same ground. Angle PAB = 62° and angle PBA = 71°. From A, the angle of elevation of the top of the turbine is 24°.

(a)

Find the distance AP.

(4)

(b)

Calculate the height of the turbine.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 angle APB = 180° − 62° − 71° = 47°	✓ 1
M1 attempt to use the sine rule: AP/sin 71° = 60/sin 47°	✓ 1
M1 correct substitution and rearrangement	✓ 1
A1 AP = 77.6 m	✓ 1
Part (b)
M1 attempt to use tan 24° = height/AP	✓ 1
A1 34.5 m	✓ 1

Exam tip: Find the third angle first so that you can pair each side with its opposite angle in the sine rule.

+ Add to test✓ In your test132IBMI3S-3-05b-07Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium6 marks

A shop entrance is raised 0.45 m above the pavement. A straight ramp rises at a constant angle of 4.8° to the horizontal.

(a)

Calculate the length of the ramp surface.

(2)

(b)

Find the horizontal distance covered by the ramp.

(2)

(c)

Space limits the horizontal distance to 5.0 m. Find the smallest angle the ramp must make with the horizontal to reach the same height.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use sin 4.8° = 0.45/L	✓ 1
A1 L = 5.38 m	✓ 1
Part (b)
M1 attempt to use tan 4.8° = 0.45/x or Pythagoras	✓ 1
A1 5.36 m	✓ 1
Part (c)
M1 attempt to use tan θ = 0.45/5.0	✓ 1
A1 θ = 5.14°	✓ 1

Common error: Using cosine instead of sine when the known side is opposite the angle.

+ Add to test✓ In your test133IBMI3S-3-05b-08Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

An offshore substation S supplies two turbines. Turbine T₁ is 6.4 km from S on a bearing of 040°. Turbine T₂ is 9.1 km from S on a bearing of 115°.

(a)

Write down the size of angle T₁ST₂.

(1)

(b)

Calculate the distance T₁T₂.

(3)

(c)

Find the bearing of T₂ from T₁, to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 115° − 40° = 75°	✓ 1
Part (b)
M1 attempt to use the cosine rule	✓ 1
M1 T₁T₂² = 6.4² + 9.1² − 2(6.4)(9.1)cos 75°	✓ 1
A1 9.68 km	✓ 1
Part (c)
M1 attempt to find angle ST₁T₂ using the sine or cosine rule	✓ 1
A1 angle ST₁T₂ = 65.3°	✓ 1
A1 back bearing of S from T₁ is 220°, so bearing = 220° − 65.3° = 155° ≈ 155°	✓ 1

Exam tip: When using the sine rule for an angle, check whether that angle could be obtuse by comparing the side lengths.

+ Add to test✓ In your test134IBMI3S-3-01r6-05Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

Three libraries A, B and C are joined by straight roads. AB = 5.2 km, AC = 7.9 km and angle BAC = 48°. Use unrounded values in later parts.

(a)

Calculate the length of the road BC, to 3 significant figures.

(3)

(b)

Calculate the area of triangle ABC, to 3 significant figures.

(2)

(c)

A cycle path is built from A to meet BC at right angles. Find the length of this path, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the cosine rule $BC^2 = 5.2^2 + 7.9^2 - 2(5.2)(7.9)\cos 48°$	✓ 1
A1 $BC^2 = 34.47$	✓ 1
A1 BC = 5.87 km	✓ 1
Part (b)
M1 attempt to use $\frac12 ab\sin C$	✓ 1
A1 15.3 km²	✓ 1
Part (c)
M1 use area $= \frac12 \times BC \times$ path length	✓ 1
A1 5.20 km	✓ 1

Exam tip: When two sides and the angle between them are known, start with the cosine rule; the sine rule needs a side and its opposite angle.

+ Add to test✓ In your test135IBMI3S-3-01r6-06Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium6 marks

A wind turbine tower TX stands vertically on level ground, with X at its base. Points A and B lie on the ground on the same side of the tower, in a straight line with X. B is 40 m further from X than A. From A the angle of elevation of the top T is 41°, and from B it is 28°. Use unrounded values in later parts.

(a)

Show that angle ATB = 13°.

(2)

(b)

Use the sine rule in triangle ATB to find the length AT, to 3 significant figures.

(2)

(c)

Find the height of the tower, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 angle TAX is an exterior angle of triangle ATB, so angle ATB = angle TAX − angle TBA	✓ 1
A1 $41° - 28° = 13°$	✓ 1
Part (b)
M1 $\frac{AT}{\sin 28°} = \frac{40}{\sin 13°}$	✓ 1
A1 AT = 83.5 m	✓ 1
Part (c)
M1 $TX = AT \sin 41°$	✓ 1
A1 54.8 m	✓ 1

Exam tip: Mark the 13° angle on your sketch; a missing angle at the top of a triangle between two sight lines is found by subtracting the two elevations.

+ Add to test✓ In your test136IBMI3S-3-01r6-07Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium8 marks

Three libraries are located on a map at A(1, 2), B(9, 2) and C(5, 8), where each unit is 1 km. A Voronoi diagram is drawn for these three sites.

(a)

Find the equation of the perpendicular bisector of AB.

(2)

(b)

Find the equation of the perpendicular bisector of AC.

(3)

(c)

Hence find the coordinates of the Voronoi vertex where these two edges meet, and the distance from this vertex to library A, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AB is (5, 2) and AB is horizontal	✓ 1
A1 $x = 5$	✓ 1
Part (b)
M1 midpoint of AC is (3, 5) and gradient of AC is $\frac{6}{4} = 1.5$	✓ 1
M1 perpendicular gradient $-\frac{2}{3}$ used in $y - 5 = -\frac23(x-3)$	✓ 1
A1 $y = -\frac23 x + 7$	✓ 1
Part (c)
M1 substitute $x = 5$ into the equation of the bisector of AC	✓ 1
A1 vertex $\left(5, \frac{11}{3}\right)$, which is (5, 3.67)	✓ 1
A1 distance to A = 4.33 km	✓ 1

Exam tip: A Voronoi vertex is equidistant from three sites, so you can check your vertex by confirming its distances to A, B and C are equal.

+ Add to test✓ In your test137IBMI3S-3-01r6-08Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium5 marks

The roof of a library is a hemispherical dome of radius 7.5 m that sits on top of a cylindrical drum of radius 7.5 m and height 4 m. The curved surfaces of the dome and of the drum are to be painted. The base of the drum and the flat circle where the dome meets the drum are not painted.

(a)

Calculate the total area to be painted, to 3 significant figures.

(3)

(b)

One litre of paint covers 8 m² and the paint is sold only in 5-litre cans costing £48.50 each. Find the number of cans required and the total cost.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 curved surface area of hemisphere $2\pi r^2$	✓ 1
M1 curved surface area of cylinder $2\pi r h$	✓ 1
A1 542 m²	✓ 1
Part (b)
M1 volume of paint needed $= 67.74$ litres, so $\frac{67.74}{5} = 13.55$ cans	✓ 1
A1 14 cans, cost £679.00	✓ 1

Common error: The number of cans must be a whole number, so round up, not to the nearest integer.

+ Add to test✓ In your test138IBMI3S-3-01r6-09Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

A chemical tank for a swimming pool is in the shape of an inverted cone with its vertex at the bottom and its axis vertical. The open top has radius 1.4 m and the tank has height 2.1 m. (1 m³ = 1000 litres.) Use unrounded values in later parts.

(a)

Calculate the capacity of the tank in litres, to 3 significant figures.

(2)

(b)

The tank is filled from empty at a rate of 35 litres per minute. Find the time needed to fill it completely, to the nearest minute.

(2)

(c)

At one time the tank holds 1500 litres. Find the depth of liquid in the tank, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use $V = \frac13 \pi r^2 h$	✓ 1
A1 4310 litres	✓ 1
Part (b)
M1 divide capacity by 35	✓ 1
A1 123 minutes, which is 123 minutes to the nearest minute	✓ 1
Part (c)
M1 radius of the liquid surface at depth $d$ is $\frac{1.4}{2.1}d = \frac23 d$ (similar triangles)	✓ 1
M1 $\frac13\pi\left(\frac23 d\right)^2 d = 1.5$	✓ 1
A1 $d = 1.48$ m	✓ 1

Exam tip: When a cone is partly filled, the liquid forms a smaller similar cone; use the ratio radius : height to link the radius to the depth.

+ Add to test✓ In your test139IBMI3S-3-01r6-10Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

The glass roof of a library atrium is a right pyramid with a square base of side 12 m. The apex is 5 m vertically above the centre of the base. The four triangular faces are made of glass. Use unrounded values in later parts.

(a)

Find the slant height of a triangular face, measured from the midpoint of a base edge to the apex, to 3 significant figures.

(2)

(b)

Find the angle between a triangular face and the base, to 3 significant figures.

(2)

(c)

Glass costs £86 per square metre. Find the cost of the glass for the four faces, to the nearest £.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 horizontal distance from centre to midpoint of an edge is 6 m, so use $\sqrt{6^2 + 5^2}$	✓ 1
A1 7.81 m	✓ 1
Part (b)
M1 attempt to use $\tan\theta = \frac56$	✓ 1
A1 39.8°	✓ 1
Part (c)
M1 area of four faces $= 4 \times \frac12 \times 12 \times$ slant height	✓ 1
A1 area = 187.4 m²	✓ 1
A1 cost = £16,120	✓ 1

Common error: The slant height of a face is not the sloping edge of the pyramid; the sloping edge runs to a corner of the base, the slant height to the midpoint of an edge.

+ Add to test✓ In your test140IBMI3S-3-02r8-02Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

Three swimming pools are modelled as sites on a map, coordinates in km: A(1, 1), B(7, 1) and C(4, 7). A resident lives at H(5, 4).

(a)

Calculate the distance from H to each pool, to 3 significant figures, and state which pool is nearest.

(3)

(b)

Find the equation of the perpendicular bisector of BC, in the form y = mx + c.

(3)

(c)

Use your answer to (b) to explain why H lies in the cell of C rather than the cell of B.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 distance formula used correctly for at least one pool	✓ 1
A1 AH = 5.00 km, BH = 3.61 km, CH = 3.16 km	✓ 1
A1 C is nearest	✓ 1
Part (b)
M1 midpoint of BC = (5.5, 4)	✓ 1
M1 gradient of BC = −2, so perpendicular gradient = ½	✓ 1
A1 y = 0.5x + 1.25	✓ 1
Part (c)
R1 at x = 5 the bisector has y = 3.75; H has y = 4, above the line on the same side as C (C: 7 > 0.5 × 4 + 1.25)	✓ 1

Exam tip: A point lies in the cell of the site on its own side of a bisector; test one point of each site against the line.

+ Add to test✓ In your test141IBMI3S-3-02r8-06Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

A ferry leaves harbour H and sails 14 km on a bearing of 062° to a wind farm W. It then sails 9 km on a bearing of 148° to a port P.

(a)

Show that angle HWP = 94°.

(2)

(b)

Find the distance HP, to 3 significant figures.

(2)

(c)

Find the bearing of P from H, to the nearest tenth of a degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of H from W = 062° + 180° = 242°	✓ 1
A1 242° − 148° = 94°	✓ 1
Part (b)
M1 HP² = 14² + 9² − 2 × 14 × 9 × cos 94°	✓ 1
A1 HP = 17.2 km	✓ 1
Part (c)
M1 sine rule: sin(∠WHP) ÷ 9 = sin 94° ÷ HP (unrounded HP)	✓ 1
A1 ∠WHP = 31.5°	✓ 1
A1 bearing 093.5°	✓ 1

Exam tip: Draw a north line at each point; angles between the legs come from back bearings (add or subtract 180°).

+ Add to test✓ In your test142IBMI3S-3-02r8-13Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Medium6 marks

A cylindrical water tank at a wind farm has internal radius 2.4 m and height 3.6 m. It is currently 35% full. A pump fills it at 15 litres per second. (1 m³ = 1000 litres.)

(a)

Calculate the capacity of the tank in m³, to 3 significant figures.

(2)

(b)

Find the volume of water, in litres, needed to fill the tank completely. Use the unrounded capacity and give your answer to 3 significant figures.

(2)

(c)

Find the time needed to fill the tank, in minutes, to 3 significant figures. Use the unrounded volume.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 2.4² × 3.6	✓ 1
A1 65.1 m³	✓ 1
Part (b)
M1 0.65 × 65 144… litres	✓ 1
A1 42 300 litres	✓ 1
Part (c)
M1 42 343… ÷ 15 ÷ 60	✓ 1
A1 47.0 minutes	✓ 1

Common error: Convert units before dividing: 1 m³ is 1000 litres and the pump rate is per second, so the time must be converted to minutes at the end.

+ Add to test✓ In your test143IBMI3S-3-02r9-05Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

Three charging stations for warehouse robots are at P(1, 1), Q(9, 1) and R(5, 7), with coordinates in metres. A fourth station S is added at (5, 3). A Voronoi diagram is drawn for the four stations.

(a)

Show that the perpendicular bisector of RS has equation y = 5.

(2)

(b)

Find the equation of the perpendicular bisector of PS. Give your answer in the form y = mx + c.

(3)

(c)

The cells of P, S and R meet at a Voronoi vertex. Find its coordinates.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 R and S have the same x-coordinate, so the bisector is horizontal	✓ 1
A1 midpoint of RS is (5, 5), so y = 5 (AG)	✓ 1
Part (b)
M1 midpoint of PS is (3, 2)	✓ 1
M1 gradient of PS = 2 ÷ 4 = 0.5, so the perpendicular gradient is −2	✓ 1
A1 y = −2x + 8	✓ 1
Part (c)
M1 solve −2x + 8 = 5	✓ 1
A1 x = 1.5, giving the vertex (1.5, 5)	✓ 1

Exam tip: After finding a vertex, check it by confirming that its distances to the three sites are equal.

+ Add to test✓ In your test144IBMI3S-3-02r9-07Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

A costume designer cuts a skirt panel from fabric. The panel is part of a circle: the region between two arcs of radii 15 cm and 70 cm, with the same centre, and with each arc subtending an angle of 270° at the centre. The shorter arc forms the waist.

(a)

Calculate the length of the waist arc. Give your answer to 3 significant figures.

(2)

(b)

Calculate the area of the panel. Give your answer to the nearest cm².

(3)

(c)

The fabric costs €14.50 per m². Using the unrounded area, find the cost of the fabric for one panel, to the nearest cent (1 m² = 10 000 cm²).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (270 ÷ 360) × 2π × 15	✓ 1
A1 70.7 cm	✓ 1
Part (b)
M1 area of large sector (270 ÷ 360) × π × 70² or equivalent	✓ 1
M1 subtract the small sector (270 ÷ 360) × π × 15²	✓ 1
A1 11 015 cm²	✓ 1
Part (c)
M1 convert to m²: 11 015.2… ÷ 10 000 = 1.1015…	✓ 1
A1 €15.97	✓ 1

Common error: Convert cm² to m² by dividing by 10 000, not by 100.

+ Add to test✓ In your test145IBMI3S-3-02r9-08Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

Two trawlers leave port O at the same time. Trawler A travels 8.4 km on a bearing of 038°. Trawler B travels 11.2 km on a bearing of 115°. Use unrounded values in later parts.

(a)

Find the size of angle AOB.

(1)

(b)

Calculate the distance AB. Give your answer to 3 significant figures.

(3)

(c)

Find the bearing of B from A, to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 115° − 38° = 77°	✓ 1
Part (b)
M1 AB² = 8.4² + 11.2² − 2 × 8.4 × 11.2 × cos 77°	✓ 1
A1 AB² = 153.7…	✓ 1
A1 AB = 12.4 km	✓ 1
Part (c)
M1 sine rule: sin(OAB) ÷ 11.2 = sin 77° ÷ AB, giving angle OAB = 61.7…°	✓ 1
M1 the bearing of O from A is 038° + 180° = 218°	✓ 1
A1 bearing = 218° − 61.7…° = 156°	✓ 1

Exam tip: Use the unrounded value of AB in the sine rule, and draw north lines at both A and O before finding the bearing.

+ Add to test✓ In your test146IBMI3S-3-02r9-09Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium5 marks

A tailor cuts triangular fabric patches. Patch 1 has two sides of length 28 cm and 40 cm with an angle of 72° between them.

(a)

Calculate the area of patch 1. Give your answer to the nearest cm².

(2)

(b)

Patch 2 has two sides of length 36 cm and 50 cm and an area of 540 cm². The angle between these sides is acute. Find the size of this angle, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × 28 × 40 × sin 72°	✓ 1
A1 533 cm²	✓ 1
Part (b)
M1 ½ × 36 × 50 × sin θ = 540	✓ 1
A1 sin θ = 0.6	✓ 1
A1 θ = 36.9°	✓ 1

Common error: The area formula needs the angle between the two given sides; the sine of an angle also has an obtuse solution, so check the condition in the question.

+ Add to test✓ In your test147IBMI3S-3-02r9-10Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium6 marks

A delivery robot climbs a straight ramp of length 3.6 m. The top of the ramp is 0.9 m higher than the bottom.

(a)

Find the angle that the ramp makes with the horizontal. Give your answer to 3 significant figures.

(2)

(b)

Find the horizontal distance between the bottom and the top of the ramp. Give your answer to 3 significant figures.

(2)

(c)

The robot can climb ramps that make an angle of at most 12° with the horizontal. A new ramp must rise 0.9 m and make an angle of exactly 12°. Find the length of the new ramp. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin θ = 0.9 ÷ 3.6	✓ 1
A1 14.5°	✓ 1
Part (b)
M1 √(3.6² − 0.9²) or 3.6 cos(14.47…°)	✓ 1
A1 3.49 m	✓ 1
Part (c)
M1 sin 12° = 0.9 ÷ length	✓ 1
A1 length = 4.33 m	✓ 1

Exam tip: Label the hypotenuse, opposite and adjacent sides before choosing sin, cos or tan.

+ Add to test✓ In your test148IBMI3S-3-02r9-11Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium5 marks

A lookout is on a cliff top 42 m above sea level. Two fishing boats, A and B, are in a straight line from the base of the cliff, both on the same side of it. The angle of depression of A from the lookout is 6.5° and the angle of depression of B is 12°. Give distances to the nearest metre.

(a)

Find the horizontal distance from the base of the cliff to boat A.

(2)

(b)

Find the horizontal distance from the base of the cliff to boat B.

(2)

(c)

Hence find the distance between the boats.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 42 ÷ tan 6.5°	✓ 1
A1 369 m	✓ 1
Part (b)
M1 42 ÷ tan 12°	✓ 1
A1 198 m	✓ 1
Part (c)
A1 369 − 198 = 171 m	✓ 1

Common error: The angle of depression equals the angle of elevation from the boat to the lookout (alternate angles); it is not measured from the vertical.

+ Add to test✓ In your test149IBMI3S-3-01r11-02Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium6 marks

A swimming pool is 25 m long and 10 m wide. The floor slopes at a constant angle from the shallow end, where the water depth is 1.2 m, to the deep end, where the depth is 2.8 m. The pool is a prism whose cross-section (a side view along the length) is a trapezium.

(a)

Find the volume of water in the full pool, in m³.

(2)

(b)

Find the angle the pool floor makes with the horizontal, correct to 3 significant figures.

(2)

(c)

The pool is filled from empty at a constant rate of 15.5 m³ per hour. Calculate the time needed to fill it, in hours, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
cross-section area = ½(1.2 + 2.8) × 25 = 50 m²	✓ 1
V = 50 × 10 = 500 m³	✓ 1
Part (b)
tan θ = (2.8 − 1.2) ÷ 25	✓ 1
θ = 3.66°	✓ 1
Part (c)
time = 500 ÷ 15.5	✓ 1
= 32.3 hours	✓ 1

Common error: The floor slope is the difference in depth over the horizontal length, not over the sloping length of the floor.

+ Add to test✓ In your test150IBMI3S-3-01r11-08Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

On a map where each unit represents 1 km, a swimming pool P is at (2, 1). A second pool Q is at an unknown point. In the Voronoi diagram for P and Q, the boundary between their regions is the line y = −2x + 10.

(a)

Write down the gradient of the line PQ.

(1)

(b)

Find the equation of the line PQ.

(2)

(c)

Find the coordinates of the point where PQ meets the boundary.

(2)

(d)

Hence find the coordinates of Q, and calculate the distance PQ, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
PQ is perpendicular to the boundary (gradient −2), so gradient of PQ = 1/2	✓ 1
Part (b)
y − 1 = 0.5(x − 2)	✓ 1
y = 0.5x	✓ 1
Part (c)
0.5x = −2x + 10	✓ 1
(4, 2)	✓ 1
Part (d)
this point is the midpoint of PQ, so Q = (6, 3)	✓ 1
PQ = √(4² + 2²) = 4.47 km	✓ 1

Common error: The boundary meets PQ at the midpoint of PQ, so the second site is found by doubling the displacement from P to that point, not by adding it once.

+ Add to test✓ In your test151IBMI3S-3-01r11-10Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

A maintenance vessel leaves harbour H and sails 12 km on a bearing of 065° to a wind turbine W. It then sails 9 km on a bearing of 150° to a second turbine X.

(a)

Show that angle HWX = 95°.

(2)

(b)

Calculate the distance HX, correct to 3 significant figures.

(3)

(c)

Find the bearing of X from H, correct to the nearest degree.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
the bearing of H from W is 065° + 180° = 245°	✓ 1
angle HWX = 245° − 150° = 95°	✓ 1
Part (b)
HX² = 12² + 9² − 2 × 12 × 9 × cos 95°	✓ 1
HX² = 243.83	✓ 1
HX = 15.6 km	✓ 1
Part (c)
sin(angle WHX) ÷ 9 = sin 95° ÷ 15.615, giving angle WHX = 35.0°	✓ 1
bearing = 065° + 35.0° = 100°	✓ 1

Common error: A bearing is measured clockwise from north, so the bearing of H from W is the bearing of W from H plus 180°.

+ Add to test✓ In your test152IBMI3S-3-01r11-11Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium6 marks

Three wind turbines T₁, T₂ and T₃ are at the corners of a triangular field. T₁T₂ = 640 m, T₂T₃ = 510 m and angle T₁T₂T₃ = 78°.

(a)

Calculate the distance T₁T₃, correct to the nearest metre.

(3)

(b)

Calculate the area of the field in hectares, correct to 3 significant figures. (1 hectare = 10 000 m².)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
T₁T₃² = 640² + 510² − 2 × 640 × 510 × cos 78°	✓ 1
T₁T₃² = 533975	✓ 1
T₁T₃ = 731 m	✓ 1
Part (b)
area = ½ × 640 × 510 × sin 78°	✓ 1
= 159634 m²	✓ 1
= 16.0 hectares	✓ 1

Exam tip: Use the sine formula for area only with two sides and the angle between them, here at T₂.

+ Add to test✓ In your test153IBMI3S-3-02r13-05Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Medium6 marks

A television camera is mounted at the top of a vertical tower of height 25 m beside a straight, level race track. A car approaches the foot of the tower along the track. The angle of depression of the car from the camera is 12°. After 1.2 seconds the angle of depression of the car is 30°.

(a)

Find the horizontal distance of the car from the foot of the tower when the angle of depression is 12°. Give your answer to 1 decimal place.

(2)

(b)

Find the horizontal distance of the car from the foot of the tower when the angle of depression is 30°. Give your answer to 1 decimal place.

(1)

(c)

Calculate the average speed of the car during the 1.2 seconds. Give your answer in m s⁻¹ to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 12° = 25/d	✓ 1
A1 117.6 m	✓ 1
Part (b)
A1 43.3 m	✓ 1
Part (c)
M1 distance travelled = 117.6 − 43.3 (use unrounded values if preferred)	✓ 1
M1 divide by 1.2	✓ 1
A1 61.9 m s⁻¹	✓ 1

Common error: The angle of depression is measured from the horizontal at the camera, so it equals the angle of elevation from the car to the camera.

+ Add to test✓ In your test154IBMI3S-3-02r13-06Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

Three apiaries A, B and C are on flat ground. B is 6.5 km from A on a bearing of 040°. C is 9.2 km from A on a bearing of 155°. Bearings are measured clockwise from north.

(a)

Show that angle BAC is 115°.

(1)

(b)

Calculate the distance BC. Give your answer to 3 significant figures.

(2)

(c)

Find the bearing of C from B, to the nearest degree.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 155° − 40° = 115° (AG)	✓ 1
Part (b)
M1 cosine rule: BC² = 6.5² + 9.2² − 2 × 6.5 × 9.2 × cos 115°	✓ 1
A1 BC = 13.3 km	✓ 1
Part (c)
M1 sine rule: sin(ABC)/9.2 = sin 115°/BC (using the unrounded BC)	✓ 1
A1 angle ABC = 38.8°	✓ 1
M1 the bearing of A from B is 220°, and C lies anticlockwise from A as seen from B, so bearing = 220° − 38.8°	✓ 1
A1 181°	✓ 1

Exam tip: To find a bearing from B back towards A, add or subtract 180° to the bearing of B from A, then apply the angle you have found.

+ Add to test✓ In your test155IBMI3S-3-02r13-07Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

Three apiaries are located at A(0, 0), B(10, 0) and C(4, 8) on a map where one unit represents 1 km. A Voronoi diagram is drawn for these three sites. A beekeeper wants to place a new hive at the Voronoi vertex, so that it is as far as possible from all three existing apiaries within the triangle ABC.

(a)

Write down the equation of the Voronoi edge between A and B.

(1)

(b)

Find the equation of the Voronoi edge between A and C. Give your answer in the form y = mx + c.

(3)

(c)

Find the coordinates of the Voronoi vertex.

(1)

(d)

Find the distance from the vertex to apiary A. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x = 5	✓ 1
Part (b)
M1 midpoint of AC = (2, 4)	✓ 1
M1 gradient of AC = 2, so the perpendicular gradient is −1/2	✓ 1
A1 y = −0.5x + 5	✓ 1
Part (c)
A1 (5, 2.5)	✓ 1
Part (d)
M1 √(5² + 2.5²)	✓ 1
A1 5.59 km	✓ 1

Exam tip: A Voronoi vertex is equidistant from three sites, so any one of the three distances gives the same answer; use the simplest.

+ Add to test✓ In your test156IBMI3S-3-02r13-08Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Medium6 marks

Two timing sensors, A and B, are placed on a map of a race circuit where one unit represents 100 m. Sensor A is at (2, 3). In the Voronoi diagram for these two sensors, the edge between their cells lies on the line x + y = 8.

(a)

Find the coordinates of sensor B.

(4)

(b)

A car is at (6, 4). Determine which sensor is nearer to the car and find that distance in metres, to the nearest metre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the line through A perpendicular to x + y = 8 has gradient 1: y = x + 1	✓ 1
M1 intersection with x + y = 8 is the midpoint (3.5, 4.5)	✓ 1
M1 B = 2 × midpoint − A	✓ 1
A1 B = (5, 6)	✓ 1
Part (b)
M1 distance to each sensor (√17 units to A and √5 units to B)	✓ 1
A1 sensor B, at 224 m	✓ 1

Common error: The sites are mirror images in the Voronoi edge, so the midpoint of the two sites lies on the edge, not on either site.

+ Add to test✓ In your test157IBMI3S-3-02r13-10Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

A team's coolant tank is in the shape of an inverted cone (point downwards) with a top radius of 0.60 m and a vertical height of 1.00 m.

(a)

Calculate the volume of the tank when it is full. Give your answer in m³ to 3 significant figures.

(2)

(b)

The tank is filled to a depth of 0.50 m. Find the volume of coolant, in m³ to 3 significant figures.

(3)

(c)

The coolant is drained at a constant rate of 4.0 litres per second from the depth of 0.50 m. Using the unrounded volume from part (b), find the time to empty the tank, to 3 significant figures. (1 m³ = 1000 litres.)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume = ⅓ × π × 0.6² × 1.0	✓ 1
A1 0.377 m³	✓ 1
Part (b)
M1 similar triangles: the surface radius is 0.30 m	✓ 1
M1 ⅓ × π × 0.3² × 0.5	✓ 1
A1 0.0471 m³	✓ 1
Part (c)
M1 volume = 47.1 litres, divided by 4.0	✓ 1
A1 11.8 s	✓ 1

Exam tip: The radius of the liquid surface scales with its depth, so the radius at half the height is half the full radius.

+ Add to test✓ In your test158IBMI3S-3-02r13-11Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Medium5 marks

Each cell of a honeycomb is a regular hexagonal prism. The hexagonal face has side length 2.9 mm and each cell has depth 11 mm. A frame contains 6500 such cells.

(a)

Calculate the area of one hexagonal face, to 3 significant figures. (A regular hexagon can be split into six equilateral triangles.)

(3)

(b)

Find the total volume of the 6500 cells in litres, using the unrounded area from part (a). Give your answer to 3 significant figures. (1 litre = 10⁶ mm³.)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 six triangles each of area ½ × 2.9² × sin 60°	✓ 1
M1 area = 6 × ½ × 2.9² × sin 60°	✓ 1
A1 21.8 mm²	✓ 1
Part (b)
M1 21.85 × 11 × 6500 mm³	✓ 1
A1 1.56 litres	✓ 1

Exam tip: State which units you are working in before converting: 1 litre is 10⁶ mm³.

+ Add to test✓ In your test159IBMI3S-3-02r15-02Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium8 marks

Three swimming pools, A(1, 2), B(9, 2) and C(5, 11), are marked on a map where each unit represents 1 km. A Voronoi diagram is drawn with the pools as sites. The Voronoi vertex V lies where the perpendicular bisectors of AB and AC meet.

(a)

Find the equation of the perpendicular bisector of AB.

(2)

(b)

Show that the equation of the perpendicular bisector of AC can be written as y − 6.5 = −(4/9)(x − 3).

(3)

(c)

Find the coordinates of V, giving each coordinate to 3 significant figures.

(2)

(d)

State what the position of V tells you about its distance from the three pools.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 midpoint of AB is (5, 2)	✓ 1
A1 AB is horizontal, so the bisector is x = 5	✓ 1
Part (b)
A1 midpoint of AC is (3, 6.5)	✓ 1
M1 gradient of AC = (11 − 2)/(5 − 1) = 9/4	✓ 1
A1 perpendicular gradient −4/9, so y − 6.5 = −(4/9)(x − 3) (AG)	✓ 1
Part (c)
M1 substitutes x = 5 into the equation of the bisector of AC	✓ 1
A1 V = (5, 5.61)	✓ 1
Part (d)
R1 V is equidistant from pools A, B and C, at a distance of about 5.39 km from each	✓ 1

Common error: The gradient of a perpendicular bisector is the negative reciprocal of the gradient of the line joining the sites, and it passes through the midpoint, not through either site.

+ Add to test✓ In your test160IBMI3S-3-02r15-04Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

Wind speed is measured at three turbines on a map where each unit represents 1 km: T₁(2, 3) with 7.2 m/s, T₂(9, 2) with 6.4 m/s and T₃(6, 8) with 8.1 m/s. The speed at any other point is estimated using nearest neighbour interpolation.

(a)

Estimate the wind speed at the point P(5, 5), showing the distances you use.

(3)

(b)

Estimate the wind speed at the point Q(8, 4).

(2)

(c)

Find the difference between the estimates at P and Q.

(1)

(d)

Give one limitation of nearest neighbour interpolation for estimating wind speed.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 distances from P: T₁ 3.61 km, T₂ 5.00 km, T₃ 3.16 km	✓ 1
A1 T₃ is nearest	✓ 1
A1 estimate 8.1 m/s	✓ 1
Part (b)
M1 compares distances from Q: T₁ 6.08 km, T₂ 2.24 km, T₃ 4.47 km	✓ 1
A1 estimate 6.4 m/s (nearest is T₂)	✓ 1
Part (c)
A1 8.1 − 6.4 = 1.7 m/s	✓ 1
Part (d)
R1 the estimate changes abruptly at a Voronoi edge, whereas real wind speed changes gradually (or: ignores the other sites, ignores terrain)	✓ 1

Common error: Compare distances (or squared distances) from the point to every site; do not assume the site that looks closest on a rough sketch is the nearest.

+ Add to test✓ In your test161IBMI3S-3-02r15-06Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium8 marks

A hydrogen storage tank at a wind farm is made from a cylinder of radius 3.2 m and height 8.5 m, with a hemisphere of radius 3.2 m attached to the top. The base of the cylinder is flat.

(a)

Calculate the volume of the tank. Give your answer to 3 significant figures.

(3)

(b)

Calculate the total outer surface area of the tank, including the flat base. Give your answer to 3 significant figures.

(3)

(c)

The outer surface is to be painted at a cost of £4.80 per m². Using the area to 3 significant figures from part (b), find the cost of painting the tank, to the nearest pound.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume of cylinder π × 3.2² × 8.5 (273.4 m³)	✓ 1
M1 volume of hemisphere (2/3) × π × 3.2³ (68.63 m³)	✓ 1
A1 total volume = 342 m³	✓ 1
Part (b)
M1 curved surface of cylinder 2π × 3.2 × 8.5 (170.9 m²) and hemisphere 2π × 3.2² (64.34 m²)	✓ 1
M1 adds the base π × 3.2² (32.17 m²)	✓ 1
A1 total area = 267 m²	✓ 1
Part (c)
M1 cost = 267 × 4.80	✓ 1
A1 £1282	✓ 1

Common error: A hemisphere on top of a cylinder adds a curved area of 2πr² but no flat circular face there; the only flat circle is the base.

+ Add to test✓ In your test162IBMI3S-3-02r15-10Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

In an open-water swimming race a swimmer starts at S and swims 400 m on a bearing of 050° to buoy P. She then swims 300 m on a bearing of 130° to buoy Q.

(a)

Show that angle SPQ is 100°.

(1)

(b)

Find the straight-line distance SQ. Give your answer to 3 significant figures.

(3)

(c)

Find the bearing of Q from S, giving your answer to the nearest degree. Use the unrounded value of SQ.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R1 the bearing of S from P is 230°; angle SPQ = 230° − 130° = 100° (AG)	✓ 1
Part (b)
M1 cosine rule SQ² = 400² + 300² − 2 × 400 × 300 × cos 100°	✓ 1
A1 SQ² = 291676	✓ 1
A1 SQ = 540 m	✓ 1
Part (c)
M1 sine rule sin(angle PSQ)/300 = sin 100°/SQ, so angle PSQ = 33.16°	✓ 1
M1 bearing = 050° + angle PSQ	✓ 1
A1 bearing = 083°	✓ 1

Exam tip: Draw the two legs with north lines at S and P, then use alternate or co-interior angles to find the interior angle between the legs.

+ Add to test✓ In your test163IBMI3S-3-02r15-11Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium6 marks

A triangular plot of land is to be developed as a small wind farm. Two straight boundaries of the plot, of lengths 420 m and 530 m, meet at an angle of 64°. 1 hectare = 10 000 m².

(a)

Find the area of the plot in hectares, giving your answer to 3 significant figures.

(2)

(b)

Each turbine needs 2.8 hectares of land. Use the area in hectares from part (a) to find the greatest number of turbines that can be built on the plot.

(2)

(c)

Find the length of the third boundary of the plot, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area = 1/2 × 420 × 530 × sin 64°	✓ 1
A1 100036 m² = 10.0 hectares	✓ 1
Part (b)
M1 10.0 ÷ 2.8 = 3.57	✓ 1
A1 3 turbines (rounded down to a whole number)	✓ 1
Part (c)
M1 cosine rule 420² + 530² − 2 × 420 × 530 × cos 64°	✓ 1
A1 third side = 512 m	✓ 1

Exam tip: Use the area formula 1/2 ab sin C only when the angle lies between the two given sides.

+ Add to test✓ In your test164IBMI3S-3-02r15-12Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

A boat travels due east in a straight line from A to B, a distance of 600 m. At A, a wind turbine T is seen on a bearing of 038°. At B, the turbine is seen on a bearing of 330°.

(a)

Show that angle ATB is 68°.

(2)

(b)

Find the distance AT, giving your answer to 3 significant figures.

(3)

(c)

Find the shortest distance from the turbine T to the line AB. Use the unrounded value of AT and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 angle TAB = 90° − 38° = 52°	✓ 1
A1 angle TBA = 330° − 270° = 60°, so angle ATB = 180° − 52° − 60° = 68° (AG)	✓ 1
Part (b)
M1 sine rule AT/sin 60° = 600/sin 68°	✓ 1
A1 correct substitution and rearrangement AT = 600 sin 60° ÷ sin 68°	✓ 1
A1 AT = 560 m	✓ 1
Part (c)
M1 distance = AT × sin 52°	✓ 1
A1 distance = 442 m	✓ 1

Common error: A bearing is measured clockwise from north; convert each bearing into an interior angle of the triangle with a sketch before using the sine rule.

+ Add to test✓ In your test165IBMI3S-3-01r17-05Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium6 marks

A planetarium has a hemispherical dome of radius 6.5 m. The curved outer surface of the dome is to be painted with two coats of paint.

(a)

Calculate the area of the curved surface of the dome, in m², correct to 3 significant figures.

(2)

(b)

One litre of paint covers 8.5 m² of surface for one coat. The paint is sold only in tins of 5 litres, and each tin costs €48. Find the minimum total cost of the paint. Use the unrounded area from part (a).

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 curved surface area of a hemisphere = 2πr² with r = 6.5	✓ 1
A1 265 m²	✓ 1
Part (b)
M1 area for two coats = 2 × 265.46 = 530.93 m²	✓ 1
M1 litres needed = 530.93 ÷ 8.5 = 62.46 litres	✓ 1
A1 number of tins = 12.49 rounded up to 13	✓ 1
A1 cost = 13 × 48 = €624	✓ 1

Common error: When items can only be bought in whole units, round the quantity up, not to the nearest integer.

+ Add to test✓ In your test166IBMI3S-3-01r17-08Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium7 marks

Two sports-science laboratories are at A(2, 6) and B. All coordinates are in km. The boundary between the region nearer to A and the region nearer to B is the line y = 1.5x − 3.5.

(a)

Find the gradient of the line AB.

(2)

(b)

Find the coordinates of B.

(3)

(c)

Calculate the distance AB, in km, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 boundary is the perpendicular bisector of AB, so gradients multiply to −1	✓ 1
A1 gradient of AB = −1/1.5 = −2/3	✓ 1
Part (b)
M1 equation of AB: y − 6 = −(2/3)(x − 2) and solve simultaneously with y = 1.5x − 3.5	✓ 1
A1 midpoint of AB is (5, 4)	✓ 1
A1 B = (8, 2) (using the midpoint: 2 × 5 − 2 and 2 × 4 − 6)	✓ 1
Part (c)
M1 AB = √((8 − 2)² + (2 − 6)²)	✓ 1
A1 7.21 km	✓ 1

Exam tip: The midpoint of a segment is the average of the endpoints, so an unknown endpoint is found by doubling the midpoint coordinates and subtracting the known endpoint.

+ Add to test✓ In your test167IBMI3S-3-01r17-11Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium7 marks

In an orienteering race, a runner goes from checkpoint A to checkpoint B, a distance of 650 m on a bearing of 065°. The runner then goes from B to checkpoint C, a distance of 480 m on a bearing of 140°.

(a)

Show that angle ABC = 105°.

(2)

(b)

Calculate the direct distance AC, in metres, correct to 3 significant figures.

(2)

(c)

Find the bearing of C from A, to the nearest degree. Give your answer as a three-figure bearing.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of A from B = 065° + 180° = 245° (or equivalent use of parallel north lines)	✓ 1
A1 angle ABC = 245° − 140° = 105° (AG)	✓ 1
Part (b)
M1 AC² = 650² + 480² − 2(650)(480)cos 105°	✓ 1
A1 AC = 902 m	✓ 1
Part (c)
M1 sine rule: sin(BAC)/480 = sin 105°/AC (use the unrounded AC)	✓ 1
A1 angle BAC = 30.9°	✓ 1
A1 bearing = 065° + 30.9° = 096°	✓ 1

Exam tip: For bearings, draw a north line at each point and work out the angle between lines using parallel lines before applying the sine or cosine rule.

+ Add to test✓ In your test168IBMI3S-3-01r17-12Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium6 marks

An astronomer wants the height of a radio mast on level ground. From a point P, the angle of elevation of the top of the mast, M, is 28°. The astronomer walks 90 m directly towards the mast to a point Q, where the angle of elevation of the top is 41°. The base of the mast, P and Q are in a straight horizontal line. The astronomer's eye height is ignored.

(a)

Show that angle PMQ = 13°.

(2)

(b)

Use the sine rule in triangle PQM to find the length QM, in metres, correct to 3 significant figures.

(2)

(c)

Find the height of the mast, in metres, correct to 3 significant figures. Use the unrounded value of QM.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 angle MQP = 180° − 41° = 139° (angles on a straight line)	✓ 1
A1 angle PMQ = 180° − 28° − 139° = 13° (AG)	✓ 1
Part (b)
M1 QM/sin 28° = 90/sin 13°	✓ 1
A1 QM = 188 m	✓ 1
Part (c)
M1 height = QM sin 41°	✓ 1
A1 123 m	✓ 1

Exam tip: An exterior angle of a triangle equals the sum of the two opposite interior angles; here 41° = 28° + 13°.

+ Add to test✓ In your test169IBMI3S-3-02r19-04Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Medium8 marks

A pharmaceutical capsule consists of a cylinder of length 14 mm and radius 4 mm, with a hemisphere of radius 4 mm attached to each end.

(a)

Calculate the volume of the capsule, to 3 significant figures.

(3)

(b)

The capsule is filled to 90% of its volume with powder of density 0.85 mg per mm³. Using the unrounded volume from (a), find the mass of powder, to 3 significant figures.

(2)

(c)

Calculate the total outer surface area of the capsule, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cylinder volume π × 4² × 14 = 703.7…	✓ 1
M1 two hemispheres make one sphere: (4/3)π × 4³ = 268.0…	✓ 1
A1 972 mm³	✓ 1
Part (b)
M1 0.90 × 971.8… × 0.85	✓ 1
A1 743 mg	✓ 1
Part (c)
M1 curved surface of cylinder 2π × 4 × 14 = 351.8…	✓ 1
M1 surface of the two hemispheres 4π × 4² = 201.0…	✓ 1
A1 553 mm²	✓ 1

Common error: The two end hemispheres together form one sphere; do not add the area of circular faces where the cylinder joins the hemispheres.

+ Add to test✓ In your test170IBMI3S-3-02r19-07Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

In a video game, a character stands on level ground in front of a castle wall of height h metres. From her first position she sees the top of the wall at an angle of elevation of 24°. She walks 35 m directly towards the wall, and now sees the top at an angle of elevation of 41°. Ignore the height of the character.

(a)

Write expressions, in terms of h, for her horizontal distance from the wall at each position.

(2)

(b)

Find the height of the wall, to 3 significant figures.

(3)

(c)

Find her horizontal distance from the wall at the second position, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 first position: h ÷ tan 24°	✓ 1
A1 second position: h ÷ tan 41°	✓ 1
Part (b)
M1 h/tan 24° − h/tan 41° = 35	✓ 1
M1 h(2.246… − 1.150…) = 35	✓ 1
A1 31.9 m	✓ 1
Part (c)
M1 31.94… ÷ tan 41° (unrounded h)	✓ 1
A1 36.7 m	✓ 1

Exam tip: When two right-angled triangles share a side, express that side twice in terms of the unknown and equate the difference to the given distance.

+ Add to test✓ In your test171IBMI3S-3-02r19-09Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

A water engineer inspects a pipeline route. Starting at S, she walks 3.2 km on a bearing of 040° to a valve A. She then walks 4.5 km on a bearing of 130° to a reservoir B.

(a)

Show that angle SAB is 90°.

(2)

(b)

Find the distance SB, to 3 significant figures.

(2)

(c)

Find the bearing of S from B, to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 north lines at S and A are parallel, so the bearing of S from A is 40° + 180° = 220°	✓ 1
R1 angle SAB = 220° − 130° = 90°	✓ 1
Part (b)
M1 Pythagoras: √(3.2² + 4.5²)	✓ 1
A1 5.52 km	✓ 1
Part (c)
M1 angle ABS = tan⁻¹(3.2/4.5) = 35.4°	✓ 1
M1 bearing of A from B is 310°, so bearing of S = 310° − 35.4°	✓ 1
A1 275°	✓ 1

Common error: The bearing back to a start point is not simply the original bearing plus 180°; use the angle in the triangle.

+ Add to test✓ In your test172IBMI3S-3-02r19-10Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

In a game, three portals X, Y and Z are placed so that XY = 120 m, XZ = 150 m and YZ = 200 m.

(a)

Calculate the size of angle YXZ, to 3 significant figures.

(3)

(b)

Calculate the area of triangle XYZ, to 3 significant figures.

(2)

(c)

Find the shortest distance from X to the line YZ, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cosine rule: cos X = (120² + 150² − 200²) ÷ (2 × 120 × 150)	✓ 1
A1 cos X = −0.0861…	✓ 1
A1 94.9°	✓ 1
Part (b)
M1 ½ × 120 × 150 × sin X (unrounded X)	✓ 1
A1 8970 m²	✓ 1
Part (c)
M1 area = ½ × 200 × h (using unrounded area 8966.6…)	✓ 1
A1 89.7 m	✓ 1

Exam tip: When all three sides are given, use the cosine rule for the largest angle first; a negative cosine tells you it is obtuse.

+ Add to test✓ In your test173IBMI3S-3-02r19-14Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 1Medium6 marks

Two points A and B are 640 m apart on a straight bank of a river. A tree C stands on the opposite bank. Angle CAB = 52° and angle CBA = 67°.

(a)

Find angle ACB.

(1)

(b)

Find the distance AC, to 3 significant figures.

(3)

(c)

The banks are straight and parallel. Find the width of the river, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 180° − 52° − 67° = 61°	✓ 1
Part (b)
M1 sine rule: AC ÷ sin 67° = 640 ÷ sin 61°	✓ 1
A1 correct substitution	✓ 1
A1 674 m	✓ 1
Part (c)
M1 width = AC × sin 52° (using unrounded AC = 673.57…)	✓ 1
A1 531 m	✓ 1

Common error: Use the sine rule with a side and its opposite angle as a matching pair: AC is opposite angle B, and AB is opposite angle C.

+ Add to test✓ In your test174IBMI3S-3-03r20-05Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

Three fish farms are located at A(1, 1), B(9, 3) and C(3, 8), where units are kilometres. A Voronoi diagram is drawn to show which farm is closest to each point of the sea.

(a)

Show that the equation of the perpendicular bisector of AB is $y=-4x+22$.

(3)

(b)

Find the coordinates of the Voronoi vertex where the regions of A, B and C meet. Give your answers to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AB is (5, 2) and gradient of AB is $\frac{2}{8}=\frac{1}{4}$	✓ 1
M1 perpendicular gradient is −4, then $y-2=-4(x-5)$	✓ 1
A1 $y=-4x+22$ (AG)	✓ 1
Part (b)
M1 midpoint of AC is (2, 4.5) and gradient of AC is $\frac{7}{2}$, so the perpendicular gradient is $-\frac{2}{7}$	✓ 1
A1 equation of the bisector of AC: $y=-\frac{2}{7}x+5.0714$	✓ 1
M1 solve $-4x+22=-\frac{2}{7}x+5.0714$	✓ 1
A1 x = 4.56, y = 3.77, so (4.56, 3.77)	✓ 1

Exam tip: Check your vertex by finding its distance to each of the three sites; the three distances must be equal.

+ Add to test✓ In your test175IBMI3S-3-03r20-06Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

Point B is 4.5 km due east of point A on a straight shoreline. A fishing boat at C is on a bearing of 052° from A and on a bearing of 330° from B.

(a)

Show that angle ACB = 82°.

(3)

(b)

Calculate the distance AC, in km, to 3 significant figures.

(2)

(c)

Find the shortest distance from the boat to the shoreline AB. Use your unrounded value of AC and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 angle BAC = 90° − 52° = 38°	✓ 1
M1 angle ABC = 330° − 270° = 60°	✓ 1
A1 angle ACB = 180° − 38° − 60° = 82° (AG)	✓ 1
Part (b)
M1 sine rule, $\frac{AC}{\sin 60°}=\frac{4.5}{\sin 82°}$	✓ 1
A1 AC = 3.94 km	✓ 1
Part (c)
M1 perpendicular distance = $AC\sin 38°$	✓ 1
A1 3.9354 × sin 38° = 2.42 km	✓ 1

Exam tip: Drawing the triangle and marking the north line at each point shows how the interior angles come from the bearings.

+ Add to test✓ In your test176IBMI3S-3-03r20-07Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

A hat designer models a hat as a cylinder of radius 14 cm and height 8 cm, with a solid hemisphere of radius 14 cm joined to its top. The base of the cylinder is open. Fabric covers the curved surface of the cylinder and the curved surface of the hemisphere.

(a)

Find the volume enclosed by the hat, in cm³, to 3 significant figures.

(3)

(b)

The fabric costs €18.50 per m². Find the cost of the fabric for one hat, giving your answer to the nearest cent. Ignore any waste.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cylinder volume $\pi(14)^2(8)$	✓ 1
M1 hemisphere volume $\frac{2}{3}\pi(14)^3$	✓ 1
A1 10673.0 cm³, so 10 700 cm³ to 3 s.f. (4926.0 + 5747.0)	✓ 1
Part (b)
M1 curved surface of cylinder $2\pi(14)(8)=703.7$ cm²	✓ 1
M1 curved surface of hemisphere $2\pi(14)^2=1231.5$ cm²	✓ 1
A1 total area 1935.2 cm² = 0.1935 m²	✓ 1
A1 cost = €3.58	✓ 1

Common error: Remember that 1 m² = 10 000 cm², not 100 cm², when converting an area before multiplying by a price per m².

+ Add to test✓ In your test177IBMI3S-3-03r20-08Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

A robotic arm has an upper segment of length 0.45 m and a lower segment of length 0.38 m, joined at an elbow. The shoulder is at S, the elbow at E and the wrist at W, so SE = 0.45 m and EW = 0.38 m.

(a)

The elbow angle SEW is 125°. Find the distance SW, in metres, to 3 significant figures.

(3)

(b)

The wrist must reach a point 0.70 m from the shoulder. Find the elbow angle SEW needed, to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cosine rule, $SW^2=0.45^2+0.38^2-2(0.45)(0.38)\cos125°$	✓ 1
A1 $SW^2=0.5431$	✓ 1
A1 SW = 0.737 m	✓ 1
Part (b)
M1 cosine rule with SW = 0.70, $0.70^2=0.45^2+0.38^2-2(0.45)(0.38)\cos E$	✓ 1
M1 rearrange, $\cos E=\frac{0.45^2+0.38^2-0.70^2}{2(0.45)(0.38)}$	✓ 1
A1 $\cos E=-0.4184$	✓ 1
A1 E = 115° (114.7°)	✓ 1

Exam tip: When all three sides are known, use the rearranged cosine rule to find an angle; the answer can be obtuse, which the sine rule cannot reveal on its own.

+ Add to test✓ In your test178IBMI3S-3-03r20-13Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

A vertical radio mast of height h metres stands on level ground at a fisheries research station. From a point P on the ground the angle of elevation of the top of the mast is 28°. From a point Q, 40 m closer to the base of the mast and in line with P, the angle of elevation is 41°.

(a)

Show that $h\left(\frac{1}{\tan28°}-\frac{1}{\tan41°}\right)=40$.

(3)

(b)

Find the value of h, to 3 significant figures.

(2)

(c)

Find the horizontal distance from P to the base of the mast, in metres, to 3 significant figures. Use the unrounded value of h.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 distance from P to the base of the mast is $\frac{h}{\tan28°}$	✓ 1
M1 distance from Q to the base is $\frac{h}{\tan41°}$	✓ 1
A1 the difference of these distances is 40, giving the required equation (AG)	✓ 1
Part (b)
M1 $h=\frac{40}{\frac{1}{\tan28°}-\frac{1}{\tan41°}}$	✓ 1
A1 h = 54.8 m (54.77)	✓ 1
Part (c)
M1 distance $=\frac{h}{\tan28°}$	✓ 1
A1 103.0 m (103 m)	✓ 1

Common error: Both elevation angles are measured from the horizontal at the observer, so the two right-angled triangles share the same vertical side, h.

+ Add to test✓ In your test179IBMI3S-3-02r22-04Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium6 marks

Two robots, A and B, are 50 m apart on level ground. A third robot R is observed from both. The angle RAB is 50° and the angle RBA is 64°.

(a)

Find the size of angle ARB.

(1)

(b)

Use the sine rule to find the distance AR, to 3 significant figures.

(3)

(c)

Find the shortest distance from R to the line AB, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 180° − 50° − 64° = 66°	✓ 1
Part (b)
M1 AR / sin 64° = 50 / sin 66°	✓ 1
A1 AR = 50 sin 64° / sin 66° (substituted)	✓ 1
A1 49.2 m	✓ 1
Part (c)
M1 AR × sin 50° (use unrounded AR or 49.2)	✓ 1
A1 37.7 m	✓ 1

Exam tip: Find the third angle first; the sine rule then needs a side and its opposite angle, so label the triangle before substituting.

+ Add to test✓ In your test180IBMI3S-3-02r22-06Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium7 marks

A fashion designer cuts a triangular panel of fabric with sides of length 40 cm, 55 cm and 70 cm.

(a)

Find the size of the largest angle of the panel, to 3 significant figures.

(3)

(b)

Find the area of the panel, to the nearest square centimetre.

(2)

(c)

The fabric costs 0.018 euros per square centimetre. Find the cost of the panel, giving your answer to the nearest cent.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cosine rule with the largest angle opposite the 70 cm side	✓ 1
A1 cos C = (40² + 55² − 70²) / (2 × 40 × 55) = −0.0625	✓ 1
A1 93.6°	✓ 1
Part (b)
M1 ½ × 40 × 55 × sin 93.6° (or the unrounded angle)	✓ 1
A1 1098 cm²	✓ 1
Part (c)
M1 area × 0.018	✓ 1
A1 €19.76	✓ 1

Exam tip: The largest angle is opposite the longest side, and a negative cosine value tells you it is obtuse.

+ Add to test✓ In your test181IBMI3S-3-02r22-09Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium7 marks

Three fish farms are located at A(0, 0), B(10, 0) and C(4, 8) on a map, where units are kilometres. A Voronoi diagram is drawn for the three farms.

(a)

Write down the equation of the edge between the regions of A and B.

(1)

(b)

Find the equation of the edge between the regions of A and C.

(3)

(c)

Find the coordinates of the Voronoi vertex where these two edges meet.

(2)

(d)

A research vessel anchors at this vertex. Find its distance from the nearest fish farm, to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x = 5	✓ 1
Part (b)
M1 midpoint of AC = (2, 4) and gradient of AC = 2	✓ 1
M1 perpendicular gradient −1/2, so y − 4 = −(1/2)(x − 2)	✓ 1
A1 y = −0.5x + 5	✓ 1
Part (c)
M1 substitute x = 5 into y = −0.5x + 5	✓ 1
A1 (5, 2.5)	✓ 1
Part (d)
A1 √(5² + 2.5²) = 5.59 km (the vertex is equidistant from A, B and C)	✓ 1

Common error: A Voronoi vertex is equidistant from three sites, so check your answer by measuring from the vertex to all three, not just one.

+ Add to test✓ In your test182IBMI3S-3-02r22-11Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium7 marks

A fish-feed silo is made from a cylinder of radius 1.5 m and height 4.0 m with a hemisphere of radius 1.5 m fixed on top.

(a)

Show that the volume of the silo is 35.3 m³, to 3 significant figures.

(3)

(b)

Feed is delivered in loads of 5.0 m³. Find the least number of loads needed to fill the silo completely.

(2)

(c)

The outside of the silo, apart from its flat base, is to be painted. Find the area to be painted, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume of the cylinder = π × 1.5² × 4.0 = 28.27 m³	✓ 1
M1 volume of the hemisphere = (2/3) × π × 1.5³ = 7.07 m³	✓ 1
A1 total 35.34 = 35.3 m³	✓ 1
Part (b)
M1 35.3 / 5.0 = 7.06...	✓ 1
A1 8 loads (must round up)	✓ 1
Part (c)
M1 2π × 1.5 × 4.0 + 2π × 1.5²	✓ 1
A1 51.8 m²	✓ 1

Common error: The surface of a hemisphere is 2πr², half of a sphere's 4πr²; do not include the circular base where the hemisphere joins the cylinder.

+ Add to test✓ In your test183IBMI3S-3-02r22-12Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium7 marks

A drone hovers at point D, 35 m vertically above a point O on level ground. Robot R1 is due east of O and robot R2 is due north of O. From D, the angle of depression of R1 is 28° and the angle of depression of R2 is 41°.

(a)

Find the distance OR1, to 3 significant figures.

(2)

(b)

Find the distance OR2, to 3 significant figures.

(2)

(c)

Find the distance between R1 and R2, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 28° = 35 / OR1	✓ 1
A1 65.8 m	✓ 1
Part (b)
M1 tan 41° = 35 / OR2	✓ 1
A1 40.3 m	✓ 1
Part (c)
M1 recognises a right angle at O (east and north)	✓ 1
M1 R1R2² = OR1² + OR2²	✓ 1
A1 77.2 m	✓ 1

Exam tip: An angle of depression is measured from the horizontal at the observer, so it equals the angle of elevation from the target to the observer; use it in a right-angled triangle with the vertical height.

+ Add to test✓ In your test184IBMI3S-3-02r22-14Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium7 marks

Three boutiques are located at P(1, 2), Q(9, 2) and R(5, 9) on a city map, where units are kilometres. A new boutique S opens at (5, 4). Each customer shops at the nearest boutique, and a Voronoi diagram is drawn for the four boutiques.

(a)

Write down the equation of the edge between the regions of S and R.

(1)

(b)

Find the equation of the edge between the regions of P and S.

(3)

(c)

These two edges meet at a Voronoi vertex. Find its coordinates.

(2)

(d)

Find the distance from this vertex to boutique P, to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 y = 6.5 (horizontal line through the midpoint (5, 6.5))	✓ 1
Part (b)
M1 midpoint of PS = (3, 3) and gradient of PS = 1/2	✓ 1
M1 perpendicular gradient −2, so y − 3 = −2(x − 3)	✓ 1
A1 y = −2x + 9	✓ 1
Part (c)
M1 −2x + 9 = 6.5	✓ 1
A1 (1.25, 6.5)	✓ 1
Part (d)
A1 √(0.25² + 4.5²) = 4.51 km	✓ 1

Exam tip: When a new site is added, only the edges between the new site and its neighbours need to be found; edges between sites that are far apart may disappear.

+ Add to test✓ In your test185IBMI3S-3-01r23-03Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

Three telecom masts are located at A(0, 0), B(10, 0) and C(4, 8), where units are kilometres. Their Voronoi diagram has exactly one vertex, V.

(a)

The perpendicular bisector of AB is x = 5. Show that the equation of the perpendicular bisector of AC is y = −0.5x + 5.

(3)

(b)

Find the coordinates of V.

(2)

(c)

A new mast is to be built at V. Calculate the distance from V to the nearest existing mast, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 midpoint of AC = (2, 4) and gradient of AC = 2	✓ 1
M1 perpendicular gradient −1/2 and y − 4 = −0.5(x − 2)	✓ 1
A1 y = −0.5x + 5 (shown, with no errors)	✓ 1
Part (b)
M1 substitutes x = 5 into the equation in (a)	✓ 1
A1 V = (5, 2.5)	✓ 1
Part (c)
M1 distance from V to A, e.g. √(5² + 2.5²)	✓ 1
A1 5.59 km (all three existing masts are this distance from V)	✓ 1

Common error: At a Voronoi vertex the three sites are equally distant, so any one of them can be used to find the distance; check with a second site to catch slips.

+ Add to test✓ In your test186IBMI3S-3-01r23-04Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

A food scientist places three sampling stations in a rectangular test orchard with 0 ≤ x ≤ 10 and 0 ≤ y ≤ 10, where units are metres. The stations are at P(0, 0), Q(6, 0) and R(0, 8). A Voronoi diagram is drawn for the stations. All trees in the orchard are assigned to the nearest station.

(a)

Find the equations of the perpendicular bisectors of PQ and PR.

(2)

(b)

Hence find the coordinates of the Voronoi vertex.

(2)

(c)

Find the area of the part of the orchard that is closest to P, and express it as a percentage of the total orchard area.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x = 3	✓ 1
A1 y = 4	✓ 1
Part (b)
M1 intersection of x = 3 and y = 4	✓ 1
A1 (3, 4)	✓ 1
Part (c)
M1 the cell of P within the orchard is the rectangle 0 ≤ x ≤ 3, 0 ≤ y ≤ 4	✓ 1
A1 area = 12 m²	✓ 1
A1 12% of the orchard	✓ 1

Exam tip: Sketch the sites and edges before finding areas; the boundary of the orchard forms part of the boundary of each cell.

+ Add to test✓ In your test187IBMI3S-3-01r23-07Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium6 marks

An ice-cream cone is a right circular cone of radius 3.0 cm with a hemisphere of ice cream of the same radius sitting on its open top face. The vertical height of the cone is 10 cm.

(a)

Calculate the volume of the cone.

(2)

(b)

Calculate the volume of the hemisphere.

(2)

(c)

The ice cream has density 0.55 g cm⁻³. The cone is hollow and is also completely filled with the same ice cream. Find the total mass of ice cream in the cone and hemisphere, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (1/3)π × 3² × 10	✓ 1
A1 94.2 cm³ = 30π	✓ 1
Part (b)
M1 (2/3)π × 3³	✓ 1
A1 56.5 cm³ = 18π	✓ 1
Part (c)
M1 adds the cone volume to the hemisphere volume (or mass of each)	✓ 1
A1 82.9 g	✓ 1

Common error: The hemisphere volume is half of 4/3 πr³; do not forget the factor 1/2.

+ Add to test✓ In your test188IBMI3S-3-01r23-08Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

A mountaineering tent is a square-based pyramid. The square base has side length 3.0 m and the perpendicular height of the apex above the centre of the base is 1.8 m. The four triangular sides are made from fabric; the base is a separate groundsheet.

(a)

Calculate the volume of air inside the tent.

(1)

(b)

Show that the slant height of each triangular face, measured from the midpoint of a base edge to the apex, is 2.34 m to 3 significant figures.

(2)

(c)

Calculate the total area of fabric needed for the four triangular faces.

(2)

(d)

The fabric costs £14.50 per square metre. Find the cost of the fabric, using the area 14.1 m². Give your answer to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 (1/3) × 3.0² × 1.8 = 5.4 m³	✓ 1
Part (b)
M1 Pythagoras with 1.8 and 1.5: √(1.8² + 1.5²)	✓ 1
A1 2.3430… = 2.34 m (shown)	✓ 1
Part (c)
M1 area of one face = ½ × 3.0 × 2.343… (unrounded slant height)	✓ 1
A1 14.06 m² → 14.1 m² to 3 s.f.	✓ 1
Part (d)
M1 14.1 × 14.50	✓ 1
A1 £204.45	✓ 1

Exam tip: The slant height is the hypotenuse of a right-angled triangle formed by the pyramid’s height and half the base edge, not the height itself.

+ Add to test✓ In your test189IBMI3S-3-01r23-11Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

A climber at point P on level ground measures the angle of elevation of the top T of a cliff as 24°. She walks 450 m directly towards the cliff to point Q and measures the angle of elevation of T as 31°. Let h be the height of T above the ground and d the horizontal distance from P to the foot of the cliff.

(a)

Write down two expressions for the horizontal distances in terms of h: d and d − 450.

(2)

(b)

Hence find h, giving your answer to the nearest metre.

(3)

(c)

Calculate the horizontal distance from P to the foot of the cliff, to the nearest metre. Use the unrounded value of h.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 d = h/tan 24°	✓ 1
A1 d − 450 = h/tan 31°	✓ 1
Part (b)
M1 subtracts: 450 = h/tan 24° − h/tan 31°	✓ 1
M1 h = 450 ÷ (1/tan 24° − 1/tan 31°)	✓ 1
A1 774 m	✓ 1
Part (c)
M1 d = h/tan 24°	✓ 1
A1 1737 m	✓ 1

Exam tip: Keep unrounded values in your calculator when a later part depends on an earlier result.

+ Add to test✓ In your test190IBMI3S-3-01r23-12Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

A rescue helicopter leaves its base H and flies 12 km on a bearing of 065° to a point R. It then flies 9 km on a bearing of 150° to a point S.

(a)

Show that angle HRS is 95°.

(2)

(b)

Calculate the distance HS, to 3 significant figures.

(2)

(c)

Calculate the bearing of S from H, to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of H from R is 065° + 180° = 245° (alternate/co-interior angles)	✓ 1
A1 angle HRS = 245° − 150° = 95° (shown)	✓ 1
Part (b)
M1 cosine rule: HS² = 12² + 9² − 2(12)(9)cos 95°	✓ 1
A1 15.6 km	✓ 1
Part (c)
M1 sine rule: sin(angle RHS)/9 = sin 95°/HS	✓ 1
A1 angle RHS = 35.0°	✓ 1
A1 bearing = 065° + 35° = 100° (ECF from earlier parts)	✓ 1

Common error: A bearing is measured clockwise from north; draw a north line at each turning point and mark the angles carefully.

+ Add to test✓ In your test191IBMI3S-3-01r23-13Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium6 marks

A triangular alpine meadow ABC has sides AB = 120 m, BC = 95 m and AC = 150 m.

(a)

Calculate the size of angle ABC, to 3 significant figures.

(2)

(b)

Calculate the area of the meadow, in m², to 3 significant figures. Use the unrounded angle.

(2)

(c)

The meadow is fertilised at a cost of €0.18 per m². Using the area 5700 m², find the cost of fertilising the whole meadow.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cosine rule: cos B = (120² + 95² − 150²)/(2 × 120 × 95)	✓ 1
A1 angle ABC = 87.7°	✓ 1
Part (b)
M1 area = ½ × 120 × 95 × sin B	✓ 1
A1 5695.3 m², which is 5700 m² to 3 s.f.	✓ 1
Part (c)
M1 5700 × 0.18	✓ 1
A1 €1026	✓ 1

Exam tip: The largest angle is opposite the longest side, which is a quick check on your cosine rule answer.

+ Add to test✓ In your test192IBMI3S-3-03r29-04Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

Calculator

Monitoring stations for a volcano are placed at A(0, 0), B(8, 2) and C(2, 8), where the coordinates are in kilometres from a reference point. The region is divided into a Voronoi diagram with the stations as sites.

(a)

Find the equation of the perpendicular bisector of [AB]. Give your answer in the form y = mx + c.

(3)

(b)

The perpendicular bisector of [AC] has equation y = −0.25x + 4.25. Find the coordinates of the Voronoi vertex where the edges from these two bisectors meet.

(2)

(c)

A new relay mast is to be built at the vertex found in (b). Find its distance from station A. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 midpoint of AB is (4, 1)	✓ 1
A1 gradient of AB is 1/4 so perpendicular gradient is −4	✓ 1
A1 y = −4x + 17	✓ 1
Part (b)
M1 sets −4x + 17 = −0.25x + 4.25 (or uses GDC to intersect)	✓ 1
A1 (3.4, 3.4)	✓ 1
Part (c)
M1 distance = √(3.4² + 3.4²)	✓ 1
A1 4.81 km	✓ 1

Exam tip: A Voronoi vertex is equidistant from three sites. If you have time, check your vertex against the third site as a quick test of your coordinates.

+ Add to test✓ In your test193IBMI3S-3-03r29-05Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams Paper 1Medium6 marks

Two ships leave port P at the same time. Ship A sails 24 km on a bearing of 040° and ship B sails 31 km on a bearing of 115°.

(a)

Calculate the distance between the ships when they stop. Give your answer to 3 significant figures.

(3)

(b)

Find the bearing on which ship A must steer to reach ship B. Give your answer to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 angle APB = 75°	✓ 1
M1 cosine rule: d² = 24² + 31² − 2(24)(31)cos 75°	✓ 1
A1 d = 33.9 km (33.94...)	✓ 1
Part (b)
M1 sine rule or cosine rule to find angle PAB	✓ 1
A1 angle PAB = 61.9°	✓ 1
A1 bearing of B from A = 220° − 61.9° = 158°	✓ 1

Common error: The bearing of B from A is not the bearing of A from P. Use the back bearing of P from A (040° + 180° = 220°) and then add or subtract the interior angle.

+ Add to test✓ In your test194IBMI3S-3-03r29-06Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams Paper 1Medium6 marks

A triangular backdrop ABC for a theatre set has AB = 8.4 m, BC = 6.3 m and angle BAC = 44°. Angle ACB is obtuse.

(a)

Find the size of angle ACB. Give your answer to the nearest 0.1°.

(3)

(b)

Hence find the area of the backdrop. Use the unrounded size of angle ACB and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sine rule: sin C / 8.4 = sin 44° / 6.3	✓ 1
A1 sin C = 0.926... so acute value is 67.9°	✓ 1
A1 obtuse angle C = 180° − 67.9° = 112.1°	✓ 1
Part (b)
A1 angle ABC = 180° − 44° − 112.15° = 23.85°	✓ 1
M1 area = ½ × 8.4 × 6.3 × sin(23.85°)	✓ 1
A1 10.7 m²	✓ 1

Exam tip: When the sine rule is used to find an angle, check whether the question tells you the angle is obtuse. If it does, subtract your calculator value from 180°.

+ Add to test✓ In your test195IBMI3S-3-03r29-08Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

Calculator

Three ferry terminals are at A(0, 0), B(8, 0) and C(2, 6), where the coordinates are in kilometres. A Voronoi diagram is drawn with the terminals as sites. The Voronoi edge between the regions of A and B is the line x = 4.

(a)

Find the equation of the perpendicular bisector of [AC].

(3)

(b)

Hence show that the Voronoi vertex where three edges meet is at (4, 2).

(2)

(c)

A fourth terminal D is built at (4, 6). Show that (4, 2) is no longer a vertex of the new Voronoi diagram.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 midpoint of AC is (1, 3)	✓ 1
A1 gradient of AC is 3, so perpendicular gradient is −1/3	✓ 1
A1 y − 3 = −(1/3)(x − 1), i.e. y = −x/3 + 10/3	✓ 1
Part (b)
M1 substitutes x = 4 into the equation of the bisector of [AC]	✓ 1
A1 y = −4/3 + 10/3 = 2, so the vertex is (4, 2) (shown)	✓ 1
Part (c)
M1 distance from (4, 2) to D is 4, and distance from (4, 2) to each of A, B and C is √20 = 4.47...	✓ 1
R1 D is nearer than A, B and C, so the point is not equidistant from its three nearest sites and cannot be a vertex	✓ 1

Exam tip: When a site is added, a point stays on its old edge or vertex only if the new site is no closer to it. Comparing distances to the new site is the quickest test.

+ Add to test✓ In your test196IBMI3S-3-03r29-12Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

Calculator

A ship is sailing due east. At point P₁ the bearing of a lighthouse L is 030°. After sailing 6.0 km due east to point P₂, the bearing of the lighthouse is 340°.

(a)

Show that angle P₁LP₂ = 50°.

(2)

(b)

Calculate the distance P₂L. Give your answer to 3 significant figures.

(3)

(c)

Hence find the shortest distance from the lighthouse to the ship's line of travel. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 angle LP₁P₂ = 90° − 30° = 60°	✓ 1
A1 angle LP₂P₁ = 340° − 270° = 70°, so angle at L = 180° − 60° − 70° = 50° (shown)	✓ 1
Part (b)
M1 sine rule: P₂L / sin 60° = 6.0 / sin 50°	✓ 1
A1 P₂L = 6.0 sin 60° / sin 50°	✓ 1
A1 6.78 km (6.783...)	✓ 1
Part (c)
M1 distance = P₂L sin 70° (or P₁L sin 60°)	✓ 1
A1 6.37 km (6.374...)	✓ 1

Common error: The shortest distance from a point to a line is the perpendicular distance, not the distance to the nearer position of the ship. Make a right-angled triangle with the line of travel.

+ Add to test✓ In your test197IBMI3S-3-01r34-02Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams Paper 1Medium7 marks

A data centre hall is a cuboid measuring 24 m by 15 m by 4.2 m. The hall contains 36 server racks, each a cuboid measuring 0.60 m by 1.10 m by 2.20 m. Give answers to 3 significant figures unless told otherwise.

(a)

Calculate the volume of the hall, in m³.

(2)

(b)

Find the volume of air in the hall when the racks are in place. Treat the racks as the only objects in the hall.

(3)

(c)

The cooling units together move air at a rate of 3.5 m³ per second. Find how many minutes the units need to move a volume of air equal to the volume found in (b). Use your unrounded answer from (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $24\times15\times4.2$	✓ 1
A1 1510 m³ (1512)	✓ 1
Part (b)
M1 volume of one rack $=0.6\times1.1\times2.2=1.452$ m³	✓ 1
M1 subtract $36\times1.452$ from the volume of the hall (use 1512, or accept 1510 for the hall volume)	✓ 1
A1 1460 m³ (1459.7…)	✓ 1
Part (c)
M1 divide the volume of air by 3.5 (and by 60 to change seconds to minutes)	✓ 1
A1 6.95 minutes (417 s)	✓ 1

Exam tip: When one solid sits inside another, subtract its volume from the container, and keep unrounded values in your calculator until the last step.

+ Add to test✓ In your test198IBMI3S-3-01r34-04Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams Paper 1Medium5 marks

Soil removed from a trench at an excavation is tipped into a heap in the shape of a cone. The heap has a base diameter of 3.6 m and a vertical height of 1.4 m.

(a)

Calculate the volume of soil in the heap, in m³, to 3 significant figures.

(3)

(b)

The soil is moved by a skip that carries up to 1.6 m³ per trip. Find the minimum number of trips needed to move all the soil.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 radius $=1.8$ m	✓ 1
M1 $V=\tfrac13\pi\times1.8^2\times1.4$	✓ 1
A1 4.75 m³ (4.7501…)	✓ 1
Part (b)
M1 divide the volume by 1.6 (2.97…; use 4.75 or the unrounded value)	✓ 1
A1 3 trips (rounded up, since a part trip is still a trip)	✓ 1

Common error: Using the diameter 3.6 m as the radius in the cone formula makes the volume four times too large.

+ Add to test✓ In your test199IBMI3S-3-01r34-06Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams Paper 1Medium7 marks

Two cyclists, Ana and Ben, leave the same camp at the same time. Ana rides 12 km on a bearing of 065°. Ben rides 9 km on a bearing of 140°. Both ride in straight lines.

(a)

Write down the angle between their two directions of travel at the camp.

(1)

(b)

Find the distance between the cyclists when they stop. Give your answer to 3 significant figures.

(3)

(c)

Find the bearing of Ben from Ana. Give your answer to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $140-65=75^\circ$	✓ 1
Part (b)
M1 cosine rule $d^2=12^2+9^2-2(12)(9)\cos75^\circ$	✓ 1
A1 $d^2=169.09\ldots$	✓ 1
A1 $d=13.0$ km (13.004…)	✓ 1
Part (c)
M1 sine rule $\dfrac{\sin A}{9}=\dfrac{\sin75^\circ}{13.004\ldots}$ with $A$ the angle at Ana	✓ 1
A1 $A=42.0^\circ$ (41.95…)	✓ 1
A1 bearing $245-41.95\ldots=203^\circ$ (bearing from Ana back to the camp is $065+180=245^\circ$)	✓ 1

Common error: Using the bearing values themselves in the cosine rule: the angle needed is the difference between the two bearings.

+ Add to test✓ In your test200IBMI3S-3-01r34-07Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams Paper 1Medium6 marks

An archaeology team surveys a buried tomb marker T from two survey points, A and B, which are 85 m apart on level ground. The angle TAB is 62° and the angle TBA is 71°.

(a)

Find the size of angle ATB.

(1)

(b)

Find the distance AT. Give your answer to 3 significant figures.

(3)

(c)

Find the shortest distance from T to the line AB. Give your answer to 3 significant figures. Use the unrounded value of AT.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $180-62-71=47^\circ$	✓ 1
Part (b)
M1 sine rule $\dfrac{AT}{\sin71^\circ}=\dfrac{85}{\sin47^\circ}$	✓ 1
A1 correct substitution and rearrangement $AT=\dfrac{85\sin71^\circ}{\sin47^\circ}$	✓ 1
A1 $AT=110$ m (109.9…)	✓ 1
Part (c)
M1 $AT\sin62^\circ$ (or $BT\sin71^\circ$ with $BT=102.6\ldots$)	✓ 1
A1 97.0 m (97.03…)	✓ 1

Exam tip: The sine rule needs a side opposite a known angle: find the third angle first so that AB is paired with angle ATB.

+ Add to test✓ In your test201IBMI3S-3-01r34-10Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams Paper 1Medium6 marks

Three data centres A(1, 3), B(7, 5) and C(6, 6) are shown on a map with coordinates in kilometres. A Voronoi diagram is drawn using the three data centres as sites.

(a)

Find the equation of the perpendicular bisector of AB. Give your answer in the form $y=mx+c$.

(3)

(b)

The perpendicular bisector of BC has equation $y=x-1$. Find the coordinates of the vertex where the edges of the Voronoi diagram meet.

(2)

(c)

State what is true about the distances from this vertex to A, B and C.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AB $=(4,4)$ and gradient of AB $=\tfrac13$	✓ 1
M1 perpendicular gradient $=-3$	✓ 1
A1 $y=-3x+16$	✓ 1
Part (b)
M1 solve $-3x+16=x-1$	✓ 1
A1 $(4.25,\,3.25)$	✓ 1
Part (c)
A1 the vertex is the same distance from all three sites ($\approx3.26$ km)	✓ 1

Exam tip: A Voronoi edge lies on the perpendicular bisector of two sites, so a vertex is equidistant from three sites. Check by finding two of the distances.

+ Add to test✓ In your test202IBMI3S-3-01r34-13Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams Paper 1Medium6 marks

Soil acidity is measured at four sampling sites in a field, with coordinates in metres. The pH values are: S₁(2, 7) pH 6.1; S₂(8, 9) pH 7.4; S₃(5, 3) pH 5.6; S₄(10, 2) pH 6.8. A new site N(7, 5) is measured and has pH 6.3. A Voronoi diagram is now drawn for all five sites.

(a)

Find the equation of the perpendicular bisector of S₃ and N. Give your answer in the form $y=mx+c$.

(3)

(b)

Without calculating any distances, use your answer to (a) to show that the point T(7.5, 3.5) is closer to N than to S₃.

(2)

(c)

T lies in the cell of N. Use nearest-neighbour interpolation to estimate the pH at T.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint $=(6,4)$ and gradient of S₃N $=1$	✓ 1
M1 perpendicular gradient $=-1$	✓ 1
A1 $y=-x+10$	✓ 1
Part (b)
M1 on the line, at $x=7.5$ the value is $y=2.5$	✓ 1
A1 T is above the line (3.5 > 2.5), the same side as N (N has $5>3$), so T is closer to N	✓ 1
Part (c)
A1 pH 6.3	✓ 1

Common error: In nearest-neighbour interpolation the estimate is simply the value at the nearest site; do not average the values at neighbouring sites.

+ Add to test✓ In your test203IBMI3S-3-03r37-03Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

The crater of a dormant volcano is shaped like an inverted cone, with its vertex (lowest point) at the bottom. At the rim the crater has radius 600 m and its depth is 150 m. After heavy rain, water in the crater has a depth of 100 m, measured from the vertex.

(a)

Use similar triangles to find the radius of the water surface.

(2)

(b)

Calculate the volume of water in the crater. Give your answer in millions of m³, to 3 significant figures.

(2)

(c)

Find the further volume of water needed to fill the crater to the rim. Use the unrounded volume from part (b). Give your answer in millions of m³, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{r}{600}=\frac{100}{150}$	✓ 1
A1 $r=400$ m	✓ 1
Part (b)
M1 $V=\frac{1}{3}\pi(400)^2(100)$	✓ 1
A1 16.8 million m³	✓ 1
Part (c)
M1 volume of full crater $\frac{1}{3}\pi(600)^2(150)=56.5$ million m³	✓ 1
M1 subtracts the volume of water already present (16.755 million m³)	✓ 1
A1 39.8 million m³	✓ 1

Common error: Using the full-size radius (600 m) instead of the radius of the water surface when finding the volume of water is a frequent slip; the water forms its own smaller cone.

+ Add to test✓ In your test204IBMI3S-3-03r37-05Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

Ash from an eruption falls to form a pile in the shape of a cone. The side of the cone makes an angle of 34° with the horizontal ground, which is the steepest angle at which the ash is stable. The volume of ash in the pile is 2.4 × 10⁶ m³. The radius of the base of the cone is r metres.

(a)

Show that the volume of the cone is $V=\frac{\pi\tan34^{\circ}}{3}r^3$.

(2)

(b)

Find the radius of the base of the pile. Give your answer in metres to 3 significant figures.

(3)

(c)

Find the height of the pile. Give your answer to the nearest metre. Use the unrounded value of r from part (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height $h=r\tan34^{\circ}$	✓ 1
A1 substitutes into $V=\frac{1}{3}\pi r^2h$ to obtain $V=\frac{\pi\tan34^{\circ}}{3}r^3$ (AG)	✓ 1
Part (b)
M1 $2.4\times10^6=\frac{\pi\tan34^{\circ}}{3}r^3$	✓ 1
M1 $r^3=\frac{3\times2.4\times10^6}{\pi\tan34^{\circ}}$ (3.398 × 10⁶)	✓ 1
A1 $r=150$ m	✓ 1
Part (c)
M1 $h=r\tan34^{\circ}$ with $r=150.3$	✓ 1
A1 101 m	✓ 1

Exam tip: When the volume is given and a length is unknown, rearrange the formula for that length first and take the cube root last, keeping the full calculator value until the end.

+ Add to test✓ In your test205IBMI3S-3-03r37-07Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

Two spotlights are fixed at points P and Q on a horizontal rail in a theatre, with PQ = 9.0 m. Both spotlights point at an actor standing at point A on the stage, where PA = 11.2 m and QA = 8.4 m. All three points may be treated as lying in one vertical plane.

(a)

Find the size of angle PAQ. Give your answer to the nearest tenth of a degree.

(3)

(b)

Calculate the area of triangle APQ. Use the unrounded angle from part (a). Give your answer in m² to 3 significant figures.

(2)

(c)

Find the shortest distance from A to the rail PQ. Use the unrounded area from part (b). Give your answer in metres to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses the cosine rule $9.0^2=11.2^2+8.4^2-2(11.2)(8.4)\cos A$	✓ 1
M1 $\cos A=\frac{115}{188.16}$	✓ 1
A1 52.3°	✓ 1
Part (b)
M1 $\frac{1}{2}(11.2)(8.4)\sin A$	✓ 1
A1 37.2 m²	✓ 1
Part (c)
M1 $\frac{1}{2}(9.0)h=37.23$	✓ 1
A1 8.27 m	✓ 1

Exam tip: When the three sides of a triangle are known and no angle is given, the cosine rule is the starting point. Rearrange to find $\cos A$ before using the inverse cosine.

+ Add to test✓ In your test206IBMI3S-3-03r37-12Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

Three ports are at A(0, 0), B(12, 0) and C(2, 8) on a map where the units are kilometres. The ports are the sites of a Voronoi diagram used to decide which port is closest to a ship at any position.

(a)

Write down the equation of the perpendicular bisector of AB.

(1)

(b)

Find the equation of the perpendicular bisector of AC. Give your answer in the form $y=mx+c$.

(3)

(c)

Find the coordinates of the Voronoi vertex where these two perpendicular bisectors meet.

(2)

(d)

A ship at this vertex is the same distance from all three ports. Find this distance. Give your answer in km to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $x=6$	✓ 1
Part (b)
M1 midpoint of AC is (1, 4)	✓ 1
M1 gradient of AC is 4, so perpendicular gradient is $-\frac{1}{4}$	✓ 1
A1 $y=-\frac{1}{4}x+4.25$ (accept $y-4=-0.25(x-1)$)	✓ 1
Part (c)
M1 substitutes $x=6$ into the equation of the bisector of AC	✓ 1
A1 (6, 2.75)	✓ 1
Part (d)
A1 $\sqrt{6^2+2.75^2}=6.60$ km	✓ 1

Exam tip: A Voronoi vertex is equidistant from three sites. You can check your vertex by calculating its distance to each site: all three must be equal.

+ Add to test✓ In your test207IBMI3S-3-01r40-06Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

An asteroid is modelled as a sphere of radius 4.50 km. Its density is 2.40 × 10³ kg m⁻³.

(a)

Calculate the volume of the asteroid in km³, correct to 3 significant figures.

(2)

(b)

Find the mass of the asteroid in kg, giving your answer in the form $a \times 10^k$ where $1 \leq a < 10$. Use the unrounded volume from part (a).

(3)

(c)

Calculate the surface area of the asteroid, in km², correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{4}{3}\pi \times 4.50^3$	✓ 1
A1 382 km³	✓ 1
Part (b)
M1 convert volume to m³: 1 km³ = 10⁹ m³, so $V \approx 3.817 \times 10^{11}$ m³	✓ 1
M1 mass = density × volume = $2400 \times 3.817 \times 10^{11}$	✓ 1
A1 $9.16 \times 10^{14}$ kg	✓ 1
Part (c)
M1 $4\pi \times 4.50^2$	✓ 1
A1 254 km²	✓ 1

Exam tip: 1 km³ is 10⁹ m³, not 10³ m³: cube the conversion factor.

+ Add to test✓ In your test208IBMI3S-3-01r40-07Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium6 marks

In an orienteering event the three control points A, B and C form a triangle. The distance AB is 60 m, angle CAB is 48° and angle ABC is 71°.

(a)

Find the size of angle ACB.

(1)

(b)

Use the sine rule to find the distance AC, correct to 3 significant figures.

(3)

(c)

Calculate the area of triangle ABC, correct to 3 significant figures. Use the unrounded value of AC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 61°	✓ 1
Part (b)
M1 $\frac{AC}{\sin 71°} = \frac{60}{\sin 61°}$	✓ 1
M1 $AC = \frac{60 \sin 71°}{\sin 61°}$	✓ 1
A1 64.9 m	✓ 1
Part (c)
M1 area = $\frac{1}{2} \times 60 \times AC \times \sin 48°$	✓ 1
A1 1450 m²	✓ 1

Exam tip: The sine rule needs a side and its opposite angle; find the third angle first so you have a matching pair.

+ Add to test✓ In your test209IBMI3S-3-01r40-09Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium6 marks

Two sports centres, P at (2, 6) and Q at (8, 2), serve a region. Coordinates are in kilometres. Each point in the region is allocated to its nearer centre, so that the two cells of a Voronoi diagram are separated by a straight edge. A straight cycle path runs along the line $y = 1$.

(a)

Find the equation of the edge between the two cells, in the form $y = mx + c$.

(4)

(b)

Find the coordinates of the point where the cycle path crosses the edge.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of PQ is (5, 4)	✓ 1
M1 gradient of PQ is $-\frac{2}{3}$	✓ 1
M1 perpendicular gradient is $\frac{3}{2}$	✓ 1
A1 $y = 1.5x - 3.5$	✓ 1
Part (b)
M1 substitute $y = 1$: $1 = 1.5x - 3.5$	✓ 1
A1 (3, 1)	✓ 1

Exam tip: Check your edge by testing the midpoint: it must satisfy your equation.

+ Add to test✓ In your test210IBMI3S-3-01r40-11Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Medium7 marks

The height $h$ metres of an observatory tower is to be found. From a point A on level ground the angle of elevation of the top of the tower is 28°. From a point B, 40 m closer to the base of the tower and in line with A, the angle of elevation is 41°.

(a)

Find the height of the tower, correct to 3 significant figures.

(5)

(b)

Find the horizontal distance from B to the base of the tower, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 horizontal distance from A is $\frac{h}{\tan 28°}$	✓ 1
M1 horizontal distance from B is $\frac{h}{\tan 41°}$	✓ 1
M1 $\frac{h}{\tan 28°} - \frac{h}{\tan 41°} = 40$	✓ 1
A1 solving: $h = \frac{40}{\frac{1}{\tan 28°} - \frac{1}{\tan 41°}}$	✓ 1
A1 54.8 m	✓ 1
Part (b)
M1 $\frac{54.77}{\tan 41°}$	✓ 1
A1 63.0 m	✓ 1

Exam tip: Two unknown distances with the same height: write both in terms of $h$, then use the given 40 m difference.

+ Add to test✓ In your test211IBMI3S-3-03r42-02Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; 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volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

Three wind farms are located at W₁(0, 0), W₂(10, 0) and W₃(4, 8) on a map, with units in kilometres. A Voronoi diagram is drawn with the wind farms as sites. The perpendicular bisector of W₁W₂ has equation x = 5.

(a)

Find the equation of the perpendicular bisector of W₁W₃. Give your answer in the form y = mx + c.

(3)

(b)

Hence find the coordinates of the Voronoi vertex where these two edges meet.

(2)

(c)

A new control centre is built at this vertex. Calculate its distance from W₁, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 midpoint of W₁W₃ = (2, 4) and gradient of W₁W₃ = 2	✓ 1
M1 perpendicular gradient = −1/2 used with the midpoint	✓ 1
A1 y = −0.5x + 5	✓ 1
Part (b)
M1 substitute x = 5 into the equation from (a) (or solve simultaneously)	✓ 1
A1 (5, 2.5)	✓ 1
Part (c)
M1 √(5² + 2.5²)	✓ 1
A1 5.59 km (5.59 km); it is the same distance from W₂ and W₃	✓ 1

Exam tip: A Voronoi vertex is equidistant from three sites, so you can check your coordinates by calculating the distance to the third site.

+ Add to test✓ In your test212IBMI3S-3-03r42-06Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; 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Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

A service boat leaves harbour H and sails 14 km on a bearing of 072° to wind farm A. It then sails 9 km on a bearing of 150° to wind farm B.

(a)

Show that angle HAB is 102°.

(2)

(b)

Calculate the distance HB, giving your answer to 3 significant figures.

(2)

(c)

Find the bearing of B from H, giving your answer to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of H from A is 072° + 180° = 252°	✓ 1
A1 252° − 150° = 102°	✓ 1
Part (b)
M1 HB² = 14² + 9² − 2 × 14 × 9 × cos 102°	✓ 1
A1 HB = 18.1 km	✓ 1
Part (c)
M1 sine rule, sin(AHB)/9 = sin 102°/HB	✓ 1
A1 angle AHB = 29.0°	✓ 1
A1 bearing = 072° + 29.0° = 101° (101°)	✓ 1

Exam tip: Draw a north line at every turning point and mark each bearing as a clockwise angle from north; angles inside the triangle then come from parallel north lines.

+ Add to test✓ In your test213IBMI3S-3-03r42-07Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; 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An open-water swimming course is a triangle of buoys A, B and C. The distance AB is 420 m, angle CAB = 48° and angle ABC = 71°.

(a)

Find the length of AC, giving your answer to the nearest metre.

(3)

(b)

Find the length of BC, giving your answer to the nearest metre.

(2)

(c)

Calculate the area enclosed by the course. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 angle ACB = 180° − 48° − 71° = 61°	✓ 1
M1 sine rule, AC/sin 71° = 420/sin 61°	✓ 1
A1 AC = 454 m	✓ 1
Part (b)
M1 sine rule, BC/sin 48° = 420/sin 61°	✓ 1
A1 BC = 357 m	✓ 1
Part (c)
M1 area = ½ × 420 × AC × sin 48° (or equivalent using the unrounded value of AC)	✓ 1
A1 70900 m² (70 900 m²)	✓ 1

Exam tip: Find the third angle first. The sine rule then needs a side and its opposite angle, which are now both known.

+ Add to test✓ In your test214IBMI3S-3-03r42-10Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; 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A swimming pool is 12.5 m wide and 24 m long. The water is 1.1 m deep at the shallow end and 2.9 m deep at the deep end, and the floor slopes at a constant angle between the two ends. The pool is a prism whose cross-section is a trapezium.

(a)

Calculate the length of the sloping floor, giving your answer to 3 significant figures.

(2)

(b)

Calculate the volume of water in the full pool.

(3)

(c)

The empty pool is filled at a constant rate of 450 litres per minute. Given that 1 m³ = 1000 litres, find the time taken to fill the pool completely. Give your answer in hours to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(24² + (2.9 − 1.1)²)	✓ 1
A1 24.1 m	✓ 1
Part (b)
M1 cross-section area = ½(1.1 + 2.9) × 24	✓ 1
A1 48 m²	✓ 1
A1 volume = 48 × 12.5 = 600 m³	✓ 1
Part (c)
M1 600 000 ÷ 450 = 1333.3 minutes (or ÷ 60 for hours)	✓ 1
A1 22.2 hours	✓ 1

Common error: A trapezium prism is not (average depth × width) alone: the trapezium area must be multiplied by the third dimension, here the width of the pool.

+ Add to test✓ In your test215IBMI3S-3-03r42-11Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; 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A decorative concrete column outside a library is a cone with its top removed (a frustum). The base has radius 0.6 m, the flat top has radius 0.3 m and the height of the column is 3 m. The column is part of a cone whose full height is H m. The volume of a cone is $\frac{1}{3}\pi r^2 h$ and the curved surface area of a cone is $\pi r l$, where l is the slant height.

(a)

Show that H = 6.

(2)

(b)

Find the volume of the column, giving your answer in m³ to 3 significant figures.

(2)

(c)

The concrete has density 2400 kg m⁻³. Using the unrounded volume, find the mass of the column to 3 significant figures.

(1)

(d)

Find the curved surface area of the column to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 similar triangles, 0.6/H = 0.3/(H − 3)	✓ 1
A1 0.6(H − 3) = 0.3H leading to H = 6	✓ 1
Part (b)
M1 volume of cone of height 6 minus volume of cone of height 3	✓ 1
A1 1.98 m³	✓ 1
Part (c)
A1 4750 kg	✓ 1
Part (d)
M1 π × 0.6 × √(6² + 0.6²) − π × 0.3 × √(3² + 0.3²)	✓ 1
A1 8.52 m²	✓ 1

Exam tip: For a frustum, subtract the small cone that was removed from the complete cone; the ratio of radii equals the ratio of heights in similar cones.

+ Add to test✓ In your test216IBMI3S-3-03r44-05Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium8 marks

A ship leaves port A and sails 18 km on a bearing of 040° to buoy B. It then sails 25 km on a bearing of 115° to port C.

(a)

Show that angle ABC = 105°.

(2)

(b)

Find the direct distance AC. Give your answer to 3 significant figures.

(3)

(c)

Find the bearing of C from A, giving your answer to the nearest degree. Use the unrounded value of AC.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of A from B is 040° + 180° = 220°, or use of alternate or co-interior angles at B	✓ 1
A1 220° − 115° = 105° (AG)	✓ 1
Part (b)
M1 use of the cosine rule: AC² = 18² + 25² − 2 × 18 × 25 × cos 105°	✓ 1
A1 AC² = 1181.9…	✓ 1
A1 34.4 km	✓ 1
Part (c)
M1 use of the sine rule: sin(BAC)/25 = sin 105°/AC	✓ 1
A1 angle BAC = 44.6°	✓ 1
A1 bearing = 040° + 44.6° = 085°	✓ 1

Common error: A bearing is measured clockwise from north, so the bearing of A from B is the bearing of B from A plus 180° when it is less than 180°.

+ Add to test✓ In your test217IBMI3S-3-03r44-06Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium8 marks

Three weather stations are at A(1, 1), B(9, 3) and C(3, 7) on a map in which one unit represents 1 km. The perpendicular bisector of AC has equation 3y + x = 14.

(a)

Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.

(3)

(b)

The two bisectors meet at a vertex V of the Voronoi diagram. Find the coordinates of V, giving each coordinate to 3 significant figures.

(2)

(c)

Find the distance from V to A. Give your answer to 3 significant figures and state its significance for station B and station C. Use the unrounded coordinates of V.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint (5, 2) and gradient of AB = 2/8 = 1/4	✓ 1
M1 perpendicular gradient −4 and substitution of the midpoint	✓ 1
A1 y = −4x + 22	✓ 1
Part (b)
M1 equating or solving simultaneously: 3(−4x + 22) + x = 14	✓ 1
A1 V = (4.73, 3.09) (exact: (52/11, 34/11))	✓ 1
Part (c)
M1 √((52/11 − 1)² + (34/11 − 1)²)	✓ 1
A1 4.27 km	✓ 1
A1 V is the same distance from B and C (4.27 km), so it is equidistant from all three stations	✓ 1

Exam tip: Give the line in the form requested, and keep fractions or full calculator values until the final line to avoid rounding drift.

+ Add to test✓ In your test218IBMI3S-3-03r44-07Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium7 marks

A volcanologist designs a steel collector to cover a gas vent. The collector is a cylinder of radius 3.5 m with a hemispherical roof of the same radius on top. The internal volume of the collector is 260 m³.

(a)

Find the height of the cylindrical part. Give your answer to 3 significant figures.

(4)

(b)

The cylinder wall and the hemispherical roof are made from steel sheet; the collector has no base. Find the total area of sheet needed. Use the unrounded height from (a) and give your answer to the nearest m².

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume of hemisphere = (2/3)π(3.5)³ = 89.8… m³	✓ 1
M1 π(3.5)²h = 260 − 89.8…	✓ 1
A1 170.2… = 38.48…h	✓ 1
A1 h = 4.42 m	✓ 1
Part (b)
M1 curved area of cylinder = 2π(3.5)(4.42…) = 97.3… m²	✓ 1
M1 area of hemisphere = 2π(3.5)² = 77.0… m²	✓ 1
A1 174 m² (174.2…)	✓ 1

Common error: A hemisphere on top of a cylinder has curved area 2πr²; do not include the flat circle where the two solids join.

+ Add to test✓ In your test219IBMI3S-3-03r44-08Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium6 marks

A column of ash rises vertically above a volcanic vent V. Two observers, at A and B, stand on level ground in a straight line with V, with B between A and V. The distance AB is 640 m. The angle of elevation of the top of the column, T, is 28° from A and 41° from B. Let h metres be the height of T above V.

(a)

Show that 640 = h/tan 28° − h/tan 41°.

(2)

(b)

Find the value of h. Give your answer to the nearest metre.

(2)

(c)

Find the horizontal distance from B to V. Give your answer to the nearest metre. Use the unrounded value of h.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 AV = h/tan 28° and BV = h/tan 41°	✓ 1
A1 AB = AV − BV, giving the given equation (AG)	✓ 1
Part (b)
M1 factorise or rearrange: h = 640/(1/tan 28° − 1/tan 41°)	✓ 1
A1 h = 876 m (876.28…)	✓ 1
Part (c)
M1 BV = 876.28…/tan 41°	✓ 1
A1 1008 m	✓ 1

Exam tip: With two angles of elevation and a known baseline, write both horizontal distances in terms of the height and subtract.

+ Add to test✓ In your test220IBMI3S-3-03r44-09Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Medium7 marks

Three buoys P, Q and R mark a triangular exclusion zone at sea. PQ = 7.4 km, PR = 9.1 km and angle QPR = 58°.

(a)

Calculate the area of the exclusion zone. Give your answer to 3 significant figures.

(2)

(b)

Find the length of QR. Give your answer to 3 significant figures.

(3)

(c)

Find the shortest distance from P to the line QR. Use the unrounded values from (a) and (b) and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area = ½ × 7.4 × 9.1 × sin 58°	✓ 1
A1 28.6 km²	✓ 1
Part (b)
M1 QR² = 7.4² + 9.1² − 2 × 7.4 × 9.1 × cos 58°	✓ 1
A1 QR² = 66.2…	✓ 1
A1 8.14 km	✓ 1
Part (c)
M1 area = ½ × QR × d, so d = 2 × 28.55…/8.136…	✓ 1
A1 7.02 km	✓ 1

Exam tip: Once the area is known, the perpendicular height to any side follows from area = ½ × base × height.

+ Add to test✓ In your test221IBMI3S-3-01r46-06Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

Three hospitals are located at A(2, 1), B(8, 5) and C(6, −3) on a map in which one unit represents 1 km. Each home is served by its nearest hospital, and the service areas form a Voronoi diagram.

(a)

Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.

(3)

(b)

The perpendicular bisector of AC has equation y = x − 5. Find the coordinates of the Voronoi vertex where the three cells meet.

(2)

(c)

A new helipad is built at this vertex. Calculate its distance from hospital A, in km, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 midpoint of AB = (5, 3)	✓ 1
A1 gradient of AB = 4/6 = 2/3, so the perpendicular gradient is −3/2	✓ 1
A1 y = −1.5x + 10.5 (accept y = −3x/2 + 21/2)	✓ 1
Part (b)
M1 equates −1.5x + 10.5 = x − 5	✓ 1
A1 vertex = (6.2, 1.2)	✓ 1
Part (c)
M1 distance = √((6.2 − 2)² + (1.2 − 1)²) = √17.68	✓ 1
A1 distance = 4.20 km	✓ 1

Exam tip: The Voronoi vertex is equidistant from all three sites, so you can check your answer by finding the distance to a second site.

+ Add to test✓ In your test222IBMI3S-3-01r46-11Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium5 marks

Three observation towers P, Q and R stand around a Formula 1 circuit. The straight-line distances between them are PQ = 410 m, PR = 275 m and QR = 520 m.

(a)

Calculate the size of angle QPR, the largest angle of triangle PQR, to the nearest 0.1°.

(3)

(b)

Calculate the area of triangle PQR, in m², to 3 significant figures. Use the unrounded angle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cos QPR = (410² + 275² − 520²) / (2 × 410 × 275)	✓ 1
A1 cos QPR = -0.1183 (negative, so the angle is obtuse)	✓ 1
A1 angle QPR = 96.8°	✓ 1
Part (b)
M1 area = ½ × 410 × 275 × sin QPR	✓ 1
A1 area = 55979 m², so 56 000 m² to 3 s.f.	✓ 1

Exam tip: Use the cosine rule to find the angle opposite the longest side first; the sign of the cosine immediately tells you whether the angle is obtuse.

+ Add to test✓ In your test223IBMI3S-3-03r46-05Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

A triangular shade sail for a mountain-hut terrace has sides of length 7.5 m, 9.2 m and 11.0 m.

(a)

Find the size of the largest angle of the sail, to the nearest tenth of a degree.

(3)

(b)

Find the area of the sail, to 3 significant figures. Use the unrounded angle from part (a).

(2)

(c)

Fabric costs €18.50 per square metre. Using the unrounded area, calculate the cost of the fabric to the nearest euro.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 largest angle is opposite 11.0 m; uses cosine rule	✓ 1
A1 correct substitution $\cos C = \dfrac{7.5^2+9.2^2-11.0^2}{2\times7.5\times9.2}$	✓ 1
A1 81.7°	✓ 1
Part (b)
M1 area = ½ × 7.5 × 9.2 × sin C	✓ 1
A1 34.1 m²	✓ 1
Part (c)
M1 34.139… × 18.50	✓ 1
A1 €632	✓ 1

Exam tip: The largest angle is always opposite the longest side, so check your answer is the biggest angle in the triangle.

+ Add to test✓ In your test224IBMI3S-3-03r46-06Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

From a point P on level ground, a climber measures the angle of elevation of the top T of a rock tower as 14°. She walks 600 m in a straight line towards the foot of the tower, to a point Q, where the angle of elevation of T is 21°. Let the height of the tower above the ground be h metres.

(a)

Show that $h = \dfrac{600}{\dfrac{1}{\tan 14°} - \dfrac{1}{\tan 21°}}$.

(3)

(b)

Calculate the value of h, to 3 significant figures.

(2)

(c)

Using the unrounded value of h, find the horizontal distance from Q to the foot of the tower, to the nearest metre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 horizontal distance from Q to the foot is $h/\tan 21°$ and from P is $h/\tan 14°$	✓ 1
M1 sets $\dfrac{h}{\tan 14°} - \dfrac{h}{\tan 21°} = 600$	✓ 1
A1 factorises and rearranges to the given expression (AG)	✓ 1
Part (b)
M1 substitutes into the expression	✓ 1
A1 427 m	✓ 1
Part (c)
M1 distance = h ÷ tan 21°	✓ 1
A1 1112 m	✓ 1

Exam tip: In a 'show that' question, every line of working must be written down and the result must match the one given.

+ Add to test✓ In your test225IBMI3S-3-03r46-07Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

Three mountain-rescue huts are at A(1, 2), B(7, 2) and C(4, 8), where one unit represents 1 km. The Voronoi diagram for these sites has a single vertex V. The edge between A and B lies on the line $x = 4$. A helicopter landing site is to be placed at V.

(a)

Find the equation of the perpendicular bisector of AC, in the form $y = mx + c$.

(3)

(b)

Find the coordinates of V.

(2)

(c)

Find the distance from V to hut A, in km.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AC = (2.5, 5)	✓ 1
M1 gradient of AC = 2, so perpendicular gradient = −½	✓ 1
A1 $y = -0.5x + 6.25$	✓ 1
Part (b)
M1 substitutes $x = 4$ into the equation from (a)	✓ 1
A1 V = (4, 4.25)	✓ 1
Part (c)
M1 $\sqrt{(4-1)^2 + (4.25-2)^2}$	✓ 1
A1 3.75 km	✓ 1

Common error: V is equidistant from all three sites; check your answer by finding the distance from V to a second hut.

+ Add to test✓ In your test226IBMI3S-3-03r46-08Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

A waffle cone has a circular opening of radius 3.0 cm and a slant height of 10.5 cm. A scoop of ice cream in the shape of a hemisphere of radius 3.0 cm sits on top of the cone.

(a)

Find the perpendicular height of the cone, to 3 significant figures.

(2)

(b)

Calculate the curved surface area of the cone, to 3 significant figures.

(2)

(c)

Calculate the total volume of the cone and the ice cream, to 3 significant figures. Use unrounded values throughout.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $h = \sqrt{10.5^2 - 3.0^2}$	✓ 1
A1 10.1 cm	✓ 1
Part (b)
M1 $A = \pi r l = \pi \times 3.0 \times 10.5$	✓ 1
A1 99.0 cm²	✓ 1
Part (c)
M1 volume of cone = $\tfrac{1}{3}\pi \times 3.0^2 \times 10.06… = 94.8…$	✓ 1
M1 volume of hemisphere = $\tfrac{2}{3}\pi \times 3.0^3 = 56.5…$	✓ 1
A1 151 cm³	✓ 1

Common error: Do not use the slant height in the volume formula of a cone; the formula needs the perpendicular height.

+ Add to test✓ In your test227IBMI3S-3-03r46-09Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium8 marks

Two telecom masts A and B are 12.4 km apart, with B due east of A. A drone D is observed from A on a bearing of 038° and from B on a bearing of 321°.

(a)

Show that angle ADB = 77°.

(2)

(b)

Find the distance AD, to 3 significant figures.

(2)

(c)

Find how far north of the line AB the drone is, to 3 significant figures. Use the unrounded value of AD.

(2)

(d)

Find the area of triangle ABD, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 angle DAB = 90° − 38° = 52° or angle DBA = 321° − 270° = 51°	✓ 1
A1 angle ADB = 180° − 52° − 51° = 77° (AG)	✓ 1
Part (b)
M1 sine rule $\dfrac{AD}{\sin 51°} = \dfrac{12.4}{\sin 77°}$	✓ 1
A1 9.89 km	✓ 1
Part (c)
M1 height = AD sin 52°	✓ 1
A1 7.79 km	✓ 1
Part (d)
M1 area = ½ × 12.4 × 7.79…	✓ 1
A1 48.3 km²	✓ 1

Exam tip: Draw a sketch with a north line at each point; the angles inside the triangle come from the bearings.

+ Add to test✓ In your test228IBMI3S-3-03r46-10Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

Rainfall stations are located at A(2, 3), B(9, 4), C(5, 10) and D(10, 11) on a map where one unit represents 1 km. The rainfall recorded in one day was A: 12 mm, B: 18 mm, C: 7 mm, D: 24 mm. The rainfall at any location is estimated using the rainfall at the nearest station.

(a)

Estimate the rainfall at P(6, 6).

(2)

(b)

A new station E(6, 8) records 15 mm. State the new estimate for the rainfall at P, with a reason.

(2)

(c)

Find the equation of the perpendicular bisector of B and E, in the form $y = mx + c$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 calculates distances from P to the stations (PA = 5, PB = 3.61, PC = 4.12, PD = 6.40)	✓ 1
A1 nearest is B, so estimate is 18 mm	✓ 1
Part (b)
M1 PE = 2, which is less than the distance to every other station	✓ 1
A1 15 mm	✓ 1
Part (c)
M1 midpoint of BE = (7.5, 6) and gradient of BE = −4/3	✓ 1
M1 perpendicular gradient = 3/4 and substitution	✓ 1
A1 $y = 0.75x + 0.375$	✓ 1

Exam tip: Nearest-neighbour interpolation only needs distances to the sites; you do not have to draw the whole Voronoi diagram.

+ Add to test✓ In your test229IBMI3S-3-03r46-11Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Medium7 marks

A wedge of cheese is a prism of length 24 cm. Its triangular cross-section has two sides of length 9.0 cm and 7.5 cm with an angle of 62° between them.

(a)

Calculate the area of the triangular cross-section, to 3 significant figures.

(2)

(b)

Hence find the volume of the wedge, to 3 significant figures.

(1)

(c)

Find the length of the third side of the triangle, to 3 significant figures.

(2)

(d)

Find the total surface area of the wedge, to 3 significant figures. Use unrounded values.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area = ½ × 9.0 × 7.5 × sin 62°	✓ 1
A1 29.8 cm²	✓ 1
Part (b)
A1 715 cm³ (using unrounded area 29.799…)	✓ 1
Part (c)
M1 cosine rule $\sqrt{9.0^2 + 7.5^2 - 2\times9.0\times7.5\cos 62°}$	✓ 1
A1 8.59 cm	✓ 1
Part (d)
M1 $2\times29.80 + (9.0+7.5+8.59…)\times24$	✓ 1
A1 662 cm²	✓ 1

Common error: The three rectangular faces together form one rectangle of width equal to the perimeter of the triangle.

+ Add to test✓ In your test230IBMI-032-06-09Volume and surface area in context Paper 1Hard8 marks

A metal bucket is the frustum of a right circular cone. It is open at the top and has a flat circular base. The top has radius 15 cm, the base has radius 10 cm, and the vertical depth is 24 cm. The bucket can be modelled as a large cone with a smaller, similar cone removed from its tip.

(a)

Show that the height of the large cone is 72 cm.

(2)

(b)

Find the capacity of the bucket in litres.

(3)

(c)

Find the total area of metal used to make the bucket.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for similar triangles: $\frac{H}{15}=\frac{H-24}{10}$	✓ 1
A1 for $10H=15H-360$, giving $H=72$	✓ 1
Part (b)
M1 for the large cone minus the small cone: $\frac{1}{3}\pi(15^2)(72)-\frac{1}{3}\pi(10^2)(48)$	✓ 1
A1 for 11 900 cm³ (11 938)	✓ 1
A1 for 11.9 litres	✓ 1
Part (c)
M1 for the slant height of the frustum, $\sqrt{24^2+5^2}=24.52$, or for the slant heights of both cones, 73.55 and 49.03	✓ 1
M1 for the curved area $\pi(15+10)(24.52)=1925$ plus the base $\pi\times10^2$	✓ 1
A1 for 2240 cm² (2239.6)	✓ 1

Common error: Taking the curved area of a frustum as $\pi R\times24$ is wrong for two reasons: it uses the vertical depth instead of a slant height, and it ignores the removed cone. Find the curved area as a difference of two cones, or use the frustum's slant height.

+ Add to test✓ In your test231IBMI3S-3-02r9-12Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Hard7 marks

A feed hopper for a fish farm is made from an open-topped cylinder of radius 0.6 m on top of a cone with the same radius and a vertical height of 0.9 m. The apex of the cone points downwards. The total volume of the hopper is 2.0 m³. Use unrounded values in later parts.

(a)

Calculate the volume of the cone. Give your answer to 3 significant figures.

(2)

(b)

Find the height of the cylinder. Give your answer to 3 significant figures.

(2)

(c)

The hopper is made of sheet metal covering the curved surface of the cylinder and the curved surface of the cone only. Find the area of sheet metal needed. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (1/3) × π × 0.6² × 0.9	✓ 1
A1 0.339 m³	✓ 1
Part (b)
M1 π × 0.6² × h = 2 − 0.3392…	✓ 1
A1 h = 1.47 m	✓ 1
Part (c)
M1 slant height of cone √(0.6² + 0.9²) = 1.0816…	✓ 1
M1 cone curved area π × 0.6 × 1.0816… = 2.039… and cylinder curved area 2 × π × 0.6 × 1.4683… = 5.535…	✓ 1
A1 7.57 m²	✓ 1

Common error: The cylinder height is found from the remaining volume, not the total volume; subtract the cone first.

+ Add to test✓ In your test232IBMI3S-3-01r11-12Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 1Hard7 marks

A substation is at P. Turbine Q is 9.2 km from P, and turbine R is 7.5 km from Q. The angle QPR is 48°. Give lengths to 3 significant figures and angles to 1 decimal place, and use unrounded values in later working.

(a)

Find the two possible sizes of angle PRQ.

(3)

(b)

Both values of angle PRQ are possible. Calculate the shorter possible distance PR.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
sin(PRQ) ÷ 9.2 = sin 48° ÷ 7.5	✓ 1
acute value: PRQ = 65.7°	✓ 1
obtuse value: PRQ = 180° − 65.7° = 114.3°	✓ 1
Part (b)
the shorter distance comes from PRQ = 114.3°	✓ 1
angle PQR = 180° − 48° − 114.3° = 17.7°	✓ 1
PR ÷ sin 17.7° = 7.5 ÷ sin 48°	✓ 1
PR = 3.07 km	✓ 1

Common error: When two sides and an angle not between them are given, the sine rule can give an acute and an obtuse angle; check whether 180° minus the acute angle also fits in the triangle.

+ Add to test✓ In your test233IBMI3S-3-01r17-07Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 1Hard7 marks

A training shot is made by casting brass into a sphere of diameter 12.0 cm. Brass has density 8.5 g cm⁻³. A solid sphere of this size would be too heavy, so a spherical cavity, concentric with the shot, is left empty inside it. The shot must have a total mass of 7.26 kg. Ignore the mass of the air in the cavity.

(a)

Calculate the mass of a solid brass sphere of diameter 12.0 cm. Give your answer in kg, correct to 3 significant figures.

(3)

(b)

Find the volume of the cavity, in cm³, so that the shot has mass 7.26 kg. Use the unrounded mass from part (a).

(2)

(c)

Find the radius of the cavity, in cm, correct to 3 significant figures. Use the unrounded volume from part (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume = (4/3)π(6.0)³	✓ 1
M1 mass = density × volume = 8.5 × 904.78	✓ 1
A1 7.69 kg (= 7690.6 g)	✓ 1
Part (b)
M1 mass to remove = 7690.6 − 7260 = 430.6 g	✓ 1
A1 cavity volume = 430.6 ÷ 8.5 = 50.7 cm³	✓ 1
Part (c)
M1 (4/3)πr³ = 50.661	✓ 1
A1 r = 2.30 cm	✓ 1

Common error: Work in grams and cubic centimetres together (density in g cm⁻³) and convert to kilograms only at the end.

+ Add to test✓ In your test234IBMI3S-3-03r37-08Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 1Hard6 marks

A surveyor stands at a point P on level ground and measures the angle of elevation of the summit S of a volcano as 11°. She walks 2.6 km in a straight line directly towards the volcano to a point Q and measures the angle of elevation of the summit as 17°. The points P, Q and the base of the volcano are in a straight line.

(a)

Show that angle PSQ is 6°.

(2)

(b)

Find the distance QS. Give your answer in km to 3 significant figures.

(2)

(c)

Find the height of the summit above the ground. Use the unrounded value of QS. Give your answer in km to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 angle SQP (interior) $=180^{\circ}-17^{\circ}=163^{\circ}$ or uses the exterior angle of the triangle	✓ 1
A1 $\widehat{PSQ}=180^{\circ}-11^{\circ}-163^{\circ}=6^{\circ}$ (AG)	✓ 1
Part (b)
M1 sine rule $\frac{QS}{\sin11^{\circ}}=\frac{2.6}{\sin6^{\circ}}$	✓ 1
A1 4.75 km	✓ 1
Part (c)
M1 height $=QS\sin17^{\circ}$ with $QS=4.746$	✓ 1
A1 1.39 km	✓ 1

Common error: Forgetting that the angle at Q inside the triangle is 180° minus the elevation angle, and using 17° there instead, makes the sine rule give a wrong distance.

+ Add to test✓ In your test235IBMI-032-06-07Volume and surface area in context Paper 2Medium10 marks

A glass entrance to a museum is a right pyramid with a square base of side 30 m. Its apex is 20 m vertically above the centre of the base. The four triangular faces are glass; the base is the floor.

(a)

Find the height of one triangular face, measured from the apex to the midpoint of a base edge.

(2)

(b)

Find the total area of glass.

(2)

(c)

Find the volume of air enclosed by the pyramid.

(2)

(d)

Find the angle between a triangular face and the floor.

(2)

(e)

Find the angle between a sloping edge of the pyramid and the floor.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\sqrt{20^2+15^2}$	✓ 1
A1 for 25 m	✓ 1
Part (b)
M1 for $4\times\frac{1}{2}\times30\times25$	✓ 1
A1 for 1500 m²	✓ 1
Part (c)
M1 for $\frac{1}{3}\times30^2\times20$	✓ 1
A1 for 6000 m³	✓ 1
Part (d)
M1 for $\tan\theta=\frac{20}{15}$	✓ 1
A1 for 53.1°	✓ 1
Part (e)
M1 for using half the diagonal of the base, $15\sqrt2=21.21$, in $\tan\theta=\frac{20}{21.21}$	✓ 1
A1 for 43.3°	✓ 1

Common error: The angle a face makes with the floor uses half the side (15 m). The angle an edge makes with the floor uses half the diagonal (21.2 m). Swapping these two lengths is a frequent mistake.

+ Add to test✓ In your test236IBMI-032-05-05Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Medium11 marks

A ferry leaves port P and sails 18 km on a bearing of 065° to a buoy A. It then turns and sails 25 km on a bearing of 150° to an island B.

(a)

Show that angle PAB = 95°.

(2)

(b)

Find the distance PB.

(3)

(c)

Find the bearing of B from P.

(4)

(d)

The ferry returns directly from B to P at an average speed of 14 km h⁻¹. Find the time taken, in minutes, to the nearest minute.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Bearing of P from A is 065° + 180° = 245° (or the angle between north at A and AP is 65° by alternate angles) M1 ; ∠PAB = 245° − 150° = 95° A1	✓ 2
Part (b)
Use of the cosine rule M1 ; PB² = 18² + 25² − 2(18)(25)cos 95° A1 ; PB = 32.1 km (32.054…) A1	✓ 3
Part (c)
sin(∠APB) / 25 = sin 95° / 32.054 M1 ; ∠APB = 51.0° A1 ; bearing = 065° + 51.0° M1 ; 116° A1	✓ 4
Part (d)
time = 32.054 / 14 × 60 M1 ; 137 minutes (2 h 17 min) A1	✓ 2

Exam tip: Draw a north line at every turning point. Bearings are measured clockwise from north, so the interior angles of the triangle come from the differences between the north lines and the paths.

+ Add to test✓ In your test237IBMI-032-05-09Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Medium10 marks

Coastguard station B is 12 km due east of coastguard station A. A fishing boat S is out at sea to the north of the line AB. The bearing of S from A is 052° and the bearing of S from B is 318°.

(a)

Find angle ASB.

(3)

(b)

Find the distances AS and BS.

(3)

(c)

The boat sails due south until it reaches the line AB. Find the distance it sails.

(2)

(d)

Find the distance of the boat from B when it reaches the line AB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
∠SAB = 90° − 52° = 38° A1 ; ∠SBA = 318° − 270° = 48° A1 ; ∠ASB = 180° − 38° − 48° = 94° A1	✓ 3
Part (b)
Sine rule, e.g. AS / sin 48° = 12 / sin 94° (or equivalent for BS) M1 ; AS = 8.94 km A1 ; BS = 7.41 km A1	✓ 3
Part (c)
distance = AS sin 38° (or BS sin 48°) M1 ; 5.50 km A1	✓ 2
Part (d)
Valid method, e.g. 12 − AS cos 38° = 12 − 7.04, or BS cos 48° M1 ; 4.96 km A1	✓ 2

Common error: A bearing of 318° is not the angle inside the triangle. Find the angle between the bearing and the direction from B to A, which has bearing 270°.

+ Add to test✓ In your test238IBMI-032-01-10Volume and surface area in context Paper 2Medium11 marks

A shipping crate is a cuboid with internal dimensions 2.4 m long, 1.5 m wide and 1.2 m high.

(a)

Find the length of the longest straight rod that fits inside the crate.

(2)

(b)

Find the angle between this rod and the base of the crate.

(2)

(c)

The crate is packed with upright cylindrical drums, each of radius 0.28 m and height 1.15 m, arranged in rows parallel to the sides. Find the maximum number of drums that fit in the crate.

(2)

(d)

Find the volume of empty space in the crate when it contains this number of drums.

(3)

(e)

Find the percentage of the crate's volume occupied by the drums.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\sqrt{2.4^2+1.5^2+1.2^2}$ M1 ; 3.07 m (3.0741…) A1	✓ 2
Part (b)
Base diagonal = $\sqrt{2.4^2+1.5^2}$ = 2.830… and tan θ = 1.2/2.830… M1 ; θ = 23.0° A1	✓ 2
Part (c)
2.4 ÷ 0.56 = 4.28… so 4 along the length and 1.5 ÷ 0.56 = 2.67… so 2 across, one layer since 1.15 < 1.2 M1 ; 8 drums A1	✓ 2
Part (d)
One drum: π(0.28)²(1.15) = 0.2832… m³ M1 ; crate volume 4.32 m³ and 4.32 − 8(0.2832…) M1 ; 2.05 m³ (2.054…) A1	✓ 3
Part (e)
2.2660… ÷ 4.32 × 100 M1 ; 52.5% A1	✓ 2

Common error: Dividing the crate volume by the drum volume gives 15.25, but drums cannot be cut up. Count how many whole diameters fit along each edge.

+ Add to test✓ In your test239IBMI-032-07-r1-05Voronoi diagrams Paper 2Medium9 marks

A rectangular park occupies the region 0 ≤ x ≤ 10, 0 ≤ y ≤ 6, where one unit represents 100 m. Two ice-cream kiosks are located at K₁(2, 1) and K₂(6, 5). Visitors always walk to the nearest kiosk.

(a)

Find the equation of the Voronoi edge separating the cells of K₁ and K₂.

(3)

(b)

Find the coordinates of the points where this edge meets the boundary of the park.

(2)

(c)

Find the area of the park that is closer to K₁, giving your answer in m².

(2)

(d)

Assuming visitors are spread evenly across the park, find the percentage of visitors expected to use K₂, and comment on your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint of K₁K₂ = (4, 3) A1	✓ 1
Gradient K₁K₂ = 1, so perpendicular gradient = −1 M1	✓ 1
y = −x + 7 A1	✓ 1
Part (b)
At y = 0: (7, 0) A1	✓ 1
At y = 6: (1, 6) A1	✓ 1
Part (c)
Cell of K₁ is the trapezium (0,0), (7,0), (1,6), (0,6): area = ½(7 + 1)(6) = 24 units² M1	✓ 1
= 24 × 10 000 = 240 000 m² (2.40 × 10⁵ m²) A1	✓ 1
Part (d)
(60 − 24)/60 × 100 = 60% A1	✓ 1
K₂ should serve more visitors than K₁ (60% vs 40%), so it may need more stock/staff R1	✓ 1

Common error: Remember to convert area units: one grid unit is 100 m, so one square unit is 100 × 100 = 10 000 m², not 100 m².

+ Add to test✓ In your test240IBMI-032-07-r2-04Voronoi diagrams Paper 2Medium9 marks

Four weather stations record daily rainfall. Their coordinates (in km) and the rainfall recorded on one day are:
S₁(2, 1): 12 mm; S₂(7, 3): 18 mm; S₃(4, 8): 9 mm; S₄(9, 9): 15 mm.
Rainfall at other locations is estimated by nearest-neighbour interpolation, using a Voronoi diagram of the stations.

(a)

Find the equation of the perpendicular bisector of S₂S₃.

(4)

(b)

The town T is at (6, 6). Determine which station's cell contains T and hence estimate the rainfall at T.

(2)

(c)

Comment on the reliability of this estimate.

(1)

(d)

A farm F at $(k, 5)$ is the same distance from S₂ and S₃. Find $k$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint (5.5, 5.5) A1	✓ 1
Gradient S₂S₃ $= -\frac{5}{3}$ M1	✓ 1
Perpendicular gradient $\frac{3}{5}$ A1	✓ 1
$y = 0.6x + 2.2$ A1	✓ 1
Part (b)
Distances from T: S₃ $\sqrt{8} = 2.83$ is the smallest (S₂ 3.16, S₄ 4.24, S₁ 6.40), so T is in the cell of S₃ R1	✓ 1
Estimated rainfall 9 mm A1	✓ 1
Part (c)
Valid comment, e.g. T is almost as close to S₂ (18 mm) as to S₃, so the estimate may be unreliable because rainfall varies continuously R1	✓ 1
Part (d)
$5 = 0.6k + 2.2$ M1	✓ 1
$k = 4.67$ A1	✓ 1

Common error: Using the gradient of S₂S₃ itself in the bisector equation. Take the negative reciprocal to get the perpendicular gradient.

+ Add to test✓ In your test241IBMI-032-07-r2-06Voronoi diagrams Paper 2Medium10 marks

A town is modelled by the square $0 \le x \le 8$, $0 \le y \le 8$, with distances in km. Its population is spread evenly at 150 people per km². There are two pharmacies, at P(1, 2) and Q(5, 4). Each resident uses the nearer pharmacy.

(a)

Find the equation of the perpendicular bisector of PQ.

(4)

(b)

Find the coordinates of the points where this bisector meets the edges $y = 8$ and $y = 0$ of the town.

(2)

(c)

Find the area of the Voronoi cell of P within the town.

(2)

(d)

Hence find the number of residents expected to use pharmacy Q.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint (3, 3) A1	✓ 1
Gradient PQ $= 0.5$ M1	✓ 1
Perpendicular gradient $-2$ A1	✓ 1
$y = -2x + 9$ A1	✓ 1
Part (b)
(0.5, 8) A1	✓ 1
(4.5, 0) A1	✓ 1
Part (c)
Trapezium with parallel sides 0.5 and 4.5 and height 8: $\frac{1}{2}(0.5 + 4.5)(8)$ M1	✓ 1
20 km² A1	✓ 1
Part (d)
Area of Q's cell $= 64 - 20 = 44$ km² M1	✓ 1
$44 \times 150 = 6600$ residents A1	✓ 1

Common error: Assuming that the cells of two sites split a region in half. This is only true when the bisector passes through the centre of a symmetric region.

+ Add to test✓ In your test242IBMI-032-02-r1-03Voronoi diagrams Paper 2Medium10 marks

A square area of farmland is modelled by the region $0 \le x \le 8$, $0 \le y \le 8$, where 1 unit represents 1 km. Two water wells are located at A(2, 2) and B(6, 4). Each point of the farmland is supplied by the nearer well.

(a)

Find the equation of the perpendicular bisector of AB.

(3)

(b)

Find the coordinates of the points where this bisector meets the boundary lines $y = 0$ and $y = 8$.

(2)

(c)

A storage shed is at F(5, 1). Show that F lies on the boundary between the two cells and interpret this result in context.

(2)

(d)

Find the area of farmland supplied by well A.

(2)

(e)

Find the percentage of the farmland supplied by well B.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint (4, 3) and gradient of AB = $\frac{2}{4} = \frac{1}{2}$ M1	✓ 1
Perpendicular gradient = −2 M1	✓ 1
$y = -2x + 11$ A1	✓ 1
Part (b)
$y = 0$: $x = 5.5$, giving (5.5, 0) A1	✓ 1
$y = 8$: $x = 1.5$, giving (1.5, 8) A1	✓ 1
Part (c)
At $x = 5$, $y = -2(5) + 11 = 1$, so F lies on the bisector (or FA = FB = $\sqrt{10}$) M1	✓ 1
F is equidistant from both wells, so either well is equally near to the shed R1	✓ 1
Part (d)
Region is the trapezium with vertices (0, 0), (5.5, 0), (1.5, 8), (0, 8): area = $\frac{5.5 + 1.5}{2} \times 8$ M1	✓ 1
= 28 km² A1	✓ 1
Part (e)
$\frac{64 - 28}{64} \times 100$ = 56.3% A1	✓ 1

Exam tip: When a Voronoi cell is clipped by a rectangular boundary, list the vertices of the cell first; the shape is often a trapezium or can be split into simple shapes.

+ Add to test✓ In your test243IBMI-032-02-r2-05Voronoi diagrams Paper 2Medium9 marks

Four weather stations record the rainfall in one month. Their positions (in km) and readings are:
P(0, 0): 42 mm; Q(6, 2): 55 mm; R(1, 7): 38 mm; S(7, 8): 61 mm.
Rainfall at other locations is estimated using nearest-neighbour interpolation, based on the Voronoi diagram of the four stations.

(a)

Find the equation of the perpendicular bisector of QR.

(3)

(b)

A farm is located at T(4, 5). Show that T lies on the boundary between the cells of Q and R.

(2)

(c)

A village is located at U(3, 3). Estimate the rainfall at U.

(3)

(d)

Comment on the reliability of using nearest-neighbour interpolation to estimate the rainfall at T.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint of QR = (3.5, 4.5) A1	✓ 1
Gradient of QR = $\frac{5}{-5} = -1$, so perpendicular gradient = 1 M1	✓ 1
$y = x + 1$ A1	✓ 1
Part (b)
TQ = TR = $\sqrt{13}$ = 3.61 km (or 5 = 4 + 1 so T is on the bisector) M1	✓ 1
TP = $\sqrt{41}$ = 6.40 km and TS = $\sqrt{18}$ = 4.24 km are both greater, so T is on the edge between Q and R R1	✓ 1
Part (c)
Distances from U: P 4.24, Q 3.16, R 4.47, S 6.40 km M1	✓ 1
Q is the nearest station A1	✓ 1
Estimated rainfall = 55 mm A1	✓ 1
Part (d)
T is equally close to Q and R, whose readings differ by 17 mm, so the estimate is ambiguous/unreliable (e.g. an average of 46.5 mm may be more sensible) R1	✓ 1

Exam tip: To decide which cell a point is in, compare its distances to all relevant sites; the smallest distance identifies the cell.

+ Add to test✓ In your test244IBMI3S-3-05-09Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Medium12 marks

A coastguard vessel leaves harbour A and travels 24 km on a bearing of 038° to a buoy B. It then travels 31 km on a bearing of 112° to a wreck C.

(a)

Show that angle ABC = 106°.

(2)

(b)

Calculate the distance AC.

(3)

(c)

Find the bearing of C from A.

(3)

(d)

The vessel returns from C directly to A at a steady 20 km h⁻¹. Calculate the time taken, in hours.

(2)

(e)

Find the shortest distance from B to the straight route AC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of A from B is 038° + 180° = 218°	✓ 1
A1 $218^\circ-112^\circ=106^\circ$ (AG)	✓ 1
Part (b)
M1 cosine rule $AC^2=24^2+31^2-2(24)(31)\cos106^\circ$	✓ 1
A1 $AC^2=1947$	✓ 1
A1 $AC=44.1$ km	✓ 1
Part (c)
M1 sine rule $\dfrac{\sin\angle BAC}{31}=\dfrac{\sin106^\circ}{44.1}$	✓ 1
A1 $\angle BAC=42.5^\circ$	✓ 1
A1 bearing = 038° + 42.5° = 080.5° (accept 080° to 081°)	✓ 1
Part (d)
M1 $44.1\div20$ (ECF from (b))	✓ 1
A1 $2.21$ hours	✓ 1
Part (e)
M1 $24\times\sin42.5^\circ$ (ECF from (c))	✓ 1
A1 $16.2$ km	✓ 1

Common error: Find the back-bearing of A from B (add or subtract 180°) before subtracting bearings to get an angle inside the triangle.

+ Add to test✓ In your test245IBMI3S-3-05-10Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Medium14 marks

Four ambulance posts are located on a map (1 unit = 1 km): A(2, 2), B(8, 2), C(5, 8) and D(5, −2). A Voronoi diagram shows the region closest to each post.

(a)

Find the equation of the perpendicular bisector of AD.

(3)

(b)

The perpendicular bisector of AB is $x=5$. Find the coordinates of the Voronoi vertex where the cells of A, B and D meet.

(3)

(c)

Find the coordinates of the Voronoi vertex where the cells of A, B and C meet.

(2)

(d)

A town T is at (5, 3). Explain why T is equally close to posts A and B, and find that distance.

(2)

(e)

A second town P is at (6, 3). Determine which post is nearest to P and its distance from P.

(2)

(f)

Find the length of the edge of the Voronoi diagram shared by the cells of A and B.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint (3.5, 0)	✓ 1
M1 gradient of AD is $-\frac43$ so perpendicular gradient is $\frac34$	✓ 1
A1 $y=\frac34(x-3.5)$ (accept $y=0.75x-2.625$)	✓ 1
Part (b)
A1 uses $x=5$	✓ 1
M1 substitutes into the equation from (a)	✓ 1
A1 (5, 1.125)	✓ 1
Part (c)
M1 gradient of AC is 2, perpendicular gradient $-\frac12$ through (3.5, 5), substituting $x=5$	✓ 1
A1 (5, 4.25)	✓ 1
Part (d)
R1 T lies on $x=5$, the perpendicular bisector of AB, between the two vertices so on their shared edge	✓ 1
A1 $\sqrt{3^2+1^2}=3.16$ km	✓ 1
Part (e)
M1 compares distances (A 4.12, B 2.24, C 5.10, D 5.10)	✓ 1
A1 post B, 2.24 km	✓ 1
Part (f)
M1 difference of the y-coordinates of the two vertices	✓ 1
A1 $4.25-1.125=3.13$ km	✓ 1

Exam tip: To place a vertex, intersect two perpendicular bisectors; the third bisector then passes through the same point, giving a quick check.

+ Add to test✓ In your test246IBMI3S-3-05-13Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Medium13 marks

A rectangular swimming pool is 25 m long and 10 m wide. The depth is 1.0 m at the shallow end and 2.4 m at the deep end, and the floor slopes uniformly between them. The side view of the pool is a trapezium.

(a)

Find the angle the sloping floor makes with the horizontal.

(2)

(b)

Calculate the volume of the pool.

(3)

(c)

The pool is filled from empty at a rate of 120 litres per second. Find the time taken, in minutes. (1 m³ = 1000 litres)

(2)

(d)

The floor and the four walls are to be tiled. Calculate the total area to be tiled.

(4)

(e)

Tiles are sold in boxes, each covering 2 m², at $38.50 per box. Find the cost of the boxes needed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\tan\theta=\dfrac{1.4}{25}$	✓ 1
A1 $3.21^\circ$	✓ 1
Part (b)
M1 trapezium area $\frac12(1.0+2.4)\times25=42.5$ m²	✓ 1
M1 multiplies by the width 10	✓ 1
A1 $425$ m³	✓ 1
Part (c)
M1 $\dfrac{425000}{120}$ seconds	✓ 1
A1 $59.0$ minutes	✓ 1
Part (d)
M1 length of sloping floor $\sqrt{25^2+1.4^2}=25.04$ m	✓ 1
M1 floor area $25.04\times10=250.4$ m²	✓ 1
M1 side walls $2\times42.5=85$ and end walls $10\times1.0+10\times2.4=34$	✓ 1
A1 $369$ m²	✓ 1
Part (e)
M1 $369\div2=184.5$ so 185 boxes (rounds up)	✓ 1
A1 $185\times38.50=$7122.50$	✓ 1

Common error: Use the sloping floor length, not the 25 m horizontal length, when finding the area of the floor to be tiled.

+ Add to test✓ In your test247IBMI3S-3-05b-10Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Medium12 marks

A retailer has three stores in a region, plotted on a map with units in km: A(2, 2), B(10, 4) and C(4, 9). Customers are assumed to shop at their nearest store, so the region is divided into a Voronoi diagram.

(a)

Find the equation of the perpendicular bisector of AB.

(3)

(b)

Find the equation of the perpendicular bisector of BC.

(3)

(c)

Find the coordinates of the Voronoi vertex where the two bisectors meet.

(2)

(d)

The retailer plans a charging point at this vertex. Find its distance from store A.

(2)

(e)

A customer lives at (6, 5). Find the distance to the store she will use.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 midpoint (6, 3)	✓ 1
M1 gradient of AB = ¼, perpendicular gradient −4	✓ 1
A1 y = −4x + 27	✓ 1
Part (b)
A1 midpoint (7, 6.5)	✓ 1
M1 gradient of BC = −5/6, perpendicular gradient 6/5	✓ 1
A1 y = 1.2x − 1.9	✓ 1
Part (c)
M1 equate −4x + 27 = 1.2x − 1.9 (or use technology)	✓ 1
A1 (5.56, 4.77)	✓ 1
Part (d)
M1 attempt to find the distance from the vertex to A	✓ 1
A1 4.51 km (equal for all three stores)	✓ 1
Part (e)
M1 attempt to find the distances to A, B and C: 5, 4.12 and 4.47	✓ 1
A1 nearest is B at 4.12 km	✓ 1

Exam tip: Every point on a Voronoi edge is equidistant from the two nearest sites; a vertex is equidistant from three.

+ Add to test✓ In your test248IBMI3S-3-05b-11Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Medium12 marks

A hospital ward is a cuboid with a rectangular floor ABCD measuring 9 m by 6 m, where AB = 9 m and BC = 6 m. The room is 3.2 m high. A sensor is placed at floor corner A and an ultraviolet lamp at ceiling corner G, directly above C.

(a)

Calculate the length of the floor diagonal AC.

(2)

(b)

Calculate the distance AG from the sensor to the lamp.

(2)

(c)

Find the angle between AG and the floor.

(2)

(d)

Fresh air enters at 0.45 m³ s⁻¹. Find the time, in minutes, to supply a volume equal to the volume of the room.

(3)

(e)

All six inside surfaces are to be sprayed with disinfectant covering 14 m² per litre. Calculate the volume of disinfectant needed.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use Pythagoras in triangle ABC	✓ 1
A1 AC = 10.8 m	✓ 1
Part (b)
M1 attempt to use Pythagoras in triangle ACG	✓ 1
A1 AG = 11.3 m	✓ 1
Part (c)
M1 attempt to use tan⁻¹(3.2/AC)	✓ 1
A1 16.5°	✓ 1
Part (d)
A1 volume = 9 × 6 × 3.2 = 172.8 m³	✓ 1
M1 172.8 ÷ 0.45 = 384 s	✓ 1
A1 6.4 minutes	✓ 1
Part (e)
M1 attempt at total surface area 2(9×6 + 9×3.2 + 6×3.2)	✓ 1
A1 204 m²	✓ 1
A1 14.6 litres	✓ 1

Common error: Using the floor diagonal as the distance to the lamp; the vertical height must be included with a second use of Pythagoras.

+ Add to test✓ In your test249IBMI3S-3-01r6-11Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Medium13 marks

Three turbines P, Q and R are positioned in a wind farm. Q is 3.7 km from P on a bearing of 040° from P. R is 5.1 km from Q on a bearing of 125° from Q. Use unrounded values in later parts.

(a)

Show that angle PQR = 95°.

(2)

(b)

Calculate the distance PR, to 3 significant figures.

(3)

(c)

Find angle QPR, to 3 significant figures.

(3)

(d)

Find the bearing of R from P, to the nearest degree.

(3)

(e)

Calculate the area of triangle PQR, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of P from Q is $040° + 180° = 220°$	✓ 1
A1 $220° - 125° = 95°$	✓ 1
Part (b)
M1 cosine rule $PR^2 = 3.7^2 + 5.1^2 - 2(3.7)(5.1)\cos 95°$	✓ 1
A1 $PR^2 = 42.99$	✓ 1
A1 PR = 6.56 km	✓ 1
Part (c)
M1 sine rule $\frac{\sin \text{QPR}}{5.1} = \frac{\sin 95°}{PR}$	✓ 1
A1 correct substitution of unrounded PR	✓ 1
A1 angle QPR = 50.8°	✓ 1
Part (d)
R1 R lies clockwise from PQ as seen from P, so the bearing is $40° + \text{angle QPR}$	✓ 1
M1 attempt to add using the unrounded angle from (c)	✓ 1
A1 91°	✓ 1
Part (e)
M1 $\frac12 (3.7)(5.1)\sin 95°$	✓ 1
A1 9.40 km²	✓ 1

Exam tip: Angle PQR is obtuse, so the other two angles must be acute; this removes any doubt about which solution to take from the sine rule.

+ Add to test✓ In your test250IBMI3S-3-02r8-03Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 2Medium13 marks

Three wind turbines are at T₁(2, 2), T₂(10, 2) and T₃(6, 9), coordinates in km. A Voronoi diagram is drawn with the turbines as sites. A maintenance depot is to be placed at the vertex of the diagram.

(a)

Find the equation of the perpendicular bisector of T₁T₂.

(2)

(b)

Find the equation of the perpendicular bisector of T₁T₃.

(3)

(c)

Hence find the coordinates of the vertex, giving exact values.

(2)

(d)

Find the distance from the depot to each turbine, giving your answer to 3 significant figures.

(2)

(e)

Find the equation of the perpendicular bisector of T₂T₃.

(3)

(f)

Show that the vertex found in (c) lies on the bisector in (e).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint (6, 2) and T₁T₂ horizontal	✓ 1
A1 x = 6	✓ 1
Part (b)
M1 midpoint (4, 5.5)	✓ 1
M1 gradient of T₁T₃ = 7/4, so perpendicular gradient = −4/7	✓ 1
A1 y − 5.5 = −(4/7)(x − 4), i.e. y = −(4/7)x + 109/14	✓ 1
Part (c)
M1 substitute x = 6	✓ 1
A1 (6, 61/14)	✓ 1
Part (d)
M1 √(4² + (61/14 − 2)²)	✓ 1
A1 4.64 km	✓ 1
Part (e)
M1 midpoint (8, 5.5)	✓ 1
M1 gradient of T₂T₃ = −7/4, so perpendicular gradient = 4/7	✓ 1
A1 y = (4/7)x + 13/14	✓ 1
Part (f)
R1 substituting x = 6 gives y = 24/7 + 13/14 = 61/14, matching the vertex	✓ 1

Exam tip: The three perpendicular bisectors of a triangle meet at one point; checking the third bisector is a useful test of your vertex.

+ Add to test✓ In your test251IBMI3S-3-02r8-12Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 2Medium13 marks

A library has a glass roof in the shape of a square-based pyramid. The base is a square of side 12 m and the vertical height is 5 m. The apex is directly above the centre of the base.

(a)

Find the volume enclosed by the roof.

(2)

(b)

Find the slant height of one triangular face (the height from the apex to the midpoint of a base edge), to 3 significant figures.

(2)

(c)

Find the total area of glass in the four triangular faces, to 3 significant figures. Use the unrounded slant height.

(2)

(d)

Glass costs £112 per m². Using the unrounded area, find the cost to the nearest pound.

(2)

(e)

Find the angle between a triangular face and the base, to 3 significant figures.

(2)

(f)

Find the length of a sloping edge of the roof (from the apex to a corner of the base), to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ⅓ × 12² × 5	✓ 1
A1 240 m³	✓ 1
Part (b)
M1 √(5² + 6²)	✓ 1
A1 7.81 m	✓ 1
Part (c)
M1 4 × ½ × 12 × 7.8102…	✓ 1
A1 187 m²	✓ 1
Part (d)
M1 187.446… × 112	✓ 1
A1 £20 994	✓ 1
Part (e)
M1 tan θ = 5 ÷ 6	✓ 1
A1 39.8°	✓ 1
Part (f)
M1 half the base diagonal = 6√2 = 8.485…	✓ 1
M1 √(5² + (6√2)²) = √97	✓ 1
A1 9.85 m	✓ 1

Common error: The slant height of a face uses half the base side (6 m); the sloping edge uses half the base diagonal (6√2 m).

+ Add to test✓ In your test252IBMI3S-3-02r9-06Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Medium12 marks

A fish-rearing tank is a cylinder of radius 0.8 m and length 4.5 m, with a hemisphere of radius 0.8 m attached to each end. Use unrounded values in later parts.

(a)

Calculate the volume of the tank. Give your answer to 3 significant figures.

(3)

(b)

Calculate the total outer surface area of the tank. Give your answer to 3 significant figures.

(3)

(c)

The outside of the tank is given two coats of paint costing €4.80 per m² of surface per coat. Find the cost of the paint, to the nearest euro.

(2)

(d)

The tank is stocked at 25 kg of fish per m³. Each fish has a mass of 0.85 kg. Find the greatest whole number of fish that can be stocked.

(2)

(e)

A model of the tank is built to a scale of 1 : 8, with the same shape. Find the volume of the model in litres (1 m³ = 1000 litres). Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cylinder volume π × 0.8² × 4.5 (= 9.048…)	✓ 1
M1 two hemispheres: (4/3) × π × 0.8³ (= 2.144…)	✓ 1
A1 11.2 m³	✓ 1
Part (b)
M1 curved area of cylinder 2 × π × 0.8 × 4.5 (= 22.62…)	✓ 1
M1 two hemispheres: 4 × π × 0.8² (= 8.042…)	✓ 1
A1 30.7 m²	✓ 1
Part (c)
M1 2 × 30.66… × 4.80	✓ 1
A1 €294	✓ 1
Part (d)
M1 25 × 11.19… ÷ 0.85 (= 329.2…)	✓ 1
A1 329 fish	✓ 1
Part (e)
M1 volume scale factor 8³ = 512 so 11.19… ÷ 512	✓ 1
A1 21.9 litres	✓ 1

Common error: Volumes scale with the cube of the length scale factor, not the square (which applies to areas).

+ Add to test✓ In your test253IBMI3S-3-01r11-03Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Medium12 marks

The atrium of a library is a vertical cylinder of radius 6 m and wall height 8 m, topped by a hemispherical glass dome of radius 6 m. The cylinder has a flat floor.

(a)

Calculate the total volume of air enclosed by the atrium, in m³, correct to the nearest m³.

(3)

(b)

Find the curved surface area of the glass dome, in m², correct to 1 decimal place.

(2)

(c)

Glass for the dome costs €185 per m². Calculate the cost of the glass, to the nearest euro.

(2)

(d)

The curved wall of the cylinder (height 8 m) is to be painted, except for a door of area 6 m². One litre of paint covers 11 m². Paint is sold only in 5-litre tins costing €48.90 each. Find the minimum cost of paint needed.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
cylinder: π × 6² × 8 = 904.8 m³	✓ 1
hemisphere: ⅔ × π × 6³ = 452.4 m³	✓ 1
total = 1357 m³	✓ 1
Part (b)
area = ½ × 4π × 6² = 2π × 6²	✓ 1
= 226.2 m²	✓ 1
Part (c)
cost = 185 × 226.19	✓ 1
= €41,846	✓ 1
Part (d)
wall area = 2π × 6 × 8 = 301.6 m²	✓ 1
area to paint = 301.6 − 6 = 295.6 m²	✓ 1
litres = 295.6 ÷ 11 = 26.87 litres	✓ 1
26.87 ÷ 5 = 5.37, so 6 tins are needed	✓ 1
cost = 6 × 48.90 = €293.40	✓ 1

Common error: When paint or materials are sold in fixed packs, round the number of packs up to a whole number, never to the nearest.

+ Add to test✓ In your test254IBMI3S-3-01r11-13Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Medium15 marks

Three wind turbine bases A, B and C stand on level ground. B is 420 m from A on a bearing of 040°. C is 380 m from A on a bearing of 092°. Give distances to 3 significant figures and angles to 1 decimal place, and use unrounded values in later working.

(a)

Show that angle BAC = 52°.

(1)

(b)

Calculate the distance BC.

(3)

(c)

Use the cosine rule to find angle ABC.

(3)

(d)

Find angle ACB, and hence find the bearing of B from C, correct to the nearest degree.

(3)

(e)

A new meteorological mast of height 95 m is built at B. Calculate the angle of elevation of the top of the mast from C, correct to 1 decimal place.

(2)

(f)

Calculate the area of triangle ABC in hectares, correct to 3 significant figures. (1 hectare = 10 000 m².)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
092° − 040° = 52°	✓ 1
Part (b)
BC² = 420² + 380² − 2 × 420 × 380 × cos 52°	✓ 1
BC² = 124281	✓ 1
BC = 353 m	✓ 1
Part (c)
cos ABC = (420² + BC² − 380²) ÷ (2 × 420 × BC)	✓ 1
cos ABC = 0.5277	✓ 1
angle ABC = 58.1°	✓ 1
Part (d)
angle ACB = 180° − 52° − 58.1° = 69.9°	✓ 1
the bearing of A from C is 092° + 180° = 272°	✓ 1
bearing of B from C = 272° + 69.9° = 342° (ECF from the angle found)	✓ 1
Part (e)
tan(angle) = 95 ÷ BC	✓ 1
angle = 15.1°	✓ 1
Part (f)
area = ½ × 420 × 380 × sin 52°	✓ 1
= 62883 m²	✓ 1
= 6.29 hectares	✓ 1

Exam tip: To find an angle with the cosine rule, rearrange first and use the inverse cosine; this avoids the ambiguity that the sine rule can bring when an angle may be obtuse.

+ Add to test✓ In your test255IBMI3S-3-01r17-06Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Medium12 marks

A railway tunnel of length 850 m is dug through a hill. Its cross-section is a rectangle of width 8 m and wall height 4 m, with a semicircle of diameter 8 m on top of the rectangle.

(a)

Calculate the area of the cross-section, in m², correct to 3 significant figures.

(3)

(b)

Find the volume of rock removed to dig the tunnel, in m³. Give your answer correct to 3 significant figures, using the unrounded cross-section area.

(2)

(c)

Once broken up, the rock occupies 1.35 times its original volume. It is carried away by lorries, each carrying 12 m³ of broken rock. Find the minimum number of lorry loads needed. Use the unrounded volume from part (b).

(3)

(d)

The two walls and the curved roof are lined with concrete; the floor is not lined. The lining costs €32.50 per m². Calculate the total cost of the lining, correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area of rectangle = 8 × 4 = 32	✓ 1
M1 area of semicircle = ½π(4)² = 25.13…	✓ 1
A1 57.1 m²	✓ 1
Part (b)
M1 volume = 57.133 × 850	✓ 1
A1 4.86 × 10⁴ m³	✓ 1
Part (c)
M1 broken volume = 1.35 × 48562.8 = 65559.8 m³	✓ 1
M1 65559.8 ÷ 12 = 5463.3	✓ 1
A1 5464 loads (rounded up)	✓ 1
Part (d)
M1 perimeter of cross-section excluding the floor = 4 + 4 + π × 4 (the semicircular arc has length π × 4)	✓ 1
M1 lined area = 20.566 × 850 = 17481 m²	✓ 1
M1 cost = lined area × 32.50	✓ 1
A1 €568,000 (3 s.f.)	✓ 1

Exam tip: Draw the cross-section and mark every length before calculating: a labelled sketch prevents errors such as including the floor in the lining.

+ Add to test✓ In your test256IBMI3S-3-01r17-09Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Medium12 marks

Three rail control centres are at A(0, 0), B(8, 0) and C(2, 6), where the coordinates are in km. A Voronoi diagram is drawn with these three points as sites.

(a)

Find the equation of the perpendicular bisector of AB.

(2)

(b)

Show that the perpendicular bisector of AC has equation y = −¹⁄₃x + ¹⁰⁄₃.

(3)

(c)

Find the coordinates of the vertex where the three cells meet.

(2)

(d)

Calculate the distance from this vertex to A, in km, correct to 3 significant figures.

(2)

(e)

A town is located at T(6, 5). By calculating distances, determine which control centre is nearest to T and state that distance, correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AB is (4, 0) and AB is horizontal	✓ 1
A1 x = 4	✓ 1
Part (b)
M1 midpoint of AC = (1, 3)	✓ 1
M1 gradient of AC = 3, so perpendicular gradient = −1/3	✓ 1
A1 y − 3 = −(1/3)(x − 1), which simplifies to y = −¹⁄₃x + ¹⁰⁄₃ (AG)	✓ 1
Part (c)
M1 substitute x = 4 into the equation from part (b)	✓ 1
A1 (4, 2)	✓ 1
Part (d)
M1 √(4² + 2²)	✓ 1
A1 4.47 km (the vertex is the same distance from B and C)	✓ 1
Part (e)
M1 attempt to find distances from T to at least two centres	✓ 1
A1 TA = 7.81 km, TB = 5.39 km, TC = 4.12 km (all three correct)	✓ 1
A1 C is nearest, at 4.12 km	✓ 1

Exam tip: A vertex of a Voronoi diagram is equidistant from three sites, which gives a quick check on your coordinates.

+ Add to test✓ In your test257IBMI3S-3-01r17-13Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Medium12 marks

Three rail stations A, B and C form a triangle. Station B is 12 km from A on a bearing of 070°. Station C is 18 km from B on a bearing of 155°.

(a)

Show that angle ABC = 95°.

(2)

(b)

Calculate the distance AC, in km, correct to 3 significant figures.

(3)

(c)

Calculate the area of triangle ABC, in km², correct to 3 significant figures.

(2)

(d)

A straight service road is to be built from B to meet AC at right angles. Find the length of this road, in km, correct to 3 significant figures. Use unrounded values from earlier parts.

(2)

(e)

Find the bearing of C from A, to the nearest degree. Give your answer as a three-figure bearing.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of A from B = 070° + 180° = 250°	✓ 1
A1 angle ABC = 250° − 155° = 95° (AG)	✓ 1
Part (b)
M1 AC² = 12² + 18² − 2(12)(18)cos 95°	✓ 1
A1 correct substitution evaluated, AC² = 505.7…	✓ 1
A1 AC = 22.5 km	✓ 1
Part (c)
M1 area = ½ × 12 × 18 × sin 95°	✓ 1
A1 108 km²	✓ 1
Part (d)
M1 area = ½ × AC × d (or d = 12 sin A or similar correct method)	✓ 1
A1 d = 9.57 km	✓ 1
Part (e)
M1 sine rule: sin(BAC)/18 = sin 95°/AC (unrounded AC)	✓ 1
A1 angle BAC = 52.9°	✓ 1
A1 bearing = 070° + 52.9° = 123°	✓ 1

Exam tip: When two sides and the included angle are known, use the cosine rule for the third side first; the sine rule is then safe for a second angle because the largest angle is already known.

+ Add to test✓ In your test258IBMI3S-3-02r19-02Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 2Medium13 marks

A water company has a pumping station P, a reservoir R and a treatment plant T on level ground. R is 8.4 km from P on a bearing of 062°. T is 5.9 km from P on a bearing of 143°.

(a)

Show that angle RPT = 81°.

(1)

(b)

Calculate the distance RT, to 3 significant figures.

(3)

(c)

Calculate the area of triangle PRT, to 3 significant figures.

(2)

(d)

Find angle PRT, to the nearest 0.1°.

(3)

(e)

The bearing of P from R is 242°. Find the bearing of T from R, to the nearest degree.

(2)

(f)

Find the shortest distance from P to the straight line RT, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 143° − 62° = 81° (bearings measured clockwise from north at P)	✓ 1
Part (b)
M1 cosine rule: RT² = 8.4² + 5.9² − 2 × 8.4 × 5.9 × cos 81°	✓ 1
A1 RT² = 89.9…	✓ 1
A1 RT = 9.48 km	✓ 1
Part (c)
M1 ½ × 8.4 × 5.9 × sin 81°	✓ 1
A1 24.5 km²	✓ 1
Part (d)
M1 sine rule: sin(PRT) ÷ 5.9 = sin 81° ÷ 9.48	✓ 1
A1 correct substitution (unrounded RT)	✓ 1
A1 37.9°	✓ 1
Part (e)
M1 T lies anticlockwise of P as seen from R, so bearing = 242° − 37.9°	✓ 1
A1 204°	✓ 1
Part (f)
M1 area = ½ × RT × h, so h = 2 × 24.47… ÷ 9.48…	✓ 1
A1 5.16 km	✓ 1

Common error: Bearings are measured clockwise from north: always sketch the north line at each point before finding an angle.

+ Add to test✓ In your test259IBMI3S-3-02r19-03Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 2Medium10 marks

In an online game, three servers A(2, 1), B(8, 1) and C(5, 7) are placed on a map, where coordinates are in km. Each player connects to the nearest server (straight-line distance), so the map is divided into a Voronoi diagram.

(a)

Find the equation of the perpendicular bisector of AB.

(2)

(b)

Find the equation of the perpendicular bisector of AC. Give your answer in the form y = mx + c.

(3)

(c)

Find the coordinates of the Voronoi vertex where the three cells meet.

(2)

(d)

A player is at (6, 3). Find which server she connects to and her distance from it, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AB is (5, 1) and AB is horizontal	✓ 1
A1 x = 5	✓ 1
Part (b)
M1 midpoint of AC is (3.5, 4)	✓ 1
M1 gradient of AC is 6/3 = 2, so perpendicular gradient is −½	✓ 1
A1 y = −0.5x + 5.75	✓ 1
Part (c)
M1 substitute x = 5 into the equation from (b)	✓ 1
A1 (5, 3.25)	✓ 1
Part (d)
M1 distances to A, B, C: 4.47, 2.83, 4.12 km (or squared distances 20, 8, 17)	✓ 1
A1 server B	✓ 1
A1 2.83 km	✓ 1

Exam tip: To decide which site a point belongs to, compare squared distances; you avoid square roots and rounding.

+ Add to test✓ In your test260IBMI3S-3-02r22-07Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Medium13 marks

A designer makes a conical felt hat. The circular rim has radius 7 cm and the slant height of the cone is 25 cm.

(a)

Find the curved surface area of the hat, to 3 significant figures.

(2)

(b)

Show that the vertical height of the hat is 24 cm.

(2)

(c)

Find the volume of the hat, to 3 significant figures.

(2)

(d)

The curved surface is cut from flat felt as a sector of a circle of radius 25 cm. Find the angle of the sector, to 3 significant figures.

(3)

(e)

The sector is cut from a full circular sheet of felt of radius 25 cm. Using your answer to (a), find the percentage of the sheet that is used, to 3 significant figures.

(2)

(f)

A cord is sewn round the rim, and 15% extra length is allowed for the overlap. Find the length of cord needed, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 7 × 25	✓ 1
A1 550 cm²	✓ 1
Part (b)
M1 height² = 25² − 7²	✓ 1
A1 576, so height = 24 cm	✓ 1
Part (c)
M1 ⅓ × π × 7² × 24	✓ 1
A1 1230 cm³	✓ 1
Part (d)
M1 arc length of sector = rim circumference = 2π × 7	✓ 1
M1 (θ / 360) × 2π × 25 = 2π × 7	✓ 1
A1 101° (100.8°)	✓ 1
Part (e)
M1 550 / (π × 25²) × 100 (ECF from (a))	✓ 1
A1 28.0%	✓ 1
Part (f)
M1 2π × 7 × 1.15	✓ 1
A1 50.6 cm	✓ 1

Exam tip: For a cone, the slant height is the hypotenuse of the right-angled triangle formed by the radius and the vertical height; use it for surface area but use the vertical height for volume.

+ Add to test✓ In your test261IBMI3S-3-03r37-04Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Medium14 marks

A navigation buoy is made from a cylinder of radius 0.65 m and height 0.90 m. A solid cone of height 1.40 m and base radius 0.65 m is fixed to the top of the cylinder, and a hemisphere of radius 0.65 m is fixed to the bottom of the cylinder.
The buoy is solid and symmetrical about a vertical axis, with the cone at the top, the cylinder in the middle and the hemisphere at the bottom.

(a)

Find the slant height of the cone.

(2)

(b)

Calculate the total volume of the buoy. Give your answer in m³ to 3 significant figures.

(4)

(c)

Calculate the total outer surface area of the buoy. Give your answer in m² to 3 significant figures. Use the unrounded slant height from part (a).

(4)

(d)

The buoy is filled with foam of density 85 kg m⁻³. Calculate the mass of foam in the buoy, to the nearest kilogram. Use the unrounded volume from part (b).

(2)

(e)

The outer surface of the buoy is painted with two coats of paint. One litre of paint covers 8 m², and paint is sold only in whole 1-litre cans costing £13.80 each. Find the cost of the paint. Use the unrounded area from part (c).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $l^2=0.65^2+1.40^2$	✓ 1
A1 $l=1.54$ m	✓ 1
Part (b)
M1 cylinder volume $\pi(0.65)^2(0.90)=1.19$	✓ 1
M1 cone volume $\frac{1}{3}\pi(0.65)^2(1.40)=0.619$	✓ 1
M1 hemisphere volume $\frac{2}{3}\pi(0.65)^3=0.575$	✓ 1
A1 2.39 m³	✓ 1
Part (c)
M1 curved area of cone $\pi rl$ (3.15) with $l=1.5435$	✓ 1
M1 curved area of cylinder $2\pi rh$ (3.68)	✓ 1
M1 curved area of hemisphere $2\pi r^2$ (2.65)	✓ 1
A1 9.48 m²	✓ 1
Part (d)
M1 mass = density × volume = 85 × 2.389	✓ 1
A1 203 kg	✓ 1
Part (e)
M1 $2\times9.482\div8=2.37$ litres, so 3 cans are needed	✓ 1
A1 3 × 13.80 = £41.40	✓ 1

Exam tip: A composite solid needs one formula per piece. Write the three volumes (or three curved areas) as separate lines before adding them, and do not include the flat faces where the pieces join.

+ Add to test✓ In your test262IBMI3S-3-03r37-10Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Medium12 marks

Two coastguard stations, A and B, are on a straight east–west coastline, with B 25 km due east of A. A distress signal is received from a ship at S, out at sea to the north of the coastline. The bearing of S from A is 038° and the bearing of S from B is 307°.

(a)

Show that angle ASB is 91°.

(3)

(b)

Find the distance AS. Give your answer in km to 3 significant figures.

(3)

(c)

Find the distance BS. Give your answer in km to 3 significant figures.

(2)

(d)

Find the shortest distance from S to the coastline. Give your answer in km to 3 significant figures.

(2)

(e)

A lifeboat leaves B and travels directly to S at a constant speed of 30 km h⁻¹. Find the time taken in minutes. Use the unrounded value of BS. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 angle SAB $=90^{\circ}-38^{\circ}=52^{\circ}$	✓ 1
M1 angle SBA $=307^{\circ}-270^{\circ}=37^{\circ}$	✓ 1
A1 $180^{\circ}-52^{\circ}-37^{\circ}=91^{\circ}$ (AG)	✓ 1
Part (b)
M1 sine rule $\frac{AS}{\sin37^{\circ}}=\frac{25}{\sin91^{\circ}}$	✓ 1
A1 correct substitution and rearrangement	✓ 1
A1 15.0 km	✓ 1
Part (c)
M1 $\frac{BS}{\sin52^{\circ}}=\frac{25}{\sin91^{\circ}}$	✓ 1
A1 19.7 km	✓ 1
Part (d)
M1 $AS\sin52^{\circ}$ or $BS\sin37^{\circ}$	✓ 1
A1 11.9 km	✓ 1
Part (e)
M1 time = $\frac{19.70}{30}\times60$	✓ 1
A1 39.4 minutes	✓ 1

Exam tip: A bearing is measured clockwise from north, so the due-west direction from B is 270°. Subtract bearings carefully to get the angle inside the triangle.

+ Add to test✓ In your test263IBMI3S-3-03r37-13Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Medium13 marks

A concert hall has three loudspeakers at A(2, 2), B(10, 2) and C(4, 8), where the coordinates are in metres from one corner of the stage area. The sites A, B and C are used to form a Voronoi diagram, so that each cell contains the points nearer to one speaker than to the others.

(a)

Show that the perpendicular bisector of AB has equation $x=6$.

(1)

(b)

Find the equation of the perpendicular bisector of AC. Give your answer in the form $y=mx+c$.

(3)

(c)

Find the coordinates of the Voronoi vertex where the bisectors in parts (a) and (b) meet.

(2)

(d)

A listener stands at P(8, 5). Find the distance from P to each of A, B and C, and hence state which speaker is nearest. Give distances in metres to 3 significant figures.

(3)

(e)

A fourth speaker is added at D(6, −2). Show that the perpendicular bisector of AD is $y=x-4$, and hence show that the new Voronoi diagram has a vertex at (6, 2).

(3)

(f)

The edge between the cells of A and B used to end at the vertex found in part (c), but now it ends at (6, 2) instead of continuing downwards. Find the length of this edge, between the vertices (6, 4) and (6, 2).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R1 AB is horizontal with midpoint (6, 2), so its perpendicular bisector is the vertical line $x=6$ (AG)	✓ 1
Part (b)
M1 midpoint of AC is (3, 5)	✓ 1
M1 gradient of AC is 3, so perpendicular gradient is $-\frac{1}{3}$	✓ 1
A1 $y=-\frac{1}{3}x+6$	✓ 1
Part (c)
M1 substitutes $x=6$ into the equation from part (b)	✓ 1
A1 (6, 4)	✓ 1
Part (d)
M1 uses the distance formula for at least two of the speakers	✓ 1
A1 PA = 6.71 m, PB = 3.61 m, PC = 5.00 m	✓ 1
A1 speaker B is nearest	✓ 1
Part (e)
M1 midpoint of AD is (4, 0) and gradient of AD is −1	✓ 1
A1 perpendicular bisector $y=x-4$ (AG)	✓ 1
A1 on $x=6$, $y=2$, so the vertex is (6, 2) (AG)	✓ 1
Part (f)
A1 2 m	✓ 1

Exam tip: To find a Voronoi vertex, solve the equations of two perpendicular bisectors together. When one bisector is vertical, just substitute its x-value into the other equation.

+ Add to test✓ In your test264IBMI3S-3-01r40-05Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Medium13 marks

A radio observatory has three dishes: A at (0, 0), B at (10, 0) and C at (4, 6), where coordinates are in kilometres. The region is divided into cells using a Voronoi diagram with A, B and C as sites. The three cell edges meet at a single vertex V. A village is at P(8, 3).

(a)

Find the equation of the edge between the cells of A and B.

(2)

(b)

Find the equation of the edge between the cells of A and C. Give your answer in the form $y = mx + c$.

(3)

(c)

Hence find the coordinates of V.

(2)

(d)

Calculate the distance from V to dish A, and explain what this shows about the distances from V to B and to C.

(3)

(e)

Determine which dish is nearest to the village P, giving the distance from P to each of the three dishes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AB is (5, 0) and AB is horizontal	✓ 1
A1 $x = 5$	✓ 1
Part (b)
M1 midpoint of AC is (2, 3)	✓ 1
M1 gradient of AC = $\frac{3}{2}$, so gradient of perpendicular bisector = $-\frac{2}{3}$	✓ 1
A1 $y = -\frac{2}{3}x + \frac{13}{3}$ (or equivalent, e.g. $2x + 3y = 13$)	✓ 1
Part (c)
M1 substitute $x = 5$ into the equation from part (b)	✓ 1
A1 V = (5, 1)	✓ 1
Part (d)
M1 $\sqrt{5^2 + 1^2}$	✓ 1
A1 5.10 km	✓ 1
R1 V is equidistant from A, B and C, so it is also 5.10 km from B and from C	✓ 1
Part (e)
M1 attempt to use the distance formula from P to the dishes	✓ 1
A1 PA = 8.54 km, PB = 3.61 km, PC = 5.00 km	✓ 1
A1 dish B is nearest, at 3.61 km	✓ 1

Exam tip: For a Voronoi edge between two sites, find the midpoint first, then use the negative reciprocal of the gradient joining the sites.

+ Add to test✓ In your test265IBMI3S-3-01r40-08Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Medium13 marks

A rail inspection vehicle leaves depot D and travels 18 km on a bearing of 072° to a signal point P. It then travels 25 km on a bearing of 152° to a second signal point Q.

(a)

Show that angle DPQ is 100°.

(2)

(b)

Calculate the direct distance DQ, correct to 3 significant figures.

(3)

(c)

Find the bearing of Q from D, correct to the nearest degree.

(4)

(d)

Calculate the shortest distance from P to the straight line DQ, correct to 3 significant figures.

(2)

(e)

The vehicle travels at 45 km h⁻¹. Find how many minutes would be saved by travelling directly from D to Q instead of via P. Use unrounded values and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of D from P is 072° + 180° = 252°	✓ 1
A1 252° − 152° = 100°	✓ 1
Part (b)
M1 $DQ^2 = 18^2 + 25^2 - 2(18)(25)\cos 100°$	✓ 1
M1 substitute correctly: $DQ^2 = 324 + 625 + 156.3$ (approximately 1105.3)	✓ 1
A1 33.2 km	✓ 1
Part (c)
M1 $\frac{\sin \angle PDQ}{25} = \frac{\sin 100°}{DQ}$	✓ 1
A1 angle PDQ = 47.8°	✓ 1
M1 bearing = 072° + 47.8°	✓ 1
A1 120°	✓ 1
Part (d)
M1 $18 \sin(\angle PDQ)$	✓ 1
A1 13.3 km	✓ 1
Part (e)
M1 $\frac{(18 + 25) - 33.25}{45} \times 60$	✓ 1
A1 13.0 minutes	✓ 1

Exam tip: In bearing problems, draw a north line at each turning point and mark the angles between it and the track before using a trigonometric rule.

+ Add to test✓ In your test266IBMI3S-3-03r44-10Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Medium14 marks

Three emergency shelters are built on a hillside at A(3, 3), B(10, 10) and C(12, 6) on a map in which one unit represents 1 km. The map is used to draw a Voronoi diagram, so that each point is assigned to its nearest shelter.

(a)

Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.

(3)

(b)

Find the equation of the perpendicular bisector of BC, giving your answer in the form y = mx + c.

(3)

(c)

Hence find the coordinates of the vertex V where the boundaries between all three regions meet.

(3)

(d)

Find the distance from V to shelter A, and state what this tells you about the distances from V to B and from V to C.

(2)

(e)

A village is located at P(9, 5). Determine which shelter is nearest to the village and find that distance, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AB = (6.5, 6.5)	✓ 1
M1 gradient of AB = 1, so perpendicular gradient = −1	✓ 1
A1 y = −x + 13	✓ 1
Part (b)
M1 midpoint of BC = (11, 8)	✓ 1
M1 gradient of BC = −2, so perpendicular gradient = 1/2	✓ 1
A1 y = 0.5x + 2.5	✓ 1
Part (c)
M1 equating: −x + 13 = 0.5x + 2.5	✓ 1
A1 x = 7	✓ 1
A1 y = 6, so V = (7, 6)	✓ 1
Part (d)
M1 √((7 − 3)² + (6 − 3)²)	✓ 1
A1 5 km, and V is also 5 km from B and from C (it is equidistant from all three shelters)	✓ 1
Part (e)
M1 calculate distances (or squared distances 40, 26 and 10) from P to all three shelters	✓ 1
A1 C is nearest	✓ 1
A1 3.16 km	✓ 1

Exam tip: Check a vertex by substituting it into the third bisector, or by showing its distances to the three sites are equal.

+ Add to test✓ In your test267IBMI3S-3-03r44-13Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Medium13 marks

A theatre auditorium has a vertical cylindrical wall of radius 11 m and height 6 m, topped by a hemispherical dome of radius 11 m. The floor is flat.

(a)

Calculate the volume of air in the auditorium. Give your answer to the nearest m³.

(3)

(b)

The curved wall and the dome are to be painted. Find the total area to be painted, to the nearest m².

(3)

(c)

The dome contains a flat circular skylight of radius 1.5 m. Find the area of the skylight as a percentage of the curved area of the dome. Give your answer to 3 significant figures. (The paint calculation below ignores the skylight.)

(2)

(d)

Two coats of paint are applied. One litre of paint covers 8 m² for one coat. Paint is sold in 5 litre tins at £34.50 per tin. Find the minimum cost of the paint needed for the wall and dome. Use the unrounded area from part (b) and give your answer in £.

(3)

(e)

A ventilation system removes air at a rate of 3.8 m³ per second. Find how many minutes it takes to remove a volume of air equal to the full volume in (a). Use the unrounded volume and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cylinder volume = π(11)²(6) = 2280.8… m³	✓ 1
M1 hemisphere volume = (2/3)π(11)³ = 2787.6… m³	✓ 1
A1 5068 m³	✓ 1
Part (b)
M1 dome area = 2π(11)² = 760.3… m²	✓ 1
M1 wall area = 2π(11)(6) = 414.7… m²	✓ 1
A1 1175 m² (1174.96…)	✓ 1
Part (c)
M1 π(1.5)²/(2π(11)²) × 100	✓ 1
A1 0.930%	✓ 1
Part (d)
M1 litres needed = 2 × 1174.96…/8 = 293.7… litres	✓ 1
M1 tins = 293.7…/5 = 58.7…, rounded up to 59 tins	✓ 1
A1 cost = 59 × 34.50 = £2035.50	✓ 1
Part (e)
M1 time = 5068.4…/3.8 = 1333.8… s, then divide by 60	✓ 1
A1 22.2 minutes	✓ 1

Exam tip: When a quantity such as tins of paint must be a whole number, round up at the end of the calculation, not earlier.

+ Add to test✓ In your test268IBMI3S-3-01r46-05Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Medium16 marks

A beehive consists of a cuboid body 46 cm wide, 38 cm deep and 52 cm high, topped by a roof in the shape of a triangular prism. The cross-section of the roof is an isosceles triangle with a horizontal base of 46 cm, whose two sloping sides each make an angle of 22° with the horizontal. The roof is 38 cm long, the same as the depth of the body. Use unrounded values in later parts and give final answers to 3 significant figures unless told otherwise.

(a)

Find the height of the roof above the top of the cuboid body.

(2)

(b)

Calculate the total volume of the hive, in cm³.

(3)

(c)

Find the length of one sloping side of the roof's triangular cross-section.

(2)

(d)

The four vertical walls, the two sloping roof panels and the two triangular gable ends are to be painted. Calculate the total area to be painted, in cm².

(3)

(e)

One tin of paint covers 0.50 m². Find the minimum number of tins that must be bought.

(2)

(f)

A straight rod is placed inside the cuboid body, running from one bottom corner to the opposite top corner. Calculate the length of the longest such rod.

(2)

(g)

Calculate the angle the rod in part (f) makes with the base of the hive.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 22° = h / 23 (half of the base is 23 cm)	✓ 1
A1 h = 9.29 cm	✓ 1
Part (b)
M1 volume of cuboid = 46 × 38 × 52 = 90896 cm³	✓ 1
M1 volume of prism = ½ × 46 × 9.29 × 38 = 8122 cm³	✓ 1
A1 total volume = 99018 cm³, so 99 000 cm³ to 3 s.f.	✓ 1
Part (c)
M1 cos 22° = 23 / s	✓ 1
A1 s = 24.81 cm	✓ 1
Part (d)
M1 vertical walls: 2 × (46 + 38) × 52 = 8736 cm²	✓ 1
M1 roof panels: 2 × 24.81 × 38 = 1885 cm² and gables: 2 × ½ × 46 × 9.29 = 427 cm²	✓ 1
A1 total area = 11049 cm², so 11 000 cm² to 3 s.f.	✓ 1
Part (e)
M1 area in m² = 1.105 (using 1 m² = 10 000 cm²) and divided by 0.5 gives 2.21	✓ 1
A1 3 tins (rounded up to a whole number)	✓ 1
Part (f)
M1 space diagonal = √(46² + 38² + 52²)	✓ 1
A1 length = 79.1 cm	✓ 1
Part (g)
M1 base diagonal = √(46² + 38²) = 59.67 cm and angle = tan⁻¹(52 / 59.67)	✓ 1
A1 angle = 41.1°	✓ 1

Common error: Painting the roof panels with the area of the triangle's base instead of the sloping length; the slope length must be found first.

+ Add to test✓ In your test269IBMI3S-3-01r46-12Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Medium14 marks

Two observation posts, P and Q, are on level ground 6.00 km apart, with Q due east of P. Both posts see a swarm of bees at S. The bearing of S from P is 038° and the bearing of S from Q is 311°. Use unrounded values in later parts and give lengths to 3 significant figures.

(a)

Find the sizes of angles SPQ and SQP.

(2)

(b)

Find the size of angle PSQ.

(1)

(c)

Calculate the distance PS, in km.

(2)

(d)

Calculate the distance QS, in km.

(2)

(e)

Find the shortest distance from S to the straight line PQ.

(2)

(f)

Find the bearing of Q from S.

(2)

(g)

A drone flies from P to S and then from S to Q at a constant speed of 12 km h⁻¹. Calculate the time taken, in minutes, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 angle SPQ = 090° − 038° = 52° (the bearing of Q from P is 090°)	✓ 1
A1 angle SQP = 311° − 270° = 41° (the bearing of P from Q is 270°)	✓ 1
Part (b)
A1 angle PSQ = 180° − 52° − 41° = 87°	✓ 1
Part (c)
M1 PS / sin 41° = 6 / sin 87°	✓ 1
A1 PS = 3.94 km	✓ 1
Part (d)
M1 QS / sin 52° = 6 / sin 87°	✓ 1
A1 QS = 4.73 km	✓ 1
Part (e)
M1 distance = PS × sin 52° (or QS × sin 41°)	✓ 1
A1 distance = 3.11 km	✓ 1
Part (f)
M1 uses alternate angles or the bearing of S from Q: 311° − 180°	✓ 1
A1 bearing = 131°	✓ 1
Part (g)
M1 total distance = PS + QS = 8.68 km	✓ 1
M1 time = 8.68 / 12 × 60	✓ 1
A1 time = 43.4 minutes	✓ 1

Common error: Writing a bearing as a two-digit number or measuring it anticlockwise. Bearings are three figures, measured clockwise from north.

+ Add to test✓ In your test270IBMI-032-00-11Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Hard13 marks

A rescue helicopter is based at B. It receives a call from a walker at P, which is 42 km from B on a bearing of 310°. A second call comes from a climber at Q, which is 35 km from B on a bearing of 025°.

(a)

Show that angle PBQ = 75°.

(1)

(b)

Find the distance PQ.

(3)

(c)

Find angle BPQ.

(3)

(d)

Find the bearing of Q from P.

(3)

(e)

The helicopter flies B → P → Q → B at a constant 210 km h⁻¹ and hovers for 12 minutes at each of P and Q. It can stay airborne for at most 60 minutes before refuelling. Determine whether the helicopter can complete the mission without refuelling.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(360° − 310°) + 25° = 75° A1	✓ 1
Part (b)
PQ² = 42² + 35² − 2(42)(35)cos 75° M1	✓ 1
PQ² = 2228 (2228.07…) A1	✓ 1
PQ = 47.2 km A1	✓ 1
Part (c)
sin BPQ / 35 = sin 75° / 47.2 (or the cosine rule) M1	✓ 1
sin BPQ = 0.716 A1	✓ 1
BPQ = 45.7° (acute, since BQ < BP) A1	✓ 1
Part (d)
Bearing of B from P = 310° − 180° = 130° M1	✓ 1
Q lies anticlockwise of the line PB as seen from P, so subtract: 130° − 45.7° (or an equivalent valid method, e.g. coordinates) M1	✓ 1
= 084.3° A1	✓ 1
Part (e)
Total distance = 42 + 47.2 + 35 = 124.2 km; flying time = 124.2 ÷ 210 h (= 35.5 minutes) M1	✓ 1
Total time = 35.5 + 24 = 59.5 minutes A1	✓ 1
59.5 minutes < 60 minutes, so yes, it can complete the mission without refuelling R1	✓ 1

Exam tip: When a comparison is close, keep at least 3 significant figures in your total before concluding. State the comparison explicitly, e.g. 59.5 < 60.

+ Add to test✓ In your test271IBMI-032-00-12Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Hard12 marks

A vertical telecommunications mast FT stands on horizontal ground, with foot F and top T. Two engineers stand at points A and B on the same horizontal ground, where B is 180 m due east of A. The mast is on a bearing of 032° from A, and angle FBA = 71°. From A, the angle of elevation of T is 26°.

(a)

Show that angle AFB = 51°.

(2)

(b)

Find the distance AF.

(2)

(c)

Find the height of the mast.

(2)

(d)

A drone flies in a straight line from A to T. Find the distance it flies.

(2)

(e)

Find the angle of elevation of T from B.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
AB points due east, so it is perpendicular to north. Hence angle FAB = 90° − 32° = 58° A1	✓ 1
Angle AFB = 180° − 58° − 71° = 51° A1	✓ 1
Part (b)
$\frac{AF}{\sin 71°} = \frac{180}{\sin 51°}$ M1	✓ 1
AF = 219 m (219.0) A1	✓ 1
Part (c)
FT = 219.0 tan 26° M1	✓ 1
= 107 m (106.8) A1	✓ 1
Part (d)
AT = 219.0 / cos 26° (or √(219.0² + 106.8²)) M1	✓ 1
= 244 m A1	✓ 1
Part (e)
$\frac{BF}{\sin 58°} = \frac{180}{\sin 51°}$ (or the cosine rule in triangle ABF) M1	✓ 1
BF = 196 m (196.4) A1	✓ 1
tan θ = 106.8 / 196.4 M1	✓ 1
θ = 28.5° A1	✓ 1

Common error: Triangle ABF lies flat on the ground, and each angle of elevation belongs to a separate vertical triangle. Do not mix horizontal and vertical angles in the same sine rule. Carry unrounded values forward to avoid rounding errors in (e).

+ Add to test✓ In your test272IBMI-032-06-12Volume and surface area in context Paper 2Hard10 marks

A drinks company is designing a closed cylindrical can that must hold 330 cm³. The can has radius r cm and height h cm.

(a)

Write down an expression for h in terms of r.

(1)

(b)

Show that the total surface area of the can is $A=2\pi r^2+\frac{660}{r}$.

(2)

(c)

Use your graphic display calculator to find the value of r that minimises A, and the minimum surface area.

(3)

(d)

In the factory, the curved surface is cut from a rectangular sheet with no waste. Each circular end is cut from a square sheet of side 2r, and the rest of each square is scrapped. For the can with minimum surface area, find the total area of metal needed, including the scrap.

(2)

(e)

Find the percentage of this metal that is scrapped.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 for $h=\frac{330}{\pi r^2}$	✓ 1
Part (b)
M1 for $A=2\pi r^2+2\pi rh$	✓ 1
A1 for substituting $h$ to give $2\pi r\times\frac{330}{\pi r^2}=\frac{660}{r}$, and hence the given result	✓ 1
Part (c)
M1 for sketching the graph of A on a GDC, or for any valid method of finding its minimum point	✓ 1
A1 for $r=3.74$ cm (3.7449…)	✓ 1
A1 for $A=264$ cm² (264.36…)	✓ 1
Part (d)
M1 for $2(2r)^2+\frac{660}{r}$, i.e. $8r^2+\frac{660}{r}$, with $r=3.7449…$	✓ 1
A1 for 288 cm² (288.43…)	✓ 1
Part (e)
M1 for a scrap area of $2r^2(4-\pi)=24.08$, divided by 288.43 and multiplied by 100	✓ 1
A1 for 8.35%	✓ 1

Exam tip: Store the unrounded value of r from your GDC and reuse it in the scrap calculations. Using r = 3.74 instead gives 8.33%, which changes the third significant figure of the final percentage.

+ Add to test✓ In your test273IBMI-032-05-07Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Hard10 marks

A park ranger is placing a marker post C on a straight footpath that starts at gate A. The footpath makes an angle of 38° with the straight fence AB, where AB = 340 m. The post must be exactly 250 m from B.

(a)

Find the two possible values of angle ACB.

(3)

(b)

Hence find the two possible distances AC.

(4)

(c)

The post is placed at the position closer to A. Find the area of triangle ABC.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
sin C / 340 = sin 38° / 250 M1 ; ∠ACB = 56.9° A1 ; or ∠ACB = 180° − 56.9° = 123.1° A1	✓ 3
Part (b)
∠ABC = 180° − 38° − 56.86° = 85.1° or 180° − 38° − 123.14° = 18.9° M1 ; AC / sin B = 250 / sin 38° (or a correct cosine-rule quadratic in AC) M1 ; AC = 405 m A1 ; AC = 131 m A1	✓ 4
Part (c)
Selects AC = 131.24 m, the shorter distance R1 ; Area = ½ × 340 × 131.24 × sin 38° M1 ; 13 700 m² (1.37 × 10⁴ m²) A1	✓ 3

Exam tip: When the sine rule is used to find an angle and the given side opposite the known angle is shorter than the other given side, check whether the obtuse angle 180° − θ also gives a valid triangle.

+ Add to test✓ In your test274IBMI-032-05-12Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Hard16 marks

An orienteering loop starts at S. Runners go 1.2 km on a bearing of 040° to checkpoint P, then 0.9 km on a bearing of 160° to checkpoint Q, and then run straight back to S.

(a)

Show that angle SPQ = 60°.

(2)

(b)

Find the distance SQ.

(3)

(c)

Find the bearing on which runners travel from Q back to S.

(4)

(d)

Find the area of ground enclosed by the loop.

(2)

(e)

A water station W is to be placed on the path SQ so that it is as close as possible to P. Find the distance PW.

(3)

(f)

One runner completes the whole loop in 22 minutes. Find her average speed in km h⁻¹.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Bearing of S from P = 040° + 180° = 220° M1 ; ∠SPQ = 220° − 160° = 60° A1	✓ 2
Part (b)
Use of the cosine rule M1 ; SQ² = 1.2² + 0.9² − 2(1.2)(0.9)cos 60° = 1.17 A1 ; SQ = 1.08 km (1.0817…) A1	✓ 3
Part (c)
sin(∠PSQ) / 0.9 = sin 60° / 1.0817 M1 ; ∠PSQ = 46.1° A1 ; bearing of Q from S = 040° + 46.1° = 086.1°, so the bearing of S from Q = 086.1° + 180° M1 ; 266° A1	✓ 4
Part (d)
½ × 1.2 × 0.9 × sin 60° M1 ; 0.468 km² A1	✓ 2
Part (e)
PW is perpendicular to SQ R1 ; PW = 1.2 sin 46.10° (or 2 × area / SQ) M1 ; 0.865 km A1	✓ 3
Part (f)
(1.2 + 0.9 + 1.0817) / (22/60) M1 ; 8.68 km h⁻¹ A1	✓ 2

Exam tip: The shortest distance from a point to a straight path is along the perpendicular. Spotting this lets you use the right-angled triangle SPW instead of setting up a harder method.

+ Add to test✓ In your test275IBMI-032-01-11Volume and surface area in context Paper 2Hard14 marks

A rectangular swimming pool is 25 m long and 10 m wide. The depth of the pool is 1.2 m at the shallow end and 2.8 m at the deep end, and the floor slopes uniformly along the length of the pool. The side walls are vertical trapeziums and the end walls are vertical rectangles.

(a)

Find the volume of the pool.

(2)

(b)

Find the length of the sloping floor, from the shallow end to the deep end.

(2)

(c)

Find the angle that the floor makes with the horizontal.

(2)

(d)

The floor and the four walls are to be tiled. Find the total area to be tiled.

(3)

(e)

Tiles cost 34.50 dollars per m², and 10% extra must be bought to allow for breakages. Find the cost of the tiles, to the nearest dollar.

(2)

(f)

The empty pool is filled until the water surface is 10 cm below the top of the walls. Water flows in at 0.6 m³ per minute. Find the time taken, in hours.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Cross-section area = ½(1.2 + 2.8)(25) = 50 m², or an equivalent method ; volume = 500 m³	✓ 2
Part (b)
$\sqrt{25^2+1.6^2}$ ; 25.1 m (25.051…)	✓ 2
Part (c)
tan θ = 1.6/25, or an equivalent trigonometric ratio using the answer to (b) ; θ = 3.66°	✓ 2
Part (d)
Floor = 25.051… × 10 = 250.5 m² ; side walls 2 × 50 = 100 m² and end walls 12 + 28 = 40 m² ; total = 391 m² (390.51…)	✓ 3
Part (e)
390.51… × 1.1 × 34.50 ; 14 820 dollars (accept 14 838 dollars from using 391 m²)	✓ 2
Part (f)
Unfilled layer = 25 × 10 × 0.1 = 25 m³, so water volume = 475 m³ ; 475 ÷ 0.6 = 791.7 minutes ; 13.2 hours	✓ 3

Exam tip: The water surface is horizontal, so the empty top layer is a 25 m × 10 m × 0.1 m cuboid, even though the floor slopes. Spotting this saves a lot of work.

+ Add to test✓ In your test276IBMI-032-01-12Volume and surface area in context Paper 2Hard12 marks

A conical funnel in a laboratory is held with its vertex pointing downwards and its outlet closed. The funnel has rim radius 40 cm and vertical height 60 cm. When the water is h cm deep, the volume of water is V cm³.

(a)

Find the capacity of the funnel in litres.

(2)

(b)

Show that $V=\frac{4\pi h^3}{27}$.

(2)

(c)

Find the depth of water when the funnel is half full.

(3)

(d)

Find the area of the inner surface of the funnel that is in contact with the water when it is half full.

(3)

(e)

The outlet is opened when the funnel is full, and water drains at a constant 250 cm³ per second. Find the time taken for the funnel to become half full.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\frac{1}{3}\pi(40^2)(60)$ = 100 531 cm³ M1 ; 101 litres A1	✓ 2
Part (b)
Similar triangles give radius of water surface r = $\frac{40}{60}h=\frac{2h}{3}$ M1 ; $V=\frac{1}{3}\pi\left(\frac{2h}{3}\right)^2h=\frac{4\pi h^3}{27}$ A1	✓ 2
Part (c)
$\frac{4\pi h^3}{27}=16000\pi$ M1 ; h³ = 108 000 M1 ; h = 47.6 cm (47.622…) A1	✓ 3
Part (d)
r = $\frac{2}{3}$(47.622…) = 31.75 cm M1 ; slant height $\sqrt{47.622^2+31.748^2}$ = 57.23 cm M1 ; πrl = 5710 cm² (5708.5…) A1	✓ 3
Part (e)
50 265.5 ÷ 250 M1 ; 201 seconds A1	✓ 2

Common error: Half the volume does not mean half the depth. Because V is proportional to h³, a half-full cone is filled to about 79% of its height.

+ Add to test✓ In your test277IBMI-032-02-r1-06Voronoi diagrams Paper 2Hard14 marks

A town is modelled by the rectangle $0 \le x \le 10$, $0 \le y \le 8$, where 1 unit represents 1 km. There are three fire stations, at A(2, 2), B(8, 2) and C(5, 6). Each point in the town is served by its nearest fire station, and a Voronoi diagram is constructed with the stations as sites.

(a)

Write down the equation of the perpendicular bisector of AB.

(1)

(b)

Show that the perpendicular bisector of AC has equation $y = -0.75x + 6.625$.

(3)

(c)

Find the coordinates of the Voronoi vertex V.

(2)

(d)

Find the area of the town served by station A.

(3)

(e)

Find the percentage of the town served by station C.

(2)

(f)

A council officer claims that V is the point in the town that is furthest from its nearest fire station. By considering the corner of the town at (0, 8), show that this claim is incorrect.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$x = 5$ A1	✓ 1
Part (b)
Gradient AC = $\frac{4}{3}$, so perpendicular gradient = $-\frac{3}{4}$ M1	✓ 1
Midpoint of AC = (3.5, 4), substituted: $y - 4 = -0.75(x - 3.5)$ M1	✓ 1
$y = -0.75x + 2.625 + 4 = -0.75x + 6.625$ A1	✓ 1
Part (c)
Substituting $x = 5$: $y = -3.75 + 6.625$ M1	✓ 1
V(5, 2.875) A1	✓ 1
Part (d)
Bisector of AC meets $x = 0$ at (0, 6.625) A1	✓ 1
Cell is the trapezium (0, 0), (5, 0), (5, 2.875), (0, 6.625): area = $\frac{6.625 + 2.875}{2} \times 5$ M1	✓ 1
= 23.75 km² (23.8 km²) A1	✓ 1
Part (e)
By symmetry B also serves 23.75 km², so C serves 80 − 2 × 23.75 = 32.5 km² M1	✓ 1
$\frac{32.5}{80} \times 100$ = 40.6% A1	✓ 1
Part (f)
Distance from (0, 8) to A = $\sqrt{40}$ = 6.32 km and to C = $\sqrt{29}$ M1	✓ 1
Nearest station to (0, 8) is C, at 5.39 km A1	✓ 1
Distance from V to its nearest station is 3.125 km; since 5.39 > 3.125 the corner is further from help than V, so the claim is incorrect R1	✓ 1

Exam tip: In a bounded region, the point furthest from every site can lie on the boundary or at a corner, not only at a Voronoi vertex. Check corners and edge intersections as well.

+ Add to test✓ In your test278IBMI-032-07-r2-05Voronoi diagrams Paper 2Hard12 marks

A rectangular county is modelled by the region $0 \le x \le 10$, $0 \le y \le 8$, with distances in km. Three towns lie at A(2, 2), B(8, 2) and C(5, 7). A waste-recycling plant is to be built in the county as far as possible from the nearest town.

(a)

Write down the equation of the perpendicular bisector of AB.

(1)

(b)

Show that the perpendicular bisector of AC has equation $y = -0.6x + 6.6$.

(3)

(c)

Find the coordinates of the Voronoi vertex V, the point equidistant from all three towns.

(2)

(d)

Find the distance from V to the nearest town.

(2)

(e)

A planner claims that V is the best location for the plant. By considering the corners of the county, show that this claim is incorrect. State a better location and its distance from the nearest town.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$x = 5$ A1	✓ 1
Part (b)
Midpoint of AC is (3.5, 4.5) A1	✓ 1
Gradient AC $= \frac{5}{3}$, so perpendicular gradient $= -0.6$ M1	✓ 1
$y - 4.5 = -0.6(x - 3.5)$ leading to $y = -0.6x + 6.6$ A1	✓ 1
Part (c)
Substitutes $x = 5$ into $y = -0.6x + 6.6$ M1	✓ 1
V(5, 3.6) A1	✓ 1
Part (d)
$\sqrt{3^2 + 1.6^2}$ or $\sqrt{11.56}$, or $7 - 3.6$ (distance to C) M1	✓ 1
3.4 km A1	✓ 1
Part (e)
Considers a top corner, e.g. (0, 8), and identifies C as its nearest town (distance to A is 6.32 km, distance to B is 10.0 km) M1	✓ 1
Distance from (0, 8) to C $= \sqrt{26} = 5.10$ km A1	✓ 1
5.10 > 3.4, so V is not the furthest point from the towns and the claim is incorrect R1	✓ 1
Plant at (0, 8) or (10, 8), 5.10 km from the nearest town A1	✓ 1

Exam tip: For a largest-empty-circle problem in a bounded region, test the Voronoi vertices, the corners of the region and the points where Voronoi edges meet the boundary. The optimum is often on the boundary.

+ Add to test✓ In your test279IBMI-032-02-r2-06Voronoi diagrams Paper 2Hard15 marks

A county is modelled by the rectangle $0 \le x \le 12$, $0 \le y \le 8$, with units in kilometres. Three towns are located at A(2, 2), B(10, 2) and C(5, 7). A waste incinerator is to be built anywhere in the county, including on its boundary. It must be at the point that is as far as possible from the nearest town.

(a)

Show that the perpendicular bisector of AC has equation $3x + 5y = 33$.

(3)

(b)

Find the equations of the perpendicular bisectors of AB and BC.

(2)

(c)

Find the coordinates of the vertex V of the Voronoi diagram and the distance from V to each town.

(3)

(d)

Find the coordinates of the three points where the edges of the Voronoi diagram meet the boundary of the county.

(3)

(e)

Determine the optimal location for the incinerator and state its distance from the nearest town. Justify your answer.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint of AC = (3.5, 4.5)	✓ 1
Gradient AC = $\frac{5}{3}$, so perpendicular gradient = $-\frac{3}{5}$	✓ 1
$y - 4.5 = -0.6(x - 3.5)$ leading to $5y = -3x + 33$, i.e. $3x + 5y = 33$	✓ 1
Part (b)
AB: $x = 6$	✓ 1
BC: midpoint (7.5, 4.5), perpendicular gradient 1, giving $y = x - 3$	✓ 1
Part (c)
Solves $x = 6$ with $3x + 5y = 33$ (or with $y = x - 3$)	✓ 1
V(6, 3)	✓ 1
Distance = $\sqrt{17}$ ≈ 4.12 km	✓ 1
Part (d)
(6, 0)	✓ 1
(11, 8)	✓ 1
(0, 6.6)	✓ 1
Part (e)
Within each Voronoi cell, the distance to that cell's town is greatest at a vertex of the cell. The candidates are therefore V, the points in (d) and the corners of the rectangle	✓ 1
Distances to the nearest town: V 4.12; (6, 0) 4.47; (11, 8) 6.08; (0, 6.6) 5.02; (0, 0) 2.83; (12, 0) 2.83; (0, 8) 5.10; (12, 8) 6.32	✓ 1
Optimal location is the corner (12, 8)	✓ 1
Distance to the nearest town (B) = $\sqrt{40}$ ≈ 6.32 km	✓ 1

Common error: The Voronoi vertex is not automatically the answer to a largest empty circle problem. The points where edges meet the boundary, and the corners of the region, must also be checked.

+ Add to test✓ In your test280IBMI3S-3-05-11Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard14 marks

A pop-up market tent is a square-based pyramid. The square base ABCD has sides of 6.0 m and the apex V is 4.0 m vertically above the centre O of the base.

(a)

Calculate the length of the diagonal AC.

(2)

(b)

Find the angle between the edge VA and the base.

(3)

(c)

Find angle AVB.

(3)

(d)

Calculate the volume of air enclosed by the tent.

(2)

(e)

Calculate the total area of the four triangular faces.

(2)

(f)

Fabric costs $8.40 per m² and 12% extra must be bought to allow for waste. Find the cost of the fabric.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $AC=\sqrt{6^2+6^2}$	✓ 1
A1 $8.49$ m	✓ 1
Part (b)
M1 identifies right-angled triangle VOA with $AO=4.24$ m	✓ 1
M1 $\tan\theta=\dfrac{4}{4.243}$	✓ 1
A1 $43.3^\circ$	✓ 1
Part (c)
M1 $VA=\sqrt{4.243^2+4^2}=5.83$ m (= VB)	✓ 1
M1 cosine rule in triangle AVB: $\cos V=\dfrac{5.83^2+5.83^2-6^2}{2\times5.83^2}$	✓ 1
A1 $61.9^\circ$	✓ 1
Part (d)
M1 $\frac13\times6^2\times4$	✓ 1
A1 $48$ m³	✓ 1
Part (e)
M1 slant height of a face $\sqrt{3^2+4^2}=5$ m	✓ 1
A1 $4\times\frac12\times6\times5=60$ m²	✓ 1
Part (f)
M1 $60\times1.12\times8.40$	✓ 1
A1 $564.48	✓ 1

Common error: The angle between a sloping edge and the base is measured to the line from the foot of the edge to the centre of the base, not to a side of the base.

+ Add to test✓ In your test281IBMI3S-3-05-12Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard12 marks

A wind turbine stands on level ground. Two observers A and B stand in a straight line with the base of the turbine, with B 45 m further away than A. The angle of elevation of the top T of the turbine is 28° from A and 19° from B. Ignore the height of the observers' eyes. Let the height of T above the ground be h metres and the distance from A to the base be x metres.

(a)

Write down two expressions for h: one in terms of x and 28°, the other in terms of x and 19°.

(2)

(b)

Show that $h=\dfrac{45}{\frac{1}{\tan19^\circ}-\frac{1}{\tan28^\circ}}$.

(3)

(c)

Calculate h.

(2)

(d)

A cable runs in a straight line from T to A. Find its length.

(2)

(e)

Observer B walks towards the turbine until the angle of elevation of T is 40°. Find the distance B walks.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $h=x\tan28^\circ$	✓ 1
A1 $h=(x+45)\tan19^\circ$	✓ 1
Part (b)
M1 $x=\dfrac{h}{\tan28^\circ}$ and $x+45=\dfrac{h}{\tan19^\circ}$	✓ 1
M1 eliminates $x$: $\dfrac{h}{\tan19^\circ}-\dfrac{h}{\tan28^\circ}=45$	✓ 1
A1 factorises and rearranges to the given result (AG)	✓ 1
Part (c)
M1 substitutes values	✓ 1
A1 $h=44.0$ m	✓ 1
Part (d)
M1 $\dfrac{h}{\sin28^\circ}$	✓ 1
A1 $93.7$ m	✓ 1
Part (e)
M1 original distance of B from base $=\dfrac{h}{\tan28^\circ}+45=127.7$ m	✓ 1
M1 new distance $=\dfrac{h}{\tan40^\circ}=52.4$ m	✓ 1
A1 $75.3$ m	✓ 1

Exam tip: When two angles of elevation are taken from points on one line, write the horizontal distance to each point in terms of the height and eliminate the unknown distance.

+ Add to test✓ In your test282IBMI3S-3-05b-14Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard13 marks

A roof is a right pyramid VABCD on a horizontal rectangular base ABCD with AB = 12 m and BC = 8 m. The apex V is 5 m vertically above the centre O of the base.

(a)

Show that AO = √52 m and find the length of edge VA.

(3)

(b)

Find the angle between edge VA and the base.

(2)

(c)

Use the cosine rule in triangle VAB to find angle AVB.

(3)

(d)

Hence find the area of triangular face VAB.

(2)

(e)

The roof is to be covered with tiles costing $85 per m². Find the cost of covering the four triangular faces.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 AC = √(12² + 8²) so AO = ½√208 = √52 (AG)	✓ 1
M1 attempt to use Pythagoras in triangle VOA	✓ 1
A1 VA = √77 = 8.77 m	✓ 1
Part (b)
M1 attempt to use tan⁻¹(5/√52)	✓ 1
A1 34.7°	✓ 1
Part (c)
M1 cos AVB = (77 + 77 − 144)/(2 × √77 × √77)	✓ 1
A1 correct substitution	✓ 1
A1 angle AVB = 86.3°	✓ 1
Part (d)
M1 ½ × VA × VB × sin(AVB)	✓ 1
A1 38.4 m²	✓ 1
Part (e)
M1 area of face VBC found, for example ½(77)sin 54.2° (or via the slant height √61)	✓ 1
A1 total area = 139 m²	✓ 1
A1 cost = $11842.16 (accept $11800 to 3 s.f.)	✓ 1

Exam tip: The two pairs of faces of a rectangular pyramid have different areas, so find both before doubling.

+ Add to test✓ In your test283IBMI3S-3-01r6-12Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Hard14 marks

Three swimming pools in a town are at P(2, 2), Q(10, 4) and R(6, 10) on a map where each unit is 1 km. A Voronoi diagram is drawn for the three pools.

(a)

Find the equation of the perpendicular bisector of PQ.

(3)

(b)

Find the equation of the perpendicular bisector of PR.

(3)

(c)

Find the coordinates of the Voronoi vertex where these two edges meet, giving each coordinate to 3 significant figures.

(3)

(d)

A resident lives at (5, 6). Find which pool is nearest.

(2)

(e)

A fourth pool S is built at (6, 3). Show that the vertex found in part (c), using its unrounded coordinates, is now closer to S than to P, and state the distance from the vertex to S, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of PQ is (6, 3)	✓ 1
M1 gradient of PQ $= \frac{2}{8} = \frac14$, so perpendicular gradient is $-4$	✓ 1
A1 $y = -4x + 27$	✓ 1
Part (b)
M1 midpoint of PR is (4, 6)	✓ 1
M1 gradient of PR is 2, so perpendicular gradient is $-\frac12$	✓ 1
A1 $y = -\frac12 x + 8$	✓ 1
Part (c)
M1 equate $-4x + 27 = -\frac12 x + 8$	✓ 1
A1 $x = \frac{38}{7} = 5.43$	✓ 1
A1 $y = \frac{37}{7} = 5.29$	✓ 1
Part (d)
M1 distances to P, Q and R found, $5$, $\sqrt{29}$ and $\sqrt{17}$	✓ 1
A1 R is nearest (distance 4.12 km)	✓ 1
Part (e)
M1 distance from vertex to P found, $\sqrt{\left(\frac{38}{7}-2\right)^2 + \left(\frac{37}{7}-2\right)^2} = 4.75$ km	✓ 1
M1 distance from vertex to S found using $\left(\frac{38}{7}, \frac{37}{7}\right)$ and (6, 3)	✓ 1
A1 distance to S = 2.36 km, which is less than 4.75 km	✓ 1

Common error: Do not use rounded coordinates when checking which site is nearer to a vertex; the distances to the competing sites can be very close.

+ Add to test✓ In your test284IBMI3S-3-01r6-13Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Hard13 marks

A swimming pool is 25 m long and 10 m wide. The depth at the shallow end is 1.2 m and the depth at the deep end is 2.8 m. The floor slopes uniformly between the two ends, so each long side wall is a trapezium and each end wall is a rectangle. Use unrounded values in later parts.

(a)

Calculate the volume of the pool when full.

(3)

(b)

Find the angle between the floor and the horizontal, to 3 significant figures.

(2)

(c)

The two long side walls, the two end walls and the floor are to be tiled. Find the total area to be tiled, to 4 significant figures.

(4)

(d)

The pool is partly emptied until 50 m³ of water remain. The water then does not reach the shallow end. Let the depth of water at the deep end be $d$ metres. Show that the water has a triangular cross-section and find $d$.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area of a long side as a trapezium $\frac12(1.2 + 2.8)\times 25$	✓ 1
A1 $50$ m²	✓ 1
A1 volume = $50 \times 10 = 500$ m³	✓ 1
Part (b)
M1 $\tan\theta = \frac{2.8 - 1.2}{25}$	✓ 1
A1 3.66°	✓ 1
Part (c)
M1 length of the sloping floor $= \sqrt{25^2 + 1.6^2}$	✓ 1
A1 floor area $= 10\times$ floor length	✓ 1
M1 sides $2 \times 50$ and ends $10 \times 1.2 + 10 \times 2.8$	✓ 1
A1 390.5 m²	✓ 1
Part (d)
M1 floor rises 1.6 m over 25 m, so the water surface meets the floor at horizontal distance $\frac{25d}{1.6}$ from the deep end	✓ 1
M1 volume $= \frac12 \times \frac{25d}{1.6} \times d \times 10$	✓ 1
M1 equate to 50 and solve, $d^2 = 0.64$	✓ 1
A1 $d = 0.8$ m, which is less than 1.6 m, consistent with the water not reaching the shallow end	✓ 1

Exam tip: In a partly filled prism with a sloping floor, the water forms a similar triangle to the whole slope; use the ratio of rise to run to link depth and length.

+ Add to test✓ In your test285IBMI3S-3-01r6-14Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Hard12 marks

A coastguard station A and a marker buoy B are 8.4 km apart. A vessel V is detected on a line from A that makes an angle of 40° with AB. Radar shows that the vessel is 6.3 km from the buoy. Use unrounded values in later parts.

(a)

Show that there are two possible values for angle AVB, and find both, to 1 decimal place.

(4)

(b)

Find the two possible distances AV, to 3 significant figures.

(4)

(c)

For the smaller value of AV, find the area of triangle ABV, to 3 significant figures.

(2)

(d)

Find the distance between the two possible positions of the vessel, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sine rule $\frac{\sin \text{AVB}}{8.4} = \frac{\sin 40°}{6.3}$	✓ 1
A1 $\sin\text{AVB} = 0.8571$	✓ 1
A1 acute value 59.0°	✓ 1
R1 the obtuse value $180° - 59.0° = 121.0°$ also satisfies the equation	✓ 1
Part (b)
M1 angle ABV found in each case using the angle sum of a triangle	✓ 1
A1 angle ABV = 81.0° and 19.0°	✓ 1
M1 $AV = \frac{6.3\sin(\text{ABV})}{\sin 40°}$ for each case	✓ 1
A1 AV = 9.68 km and 3.19 km	✓ 1
Part (c)
M1 $\frac12 (8.4)(AV)\sin 40°$ using the unrounded smaller distance	✓ 1
A1 8.61 km²	✓ 1
Part (d)
M1 subtract the two unrounded values of AV	✓ 1
A1 6.49 km	✓ 1

Common error: When the sine rule is used to find an angle, always test whether $180°$ minus the calculator value also fits the data; here the side lengths allow both.

+ Add to test✓ In your test286IBMI3S-3-02r8-04Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 2Hard12 marks

Three swimming pools are at A(3, 2), B(9, 2) and C(5, 8) in a town occupying the rectangle 0 ≤ x ≤ 12, 0 ≤ y ≤ 10, with coordinates in km. A Voronoi diagram models which pool is nearest to each point of the town.

(a)

Find the equation of the perpendicular bisector of AB.

(2)

(b)

Find the equation of the perpendicular bisector of AC.

(3)

(c)

Find the coordinates of the vertex where the edges meet.

(2)

(d)

The cell of A is the region of the town closest to A. Find its area.

(3)

(e)

Find the percentage of the town's area that is closest to pool A, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint (6, 2) and AB horizontal	✓ 1
A1 x = 6	✓ 1
Part (b)
M1 midpoint (4, 5)	✓ 1
M1 gradient of AC = 3, so perpendicular gradient = −1/3	✓ 1
A1 y = −x/3 + 19/3	✓ 1
Part (c)
M1 substitute x = 6 into the equation in (b)	✓ 1
A1 (6, 13/3)	✓ 1
Part (d)
M1 cell has vertices (0, 0), (6, 0), (6, 13/3), (0, 19/3)	✓ 1
M1 trapezium area ½(13/3 + 19/3) × 6 (or equivalent method)	✓ 1
A1 32 km²	✓ 1
Part (e)
M1 32 ÷ (12 × 10) × 100	✓ 1
A1 26.7 %	✓ 1

Exam tip: A bounded cell is found by combining the relevant bisectors with the boundary lines of the region; list its vertices before finding the area.

+ Add to test✓ In your test287IBMI3S-3-02r8-08Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 2Hard12 marks

Points A and B are on level ground in a straight line with the base of a wind turbine tower, with B between A and the tower and AB = 40 m. The angle of elevation of the top of the tower is 28° from A and 41° from B. Let T be the top of the tower.

(a)

Find the size of angle ATB.

(2)

(b)

Use the sine rule to find the length AT, to 3 significant figures.

(3)

(c)

Find the height of the tower, to 3 significant figures. Use the unrounded value of AT.

(2)

(d)

Find the horizontal distance from B to the base of the tower, to 3 significant figures.

(2)

(e)

A point D on the line is 15 m from the base of the tower, between B and the base. Find the angle of elevation of T from D, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ∠ABT = 180° − 41° = 139°	✓ 1
A1 ∠ATB = 180° − 28° − 139° = 13°	✓ 1
Part (b)
M1 AT ÷ sin 139° = 40 ÷ sin 13°	✓ 1
M1 AT = 40 sin 139° ÷ sin 13°	✓ 1
A1 117 m	✓ 1
Part (c)
M1 height = AT × sin 28°	✓ 1
A1 54.8 m	✓ 1
Part (d)
M1 distance = height ÷ tan 41°	✓ 1
A1 63.0 m	✓ 1
Part (e)
M1 tan θ = height ÷ 15	✓ 1
M1 θ = tan⁻¹(54.77… ÷ 15)	✓ 1
A1 74.7°	✓ 1

Exam tip: Keep unrounded values from earlier parts in your calculator so the final angle is not affected by rounding.

+ Add to test✓ In your test288IBMI3S-3-02r8-14Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 2Hard12 marks

A water tank is a frustum of a cone with its wider circular face (radius 3.0 m) at the bottom and its narrower circular face (radius 1.8 m) at the top. The vertical height of the tank is 4.0 m. The frustum is a complete cone with its top part removed.

(a)

Show that the cone removed has height 6 m.

(3)

(b)

Find the volume of the tank, to 3 significant figures.

(3)

(c)

Find the curved surface area of the tank, to 3 significant figures.

(3)

(d)

The tank is filled with water to a depth of 2.0 m. At that depth the radius of the water surface is 2.4 m. Find the volume of water, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 similar triangles: x ÷ (x + 4) = 1.8 ÷ 3.0, where x is the height removed	✓ 1
M1 3x = 1.8x + 7.2, i.e. 1.2x = 7.2	✓ 1
A1 x = 6	✓ 1
Part (b)
M1 large cone: ⅓π × 3² × 10	✓ 1
M1 subtract the small cone ⅓π × 1.8² × 6	✓ 1
A1 73.9 m³	✓ 1
Part (c)
M1 slant heights √(10² + 3²) and √(6² + 1.8²)	✓ 1
M1 π × 3 × 10.44… − π × 1.8 × 6.264…	✓ 1
A1 63.0 m²	✓ 1
Part (d)
M1 the cone above the water level has height 10 − 2 = 8 m and radius 2.4 m	✓ 1
M1 ⅓π × 3² × 10 − ⅓π × 2.4² × 8	✓ 1
A1 46.0 m³	✓ 1

Exam tip: Find a frustum's volume or curved area as a large cone minus a small cone, using similar triangles to get the missing height.

+ Add to test✓ In your test289IBMI3S-3-02r9-13Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard14 marks

A fishing vessel leaves harbour H and sails 25 km on a bearing of 064° to a point X. It then sails 18 km on a bearing of 150° to a point Y. Use unrounded values in later parts.

(a)

Show that angle HXY is 94°.

(2)

(b)

Calculate the distance HY. Give your answer to 3 significant figures.

(3)

(c)

Find the size of angle XHY. Give your answer to 3 significant figures.

(3)

(d)

Find the bearing of Y from H. Give your answer to the nearest degree.

(2)

(e)

Calculate the area of triangle HXY. Give your answer to 3 significant figures.

(2)

(f)

Find the shortest distance from H to the straight line through X and Y. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the bearing of H from X is 064° + 180° = 244°	✓ 1
A1 angle HXY = 244° − 150° = 94° (AG)	✓ 1
Part (b)
M1 HY² = 25² + 18² − 2 × 25 × 18 × cos 94°	✓ 1
A1 HY² = 1011.8…	✓ 1
A1 HY = 31.8 km	✓ 1
Part (c)
M1 sine rule: sin(XHY) ÷ 18 = sin 94° ÷ HY	✓ 1
A1 sin(XHY) = 0.5646… (angle XHY must be acute as it is opposite the shorter side)	✓ 1
A1 angle XHY = 34.4°	✓ 1
Part (d)
M1 add angle XHY to the bearing of X from H, 064° + 34.4…°	✓ 1
A1 098°	✓ 1
Part (e)
M1 ½ × 25 × 18 × sin 94°	✓ 1
A1 224 km²	✓ 1
Part (f)
M1 area = ½ × XY × (shortest distance), so shortest distance = 2 × 224.45… ÷ 18	✓ 1
A1 24.9 km	✓ 1

Common error: When an angle in a triangle is obtuse, the foot of the perpendicular from the opposite vertex lies outside the side, but the area formula still gives the correct perpendicular distance to the extended line.

+ Add to test✓ In your test290IBMI3S-3-02r9-14Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard14 marks

A rectangular fishing zone is described by 0 ≤ x ≤ 14 and 0 ≤ y ≤ 12, with distances in kilometres. Three monitoring buoys are at P(0, 0), Q(10, 0) and R(4, 8). A Voronoi diagram is drawn for the buoys.

(a)

Find the equation of the perpendicular bisector of PQ.

(2)

(b)

Show that the perpendicular bisector of PR has equation y = −0.5x + 5.

(3)

(c)

Find the coordinates of the Voronoi vertex where the cells of P, Q and R meet.

(2)

(d)

A fourth buoy S is placed at (10, 8). Show that the perpendicular bisector of RS is x = 7 and that the perpendicular bisector of QS is y = 4.

(2)

(e)

Hence show that (7, 4) is a Voronoi vertex, by finding its distance from R, Q and S.

(2)

(f)

The cell of S within the fishing zone is the part of the zone with x ≥ 7 and y ≥ 4. Find its area.

(2)

(g)

Find the percentage of the fishing zone that is closer to S than to any other buoy. Give your answer to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of PQ is (5, 0) and PQ is horizontal	✓ 1
A1 x = 5	✓ 1
Part (b)
M1 midpoint of PR is (2, 4) and the gradient of PR is 2	✓ 1
M1 perpendicular gradient −0.5, so y − 4 = −0.5(x − 2)	✓ 1
A1 y = −0.5x + 5 (AG)	✓ 1
Part (c)
M1 substitute x = 5 into y = −0.5x + 5	✓ 1
A1 (5, 2.5)	✓ 1
Part (d)
A1 R and S both have y = 8, midpoint (7, 8), so the bisector is x = 7	✓ 1
A1 Q and S both have x = 10, midpoint (10, 4), so the bisector is y = 4	✓ 1
Part (e)
M1 distance from (7, 4) to R, Q and S	✓ 1
A1 each distance is √(3² + 4²) = 5, so the point is equidistant from the three buoys	✓ 1
Part (f)
M1 width 14 − 7 = 7 and height 12 − 4 = 8	✓ 1
A1 56 km²	✓ 1
Part (g)
A1 56 ÷ (14 × 12) × 100 = 33.3 %	✓ 1

Exam tip: To show a point is a Voronoi vertex, show that it is the same distance from each of the three sites.

+ Add to test✓ In your test291IBMI3S-3-01r11-04Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Hard12 marks

A sensor housing on a wind turbine is a solid cone with base radius 9 cm and slant height 41 cm.

(a)

Show that the perpendicular height of the cone is 40 cm.

(2)

(b)

Calculate the volume of the cone, in cm³, correct to 1 decimal place.

(2)

(c)

Find the curved surface area of the cone, in cm², correct to 1 decimal place.

(2)

(d)

The cone is made of metal of density 2.7 g cm⁻³. Calculate its mass in kilograms, correct to 3 significant figures.

(3)

(e)

A cut parallel to the base removes a small cone from the top. The small cone has height ⁴⁰⁄₃ cm. The remaining solid is a frustum. Find the volume of the frustum, correct to the nearest cm³.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
h² = 41² − 9² = 1600	✓ 1
h = 40 cm	✓ 1
Part (b)
V = ⅓ × π × 9² × 40	✓ 1
V = 3392.9 cm³	✓ 1
Part (c)
curved area = π × 9 × 41	✓ 1
= 1159.2 cm²	✓ 1
Part (d)
mass = 2.7 × 3392.9 = 9161 g	✓ 1
convert grams to kilograms: ÷ 1000	✓ 1
mass = 9.16 kg	✓ 1
Part (e)
by similar triangles the small cone has radius 9 ÷ 3 = 3 cm	✓ 1
small cone volume = ⅓ × π × 3² × (40/3) = 125.7 cm³	✓ 1
frustum volume = 3392.9 − 125.7 = 3267 cm³	✓ 1

Exam tip: Check the units before multiplying by a density: cm³ with g cm⁻³ gives grams, and a final answer in kilograms needs a division by 1000.

+ Add to test✓ In your test292IBMI3S-3-01r11-07Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Hard14 marks

Three wind-farm control centres are located at A(1, 1), B(9, 1) and C(5, 9) on a map where each unit represents 1 km. Treat the map as extending without limit in every direction. A Voronoi diagram is drawn for the sites A, B and C.

(a)

Write down the equation of the perpendicular bisector of AB.

(1)

(b)

Show that the perpendicular bisector of AC has equation y = −0.5x + 6.5.

(2)

(c)

Find the coordinates of the Voronoi vertex.

(2)

(d)

Find the distance from the vertex to each site, and state what this shows about the vertex.

(2)

(e)

A fourth control centre is built at D(5, 2). Find the equation of the perpendicular bisector of AD, giving your answer in the form y = mx + c.

(2)

(f)

By symmetry the bisector of BD is y = 4x − 26.5, and the bisector of CD is y = 5.5. The region closest to D is a triangle bounded by these three lines. Find the coordinates of its three vertices.

(3)

(g)

Calculate the area of the region closest to D, in km².

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
x = 5	✓ 1
Part (b)
midpoint of AC = (3, 5) and gradient of AC = 2, so perpendicular gradient = −1/2	✓ 1
y − 5 = −0.5(x − 3), giving y = −0.5x + 6.5	✓ 1
Part (c)
substitute x = 5	✓ 1
vertex = (5, 4)	✓ 1
Part (d)
distance to A = √(4² + 3²) = 5 km (the same to B and C)	✓ 1
the vertex is equidistant (5 km) from the three sites	✓ 1
Part (e)
midpoint of AD = (3, 1.5) and gradient of AD = 1/4, so perpendicular gradient = −4	✓ 1
y = −4x + 13.5	✓ 1
Part (f)
AD and CD bisectors: −4x + 13.5 = 5.5 gives (2, 5.5)	✓ 1
BD and CD bisectors: 4x − 26.5 = 5.5 gives (8, 5.5)	✓ 1
AD and BD bisectors: −4x + 13.5 = 4x − 26.5 gives (5, −6.5)	✓ 1
Part (g)
base = 8 − 2 = 6 km and perpendicular height = 5.5 − (−6.5) = 12 km	✓ 1
area = ½ × 6 × 12 = 36 km²	✓ 1

Exam tip: Write down each bisector as a simple equation first; vertices then come from solving pairs of equations, and a quick equidistance check confirms each one.

+ Add to test✓ In your test293IBMI3S-3-02r13-12Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 2Hard14 marks

On a map of a circuit's infrastructure, one unit represents 1 km. Marshal posts are at A(1, 2), B(9, 2) and C(5, 8). Only points with y ≥ 0 (north of the access road along the x-axis) are considered.

(a)

Write down the equation of the Voronoi edge between A and B.

(1)

(b)

Find the equation of the Voronoi edge between A and C, in the form y = mx + c.

(3)

(c)

Find the coordinates of the Voronoi vertex for A, B and C.

(1)

(d)

Find the distance from this vertex to A. Give your answer to 3 significant figures.

(2)

(e)

A fourth post D is added at (5, 0). Show that the Voronoi edge between C and D is y = 4, and that its endpoints (the vertices where it meets the edges for A and for B) are (4.5, 4) and (5.5, 4).

(4)

(f)

The Voronoi cell of D, for y ≥ 0, is bounded by the edges for A, B and C. Find its area.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x = 5	✓ 1
Part (b)
M1 midpoint of AC = (3, 5)	✓ 1
M1 gradient of AC = 1.5, so the perpendicular gradient is −2/3	✓ 1
A1 y = −(2/3)x + 7	✓ 1
Part (c)
A1 (5, 11/3), that is (5, 3.67)	✓ 1
Part (d)
M1 √(4² + (5/3)²)	✓ 1
A1 4.33 km	✓ 1
Part (e)
A1 midpoint of CD is (5, 4) and CD is vertical, so the edge is y = 4	✓ 1
M1 edge between A and D: midpoint (3, 1), gradient of AD = −1/2, so y = 2x − 5	✓ 1
M1 edge between B and D: y = −2x + 15, and substitution of y = 4 into each edge	✓ 1
A1 x = 4.5 and x = 5.5, giving the vertices (4.5, 4) and (5.5, 4) (AG)	✓ 1
Part (f)
M1 the cell is a trapezium with vertices (4.5, 4), (5.5, 4), (7.5, 0) and (2.5, 0) (the edges meet y = 0 where 2x − 5 = 0 and −2x + 15 = 0)	✓ 1
M1 area = ½ × (1 + 5) × 4	✓ 1
A1 12 km²	✓ 1

Exam tip: To find a cell's area, first identify every vertex of the cell, then split the shape into a standard shape such as a trapezium or triangles.

+ Add to test✓ In your test294IBMI3S-3-02r13-13Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 2Hard12 marks

Three hospitals A, B and C lie on flat ground. B is 18 km from A on a bearing of 072°. C is 24 km from B on a bearing of 150° from B. Bearings are measured clockwise from north.

(a)

Show that angle ABC is 102°.

(2)

(b)

Calculate the distance AC. Give your answer to 3 significant figures.

(3)

(c)

Calculate the size of angle BAC. Give your answer to 3 significant figures.

(3)

(d)

Find the bearing of C from A, to the nearest degree.

(2)

(e)

A helicopter pad is to be built on the straight road AC at the point nearest to B. Find the shortest distance from B to the road AC. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the bearing of A from B is 72° + 180° = 252°	✓ 1
A1 252° − 150° = 102° (AG)	✓ 1
Part (b)
M1 cosine rule: AC² = 18² + 24² − 2 × 18 × 24 × cos 102°	✓ 1
M1 correct substitution evaluated, AC² = 1079.6…	✓ 1
A1 AC = 32.9 km	✓ 1
Part (c)
M1 sine rule: sin(BAC)/24 = sin 102°/AC (using the unrounded AC)	✓ 1
M1 sin(BAC) = 0.714…	✓ 1
A1 45.6°	✓ 1
Part (d)
M1 bearing = 72° + angle BAC (using the unrounded angle)	✓ 1
A1 118°	✓ 1
Part (e)
M1 distance = 18 × sin(BAC), or 2 × area of ABC ÷ AC	✓ 1
A1 12.9 km	✓ 1

Common error: Check that a sine-rule angle is acute or obtuse as appropriate: the largest angle lies opposite the longest side, so angle BAC here must be acute.

+ Add to test✓ In your test295IBMI3S-3-02r15-03Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard13 marks

An offshore wind farm has four turbines at A(2, 2), B(10, 3), C(5, 9) and D(12, 10), on a map where each unit represents 1 km. A Voronoi diagram is drawn for the region 0 ≤ x ≤ 14, 0 ≤ y ≤ 12. The operator wants to place a weather mast in this region as far as possible from the nearest turbine. For this region it can be shown that the best position is either a Voronoi vertex or a corner of the region.

(a)

Show that the equation of the perpendicular bisector of AB is y = −8x + 50.5.

(3)

(b)

The perpendicular bisector of AC is y = −(3/7)x + 7. Find the coordinates of the Voronoi vertex V₁ where the bisectors of AB and AC meet.

(3)

(c)

Find the distance from V₁ to turbine A. Use the unrounded coordinates of V₁.

(2)

(d)

The other Voronoi vertex inside the region is V₂(8.84, 7.12), to 3 significant figures. Show that the distance from V₂ to its nearest turbine, B, is 4.28 km, to 3 significant figures. The corners (0, 0) and (14, 12) are each less than 3 km from a turbine.

(2)

(e)

Find the distance from each of the corners (0, 12) and (14, 0) to its nearest turbine, and hence state the best position for the mast and its distance from the nearest turbine.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 midpoint of AB is (6, 2.5)	✓ 1
M1 gradient of AB = 1/8, so perpendicular gradient = −8	✓ 1
A1 y − 2.5 = −8(x − 6), which gives y = −8x + 50.5 (AG)	✓ 1
Part (b)
M1 equates −8x + 50.5 = −(3/7)x + 7	✓ 1
A1 x = 5.75	✓ 1
A1 y = 4.54	✓ 1
Part (c)
M1 uses the distance formula from (5.745…, 4.538…) to (2, 2)	✓ 1
A1 distance = 4.52 km	✓ 1
Part (d)
M1 distance from (8.84, 7.12) to B(10, 3) = √((1.16)² + (4.12)²)	✓ 1
A1 4.28 km (AG)	✓ 1
Part (e)
M1 nearest turbine to (0, 12) is C and to (14, 0) is B	✓ 1
A1 distance from (0, 12) to C = 5.83 km and from (14, 0) to B = 5.00 km	✓ 1
A1 the mast should be at (0, 12), 5.83 km from the nearest turbine, since this is greater than 4.52 and 4.28	✓ 1

Exam tip: When the farthest point from a set of sites must lie inside a bounded region, check the Voronoi vertices and the corners of the region; the largest of these distances gives the answer.

+ Add to test✓ In your test296IBMI3S-3-02r15-08Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard10 marks

A wind turbine rests on a concrete foundation. The foundation is a solid cylinder of radius 5.0 m and height 2.0 m, with a conical hollow cut from the top. The cone has base radius 1.5 m, height 1.2 m, and its base lies in the top face of the cylinder. Concrete has density 2400 kg per m³.

(a)

Calculate the volume of concrete in the foundation. Give your answer to 3 significant figures.

(3)

(b)

Find the mass of the foundation in tonnes (1 tonne = 1000 kg). Use the unrounded volume and give your answer to 3 significant figures.

(2)

(c)

Concrete is delivered by lorries that each carry at most 28 tonnes. Find the minimum number of lorry loads needed to supply the mass of concrete in the foundation.

(2)

(d)

A second design uses a cylinder of height 2.0 m and radius r m with the same conical hollow, and contains exactly 200 m³ of concrete. Find r, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cylinder volume π × 5.0² × 2.0 = 157.08 m³	✓ 1
M1 cone volume (1/3) × π × 1.5² × 1.2 = 2.827 m³	✓ 1
A1 volume = 154 m³	✓ 1
Part (b)
M1 mass = volume × 2400	✓ 1
A1 mass = 370 tonnes	✓ 1
Part (c)
M1 370.2 ÷ 28 = 13.2	✓ 1
A1 14 loads (whole number, rounded up)	✓ 1
Part (d)
M1 forms the equation π × r² × 2.0 − 2.827 = 200	✓ 1
M1 rearranges r² = 202.83 ÷ (2π) = 32.28	✓ 1
A1 r = 5.68 m	✓ 1

Exam tip: Subtract the volume of the hollow from the volume of the full solid, and in a reverse problem set up the volume equation before rearranging for the unknown radius.

+ Add to test✓ In your test297IBMI3S-3-02r15-13Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard13 marks

Three wind turbines A, B and C are positioned so that B is 3.2 km from A on a bearing of 065°, and C is 4.5 km from B on a bearing of 150°.

(a)

Show that angle ABC is 95°.

(2)

(b)

Find the distance AC, giving your answer to 3 significant figures.

(3)

(c)

Find the angle BAC and hence the bearing of C from A, giving the bearing to the nearest degree. Use the unrounded value of AC.

(4)

(d)

Find the area of triangle ABC, giving your answer to 3 significant figures.

(2)

(e)

Find the shortest distance from A to the line BC. Use the area from part (d) in your working and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R1 the bearing of A from B is 065° + 180° = 245°	✓ 1
A1 angle ABC = 245° − 150° = 95° (AG)	✓ 1
Part (b)
M1 cosine rule AC² = 3.2² + 4.5² − 2 × 3.2 × 4.5 × cos 95°	✓ 1
A1 AC² = 33.000	✓ 1
A1 AC = 5.74 km	✓ 1
Part (c)
M1 sine rule sin(BAC)/4.5 = sin 95°/AC	✓ 1
A1 angle BAC = 51.29°	✓ 1
M1 bearing of C from A = 065° + angle BAC (C is clockwise from B as seen from A)	✓ 1
A1 bearing = 116°	✓ 1
Part (d)
M1 area = 1/2 × 3.2 × 4.5 × sin 95°	✓ 1
A1 area = 7.17 km²	✓ 1
Part (e)
M1 area = 1/2 × BC × shortest distance, so distance = 2 × 7.173 ÷ 4.5	✓ 1
A1 shortest distance = 3.19 km	✓ 1

Exam tip: After using the sine rule to find an angle, decide whether it should be acute or obtuse by comparing the side opposite it with the other sides, then add it to or subtract it from the given bearing using a sketch.

+ Add to test✓ In your test298IBMI3S-3-01r17-10Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Hard12 marks

A rectangular survey field measures 10 km from west to east and 9 km from south to north. Axes are set up with the south-west corner at the origin, so the field is 0 ≤ x ≤ 10, 0 ≤ y ≤ 9 (units in km). Three radio telescopes are at A(1, 1), B(9, 1) and C(3, 7). A Voronoi diagram is drawn for the field with these three sites.

(a)

Show that the equation of the perpendicular bisector of AC is y = −¹⁄₃x + ¹⁴⁄₃.

(3)

(b)

The perpendicular bisector of BC is y = x − 2. Find the coordinates of the vertex where the three cells meet.

(3)

(c)

The cell of C is bounded by the two bisectors and the edges of the field. Find the area of the field that is closer to C than to A or B.

(4)

(d)

Find the percentage of the field that is closer to C than to A or B, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AC = (2, 4)	✓ 1
M1 gradient of AC = 3, so perpendicular gradient = −1/3	✓ 1
A1 y − 4 = −(1/3)(x − 2), giving y = −¹⁄₃x + ¹⁴⁄₃ (AG)	✓ 1
Part (b)
M1 perpendicular bisector of AB is x = 5	✓ 1
M1 substitute x = 5 into the equation from part (a) (or solve the two bisectors simultaneously)	✓ 1
A1 (5, 3)	✓ 1
Part (c)
M1 the cell boundary meets x = 0 at (0, ¹⁴⁄₃) and x = 10 at (10, 8)	✓ 1
M1 correct strategy for the area, for example splitting the cell into two trapezoids at x = 5, or a polygon area formula with vertices (0, ¹⁴⁄₃), (5, 3), (10, 8), (10, 9), (0, 9)	✓ 1
A1 one correct area: the left trapezoid is 155/6 = 25.83… km² and the right trapezoid is 17.5 km²	✓ 1
A1 total area = 130/3 = 43.3 km²	✓ 1
Part (d)
M1 (130/3) ÷ 90 × 100 (ECF from (c))	✓ 1
A1 48.1%	✓ 1

Exam tip: Voronoi cell boundaries are pieces of perpendicular bisectors: find the corner points first, then split the cell into shapes whose area you can calculate.

+ Add to test✓ In your test299IBMI3S-3-01r17-14Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Hard9 marks

Two radio dishes, P and Q, stand on level ground. Q is 800 m from P on a bearing of 090°. A weather balloon hovers at a point S directly above a ground point G. From P, G is on a bearing of 040°, and from Q, G is on a bearing of 330°. The angle of elevation of S from P is 25°.

(a)

Find the size of angle PGQ.

(2)

(b)

Calculate the distance PG, in metres, correct to 3 significant figures.

(2)

(c)

Find the height of the balloon above the ground, in metres, correct to 3 significant figures. Use the unrounded value of PG.

(2)

(d)

Find the angle of elevation of the balloon from Q, correct to 3 significant figures. Use unrounded values from earlier parts.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 angle GPQ = 090° − 040° = 50° and angle GQP = 330° − 270° = 60°	✓ 1
A1 angle PGQ = 180° − 50° − 60° = 70°	✓ 1
Part (b)
M1 sine rule: PG/sin 60° = 800/sin 70°	✓ 1
A1 PG = 737 m	✓ 1
Part (c)
M1 height = PG tan 25°	✓ 1
A1 344 m	✓ 1
Part (d)
M1 QG = 800 sin 50°/sin 70° (sine rule) = 652.2… m	✓ 1
M1 tan(elevation) = 343.8 ÷ 652.2 (ECF from (c))	✓ 1
A1 27.8°	✓ 1

Common error: Bearings measured from different points use different north lines; convert each bearing to an angle inside the triangle before using the sine rule.

+ Add to test✓ In your test300IBMI3S-3-02r19-05Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 2Hard12 marks

A water treatment plant has a storage tank in the shape of an inverted cone (vertex pointing down) of height 6.0 m. The circular top of the tank has radius 2.4 m. The tank is used with its axis vertical and its top level.

(a)

When the water has depth h metres, the surface of the water is a circle of radius r metres. Show that the volume V of water in the tank is V = (4π/75)h³.

(3)

(b)

Find the volume of the tank when it is full, to 3 significant figures.

(2)

(c)

Find the volume of water when the depth is 2.5 m, to 3 significant figures.

(2)

(d)

The tank contains 20.0 m³ of water. Find the depth of the water, to 3 significant figures.

(3)

(e)

Water is then drained from this 20.0 m³ at a constant 0.35 m³ per minute until the tank is empty. Find the time taken, in minutes, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 similar triangles: r/h = 2.4/6, so r = 0.4h	✓ 1
M1 substitute into V = (1/3)πr²h = (1/3)π(0.16h²)h	✓ 1
A1 V = (0.16π/3)h³ = (4π/75)h³ (shown)	✓ 1
Part (b)
M1 substitute h = 6 (or r = 2.4, h = 6 in the cone formula)	✓ 1
A1 36.2 m³	✓ 1
Part (c)
M1 (4π/75) × 2.5³	✓ 1
A1 2.62 m³	✓ 1
Part (d)
M1 set (4π/75)h³ = 20	✓ 1
M1 h³ = 119.4… and take the cube root	✓ 1
A1 4.92 m	✓ 1
Part (e)
M1 20.0 ÷ 0.35	✓ 1
A1 57.1 minutes	✓ 1

Common error: The radius of the water surface changes with depth: use similar triangles to express r in terms of h before substituting in the volume formula.

+ Add to test✓ In your test301IBMI3S-3-02r19-06Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 2Hard12 marks

A gaming company is designing a square map with 0 ≤ x ≤ 10 and 0 ≤ y ≤ 10, in km. Three spawn points are at P(1, 1), Q(7, 3) and R(3, 7). Each location on the map belongs to the cell of its nearest spawn point (a Voronoi diagram).

(a)

Show that the perpendicular bisector of PQ has equation 3x + y = 14.

(3)

(b)

The perpendicular bisector of PR has equation x + 3y = 14. Find the coordinates of the point where the bisectors of PQ and PR meet.

(3)

(c)

Show that (3.5, 3.5) is equidistant from P, Q and R, and find this distance to 3 significant figures.

(2)

(d)

The cell of P within the map is bounded by the axes and the two bisectors. Find its area, to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of PQ is (4, 2)	✓ 1
M1 gradient of PQ is 2/6 = 1/3, so perpendicular gradient is −3	✓ 1
A1 y − 2 = −3(x − 4), so 3x + y = 14 (shown)	✓ 1
Part (b)
M1 solve 3x + y = 14 and x + 3y = 14 simultaneously	✓ 1
A1 x = 3.5	✓ 1
A1 y = 3.5, so (3.5, 3.5)	✓ 1
Part (c)
M1 squared distances to P, Q and R are all 12.5	✓ 1
A1 distance = √12.5 = 3.54 km	✓ 1
Part (d)
M1 intercepts of 3x + y = 14 and x + 3y = 14 with the axes: (14/3, 0) and (0, 14/3)	✓ 1
A1 cell is the quadrilateral with vertices (0, 0), (14/3, 0), (3.5, 3.5), (0, 14/3)	✓ 1
M1 area by splitting into two triangles or using the shoelace formula (each triangle ½ × 14/3 × 3.5)	✓ 1
A1 49/3 = 16.3 km²	✓ 1

Exam tip: A Voronoi vertex is equidistant from three sites: check it by computing the three distances rather than trusting the algebra.

+ Add to test✓ In your test302IBMI3S-3-02r19-12Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context Paper 2Hard13 marks

A game designer builds a pyramid with a square base of side 40 m. The apex V is 30 m vertically above the centre of the base.

(a)

Find the length of a sloping edge, such as VA where A is a corner of the base, to 3 significant figures.

(3)

(b)

Find the angle between a sloping edge and the base, to the nearest 0.1°.

(2)

(c)

Find the angle between a triangular face and the base, to the nearest 0.1°.

(2)

(d)

Calculate the volume of the pyramid.

(2)

(e)

The four triangular faces and the base are to be covered with a surface texture. Calculate the total area, to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 half the base diagonal = 40√2 ÷ 2 = 28.28…	✓ 1
M1 Pythagoras: √(28.28…² + 30²)	✓ 1
A1 41.2 m	✓ 1
Part (b)
M1 tan θ = 30 ÷ 28.28…	✓ 1
A1 46.7°	✓ 1
Part (c)
M1 tan φ = 30 ÷ 20 (half the side of the base)	✓ 1
A1 56.3°	✓ 1
Part (d)
M1 (1/3) × 40² × 30	✓ 1
A1 16 000 m³	✓ 1
Part (e)
M1 slant height of a face = √(20² + 30²) = 36.06…	✓ 1
M1 area of one face = ½ × 40 × 36.06… = 721.1…, four faces 2884.4…	✓ 1
M1 add the base 40² = 1600	✓ 1
A1 4480 m²	✓ 1

Common error: The slant height of a face (to the midpoint of a base edge) is different from the length of a sloping edge (to a corner).

+ Add to test✓ In your test303IBMI3S-3-03r20-11Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Hard15 marks

A harbour is at H. A fishing buoy B is 8.4 km from H on a bearing of 072°, and a second buoy C is 11.6 km from H on a bearing of 145°.

(a)

Find angle BHC.

(1)

(b)

Calculate the distance BC, in km, to 3 significant figures.

(3)

(c)

Find the size of angle HBC, to the nearest 0.1°. Use the cosine rule and your unrounded value of BC.

(3)

(d)

Find the bearing of C from B, to the nearest degree.

(3)

(e)

Find the area of triangle HBC, in km², to 3 significant figures.

(2)

(f)

A patrol boat sails along the straight line from B to C. Find the shortest distance between H and this line, in km, to 3 significant figures. Use the unrounded area and BC.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 145° − 72° = 73°	✓ 1
Part (b)
M1 cosine rule, $BC^2=8.4^2+11.6^2-2(8.4)(11.6)\cos73°$	✓ 1
A1 $BC^2=148.14$	✓ 1
A1 BC = 12.17 km	✓ 1
Part (c)
M1 $\cos(HBC)=\frac{8.4^2+BC^2-11.6^2}{2(8.4)(BC)}$	✓ 1
A1 $\cos(HBC)=0.4115$	✓ 1
A1 angle HBC = 65.7°	✓ 1
Part (d)
M1 bearing of H from B is 072° + 180° = 252°	✓ 1
M1 C lies anticlockwise of H as seen from B, so subtract: 252° − 65.7°	✓ 1
A1 186.3° so 186°	✓ 1
Part (e)
M1 $\frac{1}{2}(8.4)(11.6)\sin73°$	✓ 1
A1 46.59 km² (46.6)	✓ 1
Part (f)
M1 area = $\frac{1}{2}\times BC\times$ perpendicular distance	✓ 1
M1 $d=\frac{2\times46.59}{12.17}$ (unrounded values)	✓ 1
A1 d = 7.66 km	✓ 1

Exam tip: If two sides and the included angle are given, start with the cosine rule; once all three sides are known, the cosine rule again avoids the ambiguity of an obtuse angle.

+ Add to test✓ In your test304IBMI3S-3-03r20-12Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Hard14 marks

Four soil-monitoring robots are parked at A(0, 0), B(10, 0), C(5, 4) and D(5, 12), where coordinates are in metres. A Voronoi diagram is drawn for these four sites. The cell containing C is a triangle.

(a)

Find the equation of the perpendicular bisector of AC. Give your answer in the form $y=mx+c$.

(3)

(b)

Find the equation of the perpendicular bisector of CD.

(2)

(c)

Hence find the coordinates of the vertex where the bisectors of AC and CD meet.

(3)

(d)

By symmetry, the other vertex on the line $y=8$ is (12.3, 8). Show that the third vertex of the cell of C is (5, −1.125).

(2)

(e)

Find the area of the cell of C, in m², to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AC is (2.5, 2) and gradient of AC is $\frac{4}{5}$	✓ 1
M1 perpendicular gradient is $-\frac{5}{4}$, so $y-2=-1.25(x-2.5)$	✓ 1
A1 $y=-1.25x+5.125$	✓ 1
Part (b)
M1 midpoint of CD is (5, 8) and CD is vertical	✓ 1
A1 $y=8$	✓ 1
Part (c)
M1 substitute $y=8$ into the equation from (a)	✓ 1
M1 $8=-1.25x+5.125$ so $x=-2.3$	✓ 1
A1 (−2.3, 8)	✓ 1
Part (d)
M1 the bisectors of AC and BC meet on the line $x=5$ by symmetry, so substitute $x=5$ into $y=-1.25x+5.125$	✓ 1
A1 $y=-6.25+5.125=-1.125$, giving (5, −1.125) (AG)	✓ 1
Part (e)
M1 base of triangle along $y=8$ is $12.3-(-2.3)=14.6$	✓ 1
M1 perpendicular height is $8-(-1.125)=9.125$	✓ 1
M1 area $=\frac{1}{2}\times14.6\times9.125$	✓ 1
A1 66.6 m² (66.6125)	✓ 1

Exam tip: For a triangle with a horizontal base, subtract the y-coordinates for the height; this avoids using the determinant or shoelace formula.

+ Add to test✓ In your test305IBMI3S-3-03r20-14Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Hard13 marks

A shop has a glass atrium in the shape of a right pyramid with square base ABCD of side 14 m. The vertex V is 9 m vertically above the centre M of the base. The four triangular faces are made of glass.

(a)

Find the length of the diagonal AC, in metres, to 3 significant figures.

(2)

(b)

Find the length of the edge VA, in metres, to 3 significant figures. Use the unrounded value of AC.

(3)

(c)

Find the angle between the edge VA and the base ABCD, to the nearest 0.1°.

(2)

(d)

Let N be the midpoint of AB. Find the length VN, and hence find the total area of glass in the four faces, in m², to 3 significant figures.

(4)

(e)

The glass costs €285 per m². Using the unrounded area, find the cost of the glass, to the nearest euro.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 Pythagoras, $AC=\sqrt{14^2+14^2}$	✓ 1
A1 AC = 19.80 m (19.8)	✓ 1
Part (b)
M1 $AM=\frac{AC}{2}=9.899$ and triangle VMA is right-angled at M	✓ 1
M1 $VA=\sqrt{9^2+9.899^2}$	✓ 1
A1 VA = 13.4 m (13.38)	✓ 1
Part (c)
M1 angle VAM, $\tan(VAM)=\frac{9}{9.899}$	✓ 1
A1 42.3°	✓ 1
Part (d)
M1 $MN=7$ and triangle VMN is right-angled at M	✓ 1
A1 $VN=\sqrt{9^2+7^2}=11.402$ m	✓ 1
M1 area of one face $=\frac{1}{2}(14)(VN)$, then multiply by 4	✓ 1
A1 319.2 m² (319)	✓ 1
Part (e)
M1 multiply the unrounded area by 285	✓ 1
A1 €90986	✓ 1

Exam tip: In a right pyramid, the height, half the base diagonal and the edge form one right-angled triangle, and the height, half the base side and the slant height form another.

+ Add to test✓ In your test306IBMI3S-3-02r22-10Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Hard12 marks

Three weather stations are located at A(0, 0), B(12, 0) and C(4, 10) on a map of an island, where units are kilometres. A Voronoi diagram is drawn for the three stations.

(a)

Write down the equation of the edge between the regions of A and B.

(1)

(b)

Find the equation of the edge between the regions of A and C.

(3)

(c)

Find the coordinates of the Voronoi vertex V where these two edges meet.

(2)

(d)

Show that V lies on the edge between the regions of B and C.

(2)

(e)

Find the distance from V to station A, to 3 significant figures.

(2)

(f)

A new station is to be built at V. Explain why V is a sensible choice if the aim is to place it as far as possible from all three existing stations.

(1)

(g)

A sensor is at R(3, 6). Determine which of the three stations is nearest to R and find that distance, to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x = 6	✓ 1
Part (b)
M1 midpoint of AC = (2, 5) and gradient of AC = 5/2	✓ 1
M1 perpendicular gradient −2/5, so y − 5 = −(2/5)(x − 2)	✓ 1
A1 y = −0.4x + 5.8	✓ 1
Part (c)
M1 substitute x = 6 into the equation from (b)	✓ 1
A1 V = (6, 3.4)	✓ 1
Part (d)
M1 midpoint of BC = (8, 5), gradient of BC = −5/4, so perpendicular gradient 4/5: y = 0.8x − 1.4	✓ 1
A1 at x = 6, y = 0.8 × 6 − 1.4 = 3.4, which matches V	✓ 1
Part (e)
M1 √(6² + 3.4²)	✓ 1
A1 6.90 km	✓ 1
Part (f)
R1 V is equidistant (6.90 km) from A, B and C, and moving away from V in any direction brings you closer to at least one of them	✓ 1
Part (g)
A1 AR = 6.71, BR = 10.8, CR = √17 = 4.12; C is nearest at 4.12 km (all three distances compared)	✓ 1

Exam tip: After finding a vertex from two edges, always check it against the third edge or by measuring to all three sites; it is a quick way to catch an arithmetic slip.

+ Add to test✓ In your test307IBMI3S-3-02r22-13Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Hard13 marks

A trawler leaves port P and sails 15 km on a bearing of 038° to a point T. It then sails 22 km on a bearing of 125° to a point Q.

(a)

Show that angle PTQ is 93°.

(2)

(b)

Find the distance PQ, to 3 significant figures.

(3)

(c)

Find the bearing of Q from P, to the nearest degree. Use the unrounded value of PQ.

(4)

(d)

A rescue vessel leaves P and travels in a straight line directly to Q at 45 km per hour. Find the time taken, in minutes, to 3 significant figures. Use the unrounded value of PQ.

(2)

(e)

The trawler's fuel use is 4.2 litres per kilometre. Find the fuel used on the round trip P to T to Q and back to P, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the bearing of P from T is 038° + 180° = 218°	✓ 1
A1 218° − 125° = 93°	✓ 1
Part (b)
M1 PQ² = 15² + 22² − 2 × 15 × 22 × cos 93°	✓ 1
A1 PQ² = 743.5...	✓ 1
A1 27.3 km	✓ 1
Part (c)
M1 sin(TPQ) / 22 = sin 93° / PQ	✓ 1
A1 angle TPQ = 53.7°	✓ 1
M1 bearing = 038° + 53.7°	✓ 1
A1 092°	✓ 1
Part (d)
M1 PQ / 45 × 60	✓ 1
A1 36.4 minutes	✓ 1
Part (e)
M1 (15 + 22 + PQ) × 4.2	✓ 1
A1 270 litres	✓ 1

Common error: For bearings problems, mark the north line at each point and find the interior angle from the back bearing; measuring from the wrong north line gives the wrong triangle.

+ Add to test✓ In your test308IBMI3S-3-01r23-05Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard15 marks

A rural area is served by four mobile masts at A(0, 0), B(8, 0), C(7, 6) and D(0, 4), where units are kilometres. The area is divided into Voronoi cells. The vertex V₁ is where the cells of A, B and D meet, and the vertex V₂ is where the cells of B, C and D meet.

(a)

Write down the equation of the perpendicular bisector of AB and of AD.

(2)

(b)

Hence show that V₁ is at (4, 2), and verify that C is further from V₁ than A is.

(3)

(c)

Find the equation of the perpendicular bisector of BD, and of the perpendicular bisector of BC.

(3)

(d)

Find the coordinates of V₂, giving each coordinate to 3 significant figures.

(3)

(e)

Calculate the length of the Voronoi edge between V₁ and V₂, to 3 significant figures. Use the unrounded coordinates of V₂.

(2)

(f)

A new mast is to be placed at V₂. Find the distance from V₂ to the nearest existing mast, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x = 4	✓ 1
A1 y = 2	✓ 1
Part (b)
M1 intersection of x = 4 and y = 2	✓ 1
A1 V₁ = (4, 2)	✓ 1
R1 AV₁ = √20 ≈ 4.47 km but CV₁ = 5 km, so C is further	✓ 1
Part (c)
A1 BD: midpoint (4, 2), gradient of BD = −1/2, so y = 2x − 6	✓ 1
M1 midpoint of BC (7.5, 3) and gradient of BC = −6 so perpendicular gradient 1/6	✓ 1
A1 BC: y − 3 = (x − 7.5)/6, i.e. y = x/6 + 1.75	✓ 1
Part (d)
M1 solves the equations of the bisectors of BD and BC simultaneously (GDC)	✓ 1
A1 x = 4.23	✓ 1
A1 y = 2.45	✓ 1
Part (e)
M1 uses distance formula with (4, 2) and (x, y) of V₂ (x = 93/22, y = 27/11)	✓ 1
A1 0.508 km	✓ 1
Part (f)
M1 distance from V₂ to B, C or D	✓ 1
A1 4.50 km	✓ 1

Common error: Check each candidate vertex against the other sites: a point where three bisectors meet is only a Voronoi vertex if no other site is closer.

+ Add to test✓ In your test309IBMI3S-3-01r23-09Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard13 marks

A yoghurt pot is shaped as a frustum of a cone: a cone with its top removed. The circular top of the pot has radius 4.0 cm, the circular base has radius 2.5 cm, and the vertical height of the pot is 9.0 cm. The pot is open at the top. The pot is the part that remains when a small cone, with radius 2.5 cm at its widest, is cut off the pointed end of a large cone whose radius is 4.0 cm.

(a)

Let H cm be the height of the large cone and (H − 9) cm the height of the small cone removed. By using similar triangles, show that H = 24.

(3)

(b)

Find the volume of the pot, to the nearest cm³.

(3)

(c)

The curved surface area of a cone is πrl, where l is the slant height. Find the slant height of the large cone and of the small cone.

(2)

(d)

Hence find the external surface area of the pot, including its circular base but not its open top, to 3 significant figures. Use the unrounded slant heights.

(3)

(e)

The plastic costs €0.0035 per cm² of external surface area. Find the cost of the plastic for 5000 pots, using 206 cm² per pot. Give your answer to the nearest euro.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sets up ratio 4/H = 2.5/(H − 9)	✓ 1
M1 4(H − 9) = 2.5H, so 1.5H = 36	✓ 1
A1 H = 24 (shown)	✓ 1
Part (b)
M1 volume of large cone (1/3)π(4)²(24) = 128π and small cone (1/3)π(2.5)²(15) = 31.25π	✓ 1
M1 subtracts the volumes	✓ 1
A1 96.75π = 304 cm³	✓ 1
Part (c)
A1 large cone: √(4² + 24²) = 24.3 cm	✓ 1
A1 small cone: √(2.5² + 15²) = 15.2 cm	✓ 1
Part (d)
M1 curved area = π(4)(24.33…) − π(2.5)(15.21…)	✓ 1
A1 curved area = 186.3 cm²	✓ 1
A1 adds base π(2.5)² = 19.6 cm²: total 206 cm²	✓ 1
Part (e)
M1 5000 × 206 × 0.0035	✓ 1
A1 €3605	✓ 1

Common error: The frustum’s curved area is the difference of two cone areas: both slant heights are needed, not the height of the pot.

+ Add to test✓ In your test310IBMI3S-3-01r23-14Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard13 marks

Two transmitter towers A and B stand on level ground, with B 8.0 km due east of A. A hiker at H receives signals from both towers. From A, the bearing of H is 052°. From B, the bearing of H is 313°. A straight road runs due east through A and B.

(a)

Find the size of angle HAB and of angle HBA.

(2)

(b)

Show that angle AHB = 99°.

(2)

(c)

Use the sine rule to find the distance AH, to 3 significant figures.

(2)

(d)

Find the distance BH, to 3 significant figures.

(2)

(e)

The hiker walks by the shortest route to the road. Calculate the length of this route, to 3 significant figures, using the unrounded value of AH.

(2)

(f)

The hiker walks at 3.5 km h⁻¹. Calculate the time taken to reach the road, to the nearest minute. Use the unrounded distance.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 angle HAB = 90° − 52° = 38°	✓ 1
A1 angle HBA = 313° − 270° = 43°	✓ 1
Part (b)
M1 angles in a triangle sum to 180°	✓ 1
A1 180 − 38 − 43 = 99° (shown)	✓ 1
Part (c)
M1 AH/sin 43° = 8/sin 99°	✓ 1
A1 AH = 5.52 km	✓ 1
Part (d)
M1 BH/sin 38° = 8/sin 99°	✓ 1
A1 BH = 4.99 km	✓ 1
Part (e)
M1 perpendicular distance = AH × sin 38°	✓ 1
A1 3.40 km	✓ 1
Part (f)
M1 time = distance ÷ speed (in hours)	✓ 1
M1 multiplies by 60 for minutes	✓ 1
A1 58 minutes	✓ 1

Common error: Convert bearings into angles inside the triangle by sketching north lines at both towers; the bearing from B must be measured from B’s north line.

+ Add to test✓ In your test311IBMI3S-3-03r29-11Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams Paper 2Hard13 marks

Calculator

Ash-fall sensors are installed at A(2, 2), B(8, 2) and C(5, 7) on a rectangular island, where coordinates are in kilometres. The island occupies the region 0 ≤ x ≤ 10, 0 ≤ y ≤ 8. A Voronoi diagram is drawn for the island with the sensors as sites. The recorded ash depths are A: 14 mm, B: 9 mm and C: 22 mm.

(a)

Write down the equation of the perpendicular bisector of [AB].

(2)

(b)

Find the equation of the perpendicular bisector of [AC]. Give your answer in the form y = mx + c.

(3)

(c)

Hence find the coordinates of the Voronoi vertex inside the island where the edges from A, B and C meet.

(2)

(d)

Use nearest-neighbour interpolation to estimate the ash depth at the point (7, 5).

(2)

(e)

The bisector of [BC] is y = 0.6x + 0.6. The region of the island closer to C than to A or B is a polygon. Find its area.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 A and B have the same y-coordinate, so the bisector is vertical through the midpoint	✓ 1
A1 x = 5	✓ 1
Part (b)
A1 midpoint of AC is (3.5, 4.5)	✓ 1
A1 gradient of AC is 5/3, so perpendicular gradient is −3/5	✓ 1
A1 y = −0.6x + 6.6	✓ 1
Part (c)
M1 substitutes x = 5 into the equation from (b)	✓ 1
A1 (5, 3.6)	✓ 1
Part (d)
M1 compares squared distances to the sensors: A 34, B 10, C 8 (or equivalent)	✓ 1
A1 C is nearest, so 22 mm	✓ 1
Part (e)
M1 identifies the vertices (5, 3.6), (0, 6.6), (10, 6.6) and the top corners (0, 8) and (10, 8)	✓ 1
A1 rectangle 10 × 1.4 = 14	✓ 1
M1 triangle ½ × 10 × (6.6 − 3.6)	✓ 1
A1 total area 29 km²	✓ 1

Exam tip: To find a region's area, split it into shapes you can measure, such as a rectangle and a triangle. Find each vertex by intersecting edges or by meeting the boundary of the region.

+ Add to test✓ In your test312IBMI3S-3-01r34-08Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams Paper 2Hard12 marks

A cycling club plans a triangular race course ABC. Riders start at A and ride to B, then to C, then back to A, all in straight lines. AB is 14.0 km on a bearing of 072°. BC is 11.5 km on a bearing of 158°. In every part, use unrounded values from earlier parts in later working unless told otherwise.

(a)

Show that angle ABC is 94°.

(2)

(b)

Find the length of AC, in km, to 3 significant figures.

(3)

(c)

Find the size of angle BAC, to 3 significant figures.

(3)

(d)

Find the bearing of C from A, to the nearest degree.

(2)

(e)

A cyclist rides the whole course, from A round to A, at an average speed of 32 km h⁻¹. Find the time taken, in minutes, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of A from B is $072+180=252^\circ$	✓ 1
A1 $252-158=94^\circ$ (or equivalent angle reasoning with alternate or co-interior angles)	✓ 1
Part (b)
M1 cosine rule $AC^2=14^2+11.5^2-2(14)(11.5)\cos94^\circ$	✓ 1
A1 $AC^2=350.7\ldots$	✓ 1
A1 $AC=18.7$ km (18.727…)	✓ 1
Part (c)
M1 sine rule $\dfrac{\sin BAC}{11.5}=\dfrac{\sin94^\circ}{AC}$	✓ 1
A1 correct substitution with $AC=18.727\ldots$	✓ 1
A1 $BAC=37.8^\circ$ (37.78…; acute, since 94° is the obtuse angle of the triangle)	✓ 1
Part (d)
M1 $072+37.78\ldots$	✓ 1
A1 $110^\circ$ (109.78…; ECF from (c))	✓ 1
Part (e)
M1 total distance $14+11.5+18.727\ldots=44.227\ldots$ km divided by 32, then multiplied by 60	✓ 1
A1 82.9 minutes (82.93…; ECF from (b))	✓ 1

Common error: Reading the angle at B straight from the two bearings. Draw a north line at each point and use the rule that the bearing back is the forward bearing plus or minus 180°.

+ Add to test✓ In your test313IBMI3S-3-01r34-12Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams Paper 2Hard14 marks

A city has three bike-hire hubs: A(2, 2), B(10, 2) and C(6, 10), with coordinates in kilometres. A Voronoi diagram is drawn for the hubs.

(a)

Show that the equation of the edge between the cells of A and B is $x=6$.

(2)

(b)

Find the equation of the perpendicular bisector of BC. Give your answer in the form $y=mx+c$.

(3)

(c)

Find the coordinates of the Voronoi vertex where the three edges meet.

(2)

(d)

A customer is at (9, 6). Find the distance from the customer to hub B, and determine which hub is nearest. Give the distance to 3 significant figures.

(3)

(e)

A fourth hub, D(10, 8), is added. Find the equation of the perpendicular bisector of CD. Give your answer in the form $y=mx+c$.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AB $=(6,2)$ and AB is horizontal	✓ 1
A1 the perpendicular bisector is vertical through the midpoint, $x=6$	✓ 1
Part (b)
M1 midpoint of BC $=(8,6)$	✓ 1
M1 gradient of BC $=-2$ so perpendicular gradient $=\tfrac12$	✓ 1
A1 $y=\tfrac12x+2$	✓ 1
Part (c)
M1 substitute $x=6$ into $y=\tfrac12x+2$	✓ 1
A1 $(6,5)$	✓ 1
Part (d)
M1 distances from the customer to the hubs, using the distance formula	✓ 1
A1 hub B at 4.12 km (to A: 8.06 km, to C: 5 km)	✓ 1
A1 hub B is nearest	✓ 1
Part (e)
M1 midpoint of CD $=(8,9)$	✓ 1
M1 gradient of CD $=-\tfrac12$	✓ 1
M1 perpendicular gradient $=2$	✓ 1
A1 $y=2x-7$	✓ 1

Exam tip: To find a perpendicular bisector, do three things in order: midpoint, gradient of the segment, negative reciprocal. Substitute the midpoint into $y=mx+c$ to find $c$.

+ Add to test✓ In your test314IBMI3S-3-03r37-09Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard14 marks

A cargo ship leaves port A and sails 40 km on a bearing of 065° to a buoy B. It then sails 55 km on a bearing of 150° to port C.

(a)

Show that angle ABC is 95°.

(2)

(b)

Find the distance AC. Give your answer in km to 3 significant figures.

(3)

(c)

Find the bearing of C from A. Give your answer to the nearest degree.

(4)

(d)

A reef lies along the straight line AC. Find the shortest distance from the buoy B to this line. Give your answer in km to 3 significant figures.

(3)

(e)

The ship returns directly from C to A at a constant speed of 18 km h⁻¹. Find the time taken. Use the unrounded value of AC. Give your answer in hours to 2 decimal places.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the bearing of A from B is $065^{\circ}+180^{\circ}=245^{\circ}$ (or uses co-interior angles with the north lines)	✓ 1
A1 $245^{\circ}-150^{\circ}=95^{\circ}$ (AG)	✓ 1
Part (b)
M1 cosine rule $AC^2=40^2+55^2-2(40)(55)\cos95^{\circ}$	✓ 1
A1 $AC^2=5008.5$	✓ 1
A1 $AC=70.8$ km	✓ 1
Part (c)
M1 sine rule $\frac{\sin\widehat{BAC}}{55}=\frac{\sin95^{\circ}}{AC}$	✓ 1
A1 $\widehat{BAC}=50.7^{\circ}$ (acute, since the largest angle is at B)	✓ 1
M1 bearing $=065^{\circ}+\widehat{BAC}$	✓ 1
A1 116° (accept 115.7°)	✓ 1
Part (d)
M1 uses $\frac{1}{2}(40)(55)\sin95^{\circ}=\frac{1}{2}(AC)h$ or $h=40\sin\widehat{BAC}$	✓ 1
A1 correct substitution (area 1095.8 km², or $40\sin50.7^{\circ}$)	✓ 1
A1 31.0 km	✓ 1
Part (e)
M1 time = $\frac{70.77}{18}$	✓ 1
A1 3.93 hours	✓ 1

Exam tip: In a bearings problem, draw a north line at each turning point and find the interior angle using the fact that angles on a straight line sum to 180° or that co-interior angles add to 180°.

+ Add to test✓ In your test315IBMI3S-3-01r40-12Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Hard15 marks

A network of four weather stations is at A(0, 0), B(8, 0), C(2, 6) and D(8, 8), where coordinates are in kilometres. A Voronoi diagram is drawn with the stations as sites. Vertex V₁ is where the cell edges of A, B and C meet, and vertex V₂ is where the cell edges of B, C and D meet.

(a)

Find the equation of the edge between the cells of A and C, giving your answer in the form $y = mx + c$.

(3)

(b)

Find the coordinates of V₁.

(3)

(c)

Find the equation of the edge between the cells of B and C.

(3)

(d)

Find the coordinates of V₂.

(3)

(e)

Calculate the length of the line segment V₁V₂, correct to 3 significant figures.

(2)

(f)

State which two stations have cells that share the edge V₁V₂.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AC is (1, 3)	✓ 1
M1 gradient of AC is 3, so perpendicular gradient is $-\frac{1}{3}$	✓ 1
A1 $y = -\frac{1}{3}x + \frac{10}{3}$	✓ 1
Part (b)
M1 edge between A and B is $x = 4$	✓ 1
M1 substitute into the equation from part (a)	✓ 1
A1 V₁ = (4, 2)	✓ 1
Part (c)
M1 midpoint of BC is (5, 3)	✓ 1
M1 gradient of BC is $-1$, so perpendicular gradient is 1	✓ 1
A1 $y = x - 2$	✓ 1
Part (d)
A1 edge between B and D is $y = 4$	✓ 1
M1 substitute $y = 4$ into $y = x - 2$	✓ 1
A1 V₂ = (6, 4)	✓ 1
Part (e)
M1 $\sqrt{(6-4)^2 + (4-2)^2}$	✓ 1
A1 2.83 km	✓ 1
Part (f)
A1 B and C	✓ 1

Exam tip: After finding a vertex, check it is the same distance from each of its three sites, and no other site is closer.

+ Add to test✓ In your test316IBMI3S-3-01r40-13Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Hard13 marks

A conical heap of rail ballast has a circular base of radius $r$ metres. The sloping side makes an angle of 34° with the horizontal ground. The heap has volume 90 m³.

(a)

Show that the height of the heap is $h = r \tan 34°$.

(1)

(b)

Hence find the value of $r$, correct to 3 significant figures.

(3)

(c)

Find the height of the heap, correct to 3 significant figures.

(2)

(d)

Find the slant height of the cone, correct to 3 significant figures.

(2)

(e)

Calculate the curved surface area of the heap, correct to 3 significant figures. Use unrounded values.

(2)

(f)

A tarpaulin covering the curved surface must be 20% larger than the curved surface area to allow for overlaps. The tarpaulin costs €6.40 per m². Find the cost of the tarpaulin, to the nearest euro.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R1 right-angled triangle with the height opposite the 34° angle and the radius adjacent, so $\tan 34° = \frac{h}{r}$	✓ 1
Part (b)
M1 $V = \frac{1}{3}\pi r^2 h$	✓ 1
M1 $90 = \frac{1}{3}\pi r^3 \tan 34°$	✓ 1
A1 $r = 5.03$ m	✓ 1
Part (c)
M1 $h = 5.032 \times \tan 34°$	✓ 1
A1 3.39 m	✓ 1
Part (d)
M1 $l = \frac{r}{\cos 34°}$ (or Pythagoras)	✓ 1
A1 6.07 m	✓ 1
Part (e)
M1 curved area = $\pi r l$	✓ 1
A1 96.0 m²	✓ 1
Part (f)
M1 area needed = 1.2 × 95.95	✓ 1
M1 cost = area × 6.40	✓ 1
A1 €737	✓ 1

Exam tip: When the volume is given and the radius is unknown, substitute the height in terms of $r$ first so only one unknown remains.

+ Add to test✓ In your test317IBMI3S-3-01r40-14Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context Paper 2Hard13 marks

A sports scientist marks three points A, B and C on a field. The distance AB is 95 m, the distance BC is 70 m and angle BAC is 36°. The scientist wants to find the possible positions of C.

(a)

Use the sine rule to find the two possible sizes of angle ACB, correct to 3 significant figures.

(3)

(b)

For each possible triangle find the length of AC, correct to 3 significant figures.

(4)

(c)

A later measurement shows that AC is greater than 100 m. State the size of angle ACB in this case.

(1)

(d)

For the triangle in part (c), calculate the area, correct to 3 significant figures. Use unrounded values.

(2)

(e)

For the other triangle, use the cosine rule to verify that BC is 70 m, correct to 2 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{\sin C}{95} = \frac{\sin 36°}{70}$	✓ 1
A1 $C = 52.9°$	✓ 1
A1 $C = 127°$ (the obtuse angle $180° - 52.9°$)	✓ 1
Part (b)
M1 angle ABC = $180° - 36° - C$ (91.1° or 16.9°)	✓ 1
M1 sine rule: $AC = \frac{70 \sin(ABC)}{\sin 36°}$	✓ 1
A1 119 m	✓ 1
A1 34.6 m	✓ 1
Part (c)
A1 $C = 52.9°$	✓ 1
Part (d)
M1 area = $\frac{1}{2} \times 95 \times AC \times \sin 36°$	✓ 1
A1 3320 m²	✓ 1
Part (e)
M1 $BC^2 = 95^2 + 34.6^2 - 2(95)(34.6)\cos 36°$	✓ 1
M1 $BC^2 = 9025 + 1197 - 5319$ (approximately 4903)	✓ 1
A1 $BC = 70$ m	✓ 1

Common error: The sine rule can give two angles between 0° and 180°; always check whether the obtuse angle also makes a valid triangle.

+ Add to test✓ In your test318IBMI3S-3-03r42-04Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; 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Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; 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volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard12 marks

A lakeside park is the rectangle 0 ≤ x ≤ 10, 0 ≤ y ≤ 8, where 1 unit represents 100 m. Three refreshment kiosks are at A(2, 2), B(8, 2) and C(5, 7). A Voronoi diagram is drawn for the kiosks, showing which kiosk is nearest to each point of the park.

(a)

Write down the equation of the perpendicular bisector of AB.

(1)

(b)

Find the equation of the perpendicular bisector of AC, in the form y = mx + c.

(3)

(c)

Find the coordinates of the Voronoi vertex where the edges of cells A, B and C meet.

(2)

(d)

The cell of kiosk A is the region of the park with x ≤ 5 that lies below the line found in (b). Find the area of this cell, in square units.

(4)

(e)

By symmetry, the cell of B has the same area as the cell of A. Find the area of the cell of kiosk C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x = 5	✓ 1
Part (b)
A1 midpoint of AC = (3.5, 4.5) and gradient of AC = 5/3	✓ 1
M1 perpendicular gradient = −3/5 used with the midpoint	✓ 1
A1 y = −0.6x + 6.6	✓ 1
Part (c)
M1 substitute x = 5 into the equation from (b)	✓ 1
A1 (5, 3.6)	✓ 1
Part (d)
M1 correct description of the cell as a trapezium (or an integral of y = −0.6x + 6.6 from 0 to 5)	✓ 1
A1 parallel sides 6.6 (at x = 0) and 3.6 (at x = 5)	✓ 1
M1 area = ½(6.6 + 3.6) × 5	✓ 1
A1 25.5 square units	✓ 1
Part (e)
M1 total area 10 × 8 = 80 and 80 − 2 × 25.5	✓ 1
A1 29 square units	✓ 1

Exam tip: Use any symmetry in the arrangement of sites to reduce the work: the cells of two mirror-image sites have equal area.

+ Add to test✓ In your test319IBMI3S-3-03r42-08Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; 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Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard12 marks

A maintenance helicopter flies from wind farm A to wind farm B, a distance of 25 km on a bearing of 038°. It then flies 40 km on a bearing of 125° to wind farm C.

(a)

Show that angle ABC = 93°.

(2)

(b)

Calculate the direct distance AC, giving your answer to 3 significant figures.

(3)

(c)

Find the bearing of C from A, giving your answer to the nearest 0.1°.

(3)

(d)

Find the shortest distance from B to the straight line AC, giving your answer to 3 significant figures.

(2)

(e)

On a particular day the helicopter returns directly from C to A instead of repeating the route via B. It burns 4.0 litres of fuel per kilometre. Calculate the fuel saved on the direct route compared with the route C to B to A, to the nearest litre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of A from B is 038° + 180° = 218°	✓ 1
A1 218° − 125° = 93°	✓ 1
Part (b)
M1 AC² = 25² + 40² − 2 × 25 × 40 × cos 93°	✓ 1
A1 AC² = 2329.7	✓ 1
A1 AC = 48.3 km	✓ 1
Part (c)
M1 sine rule, sin(BAC)/40 = sin 93°/AC	✓ 1
A1 angle BAC = 55.9°	✓ 1
A1 bearing = 038° + 55.9° = 93.9°	✓ 1
Part (d)
M1 25 sin(BAC) (or ½ × 25 × 40 × sin 93° = ½ × AC × h)	✓ 1
A1 20.7 km	✓ 1
Part (e)
M1 (25 + 40 − AC) × 4.0 (extra distance 16.7 km)	✓ 1
A1 67 litres	✓ 1

Common error: A bearing is measured clockwise from north. After using the sine rule to find an angle inside the triangle, add or subtract it from the known bearing and check against a sketch.

+ Add to test✓ In your test320IBMI3S-3-03r42-13Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard12 marks

The glass roof of a library atrium is a right pyramid with a square base of side 12 m and perpendicular height 9 m. The four triangular faces are glass. The volume of a pyramid is $\frac{1}{3} \times$ base area $\times$ height.

(a)

Calculate the slant height of a triangular face, measured from the midpoint of a base edge to the apex. Give your answer to 3 significant figures.

(2)

(b)

Calculate the total area of glass, to 3 significant figures.

(2)

(c)

Calculate the volume enclosed by the roof.

(2)

(d)

Find the angle between a triangular face and the base, to the nearest 0.1°.

(2)

(e)

Find the angle between a sloping edge (from a corner of the base to the apex) and the base, to the nearest 0.1°.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(9² + 6²)	✓ 1
A1 10.8 m	✓ 1
Part (b)
M1 4 × ½ × 12 × slant height (using the unrounded value)	✓ 1
A1 260 m² (260 m²)	✓ 1
Part (c)
M1 ⅓ × 12² × 9	✓ 1
A1 432 m³	✓ 1
Part (d)
M1 tan θ = 9/6	✓ 1
A1 56.3°	✓ 1
Part (e)
M1 diagonal of base = √(12² + 12²)	✓ 1
A1 half-diagonal = 8.49 m	✓ 1
M1 tan θ = 9/half-diagonal	✓ 1
A1 46.7°	✓ 1

Common error: The angle between a sloping edge and the base uses the half-diagonal of the base, not the half-side, which gives the angle of a face instead.

+ Add to test✓ In your test321IBMI3S-3-03r44-11Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Hard13 marks

A volcano is modelled as a frustum: a solid cone of base radius 3.2 km and height 1.6 km with the top cone of height 0.4 km removed (the cut is parallel to the base). Lengths are in kilometres.

(a)

Calculate the volume of the complete cone, giving your answer to 3 significant figures.

(2)

(b)

Find the radius of the top circular face of the frustum.

(2)

(c)

Find the volume of the frustum, giving your answer to 3 significant figures.

(3)

(d)

Write the volume of the frustum in m³ in the form a × 10^k, where 1 ≤ a < 10 and k is an integer.

(2)

(e)

The rock has density 2600 kg m⁻³. Find the mass of the frustum, giving your answer in the form a × 10^k kg with a to 3 significant figures. Use the unrounded volume from (c).

(2)

(f)

A student models the volcano as the complete cone instead of the frustum. Calculate the percentage error in the volume that results from this simplification. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 V = ⅓π(3.2)²(1.6)	✓ 1
A1 17.2 km³ (17.157…)	✓ 1
Part (b)
M1 similar triangles: r/0.4 = 3.2/1.6	✓ 1
A1 r = 0.8 km	✓ 1
Part (c)
M1 volume of removed cone = ⅓π(0.8)²(0.4) = 0.268… km³	✓ 1
M1 subtraction 17.157… − 0.268…	✓ 1
A1 16.9 km³	✓ 1
Part (d)
M1 1 km³ = 10⁹ m³	✓ 1
A1 1.69 × 10¹⁰ m³	✓ 1
Part (e)
M1 mass = density × volume = 2600 × 1.6889… × 10¹⁰	✓ 1
A1 4.39 × 10¹³ kg	✓ 1
Part (f)
M1 percentage error = \|17.157… − 16.889…\|/16.889… × 100	✓ 1
A1 1.59%	✓ 1

Common error: Percentage error is measured relative to the exact (frustum) value, not the approximate (full cone) value.

+ Add to test✓ In your test322IBMI3S-3-03r44-12Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Hard13 marks

Two ships leave port P at the same time. Ship A sails on a bearing of 035° at a constant speed of 16 km h⁻¹. Ship B sails on a bearing of 148° at a constant speed of 12 km h⁻¹. A lighthouse L stands 20 km from P on a bearing of 100°.

(a)

Find the size of angle APB.

(2)

(b)

Find the distance between the ships after 2.5 hours. Give your answer to 3 significant figures.

(3)

(c)

Find the bearing of ship B from ship A at this time, giving your answer to the nearest degree. Use the unrounded distance from (b).

(3)

(d)

The ships are travelling at constant speeds on constant bearings. Find the time, in hours, after leaving port at which they are 100 km apart. Give your answer to 3 significant figures.

(2)

(e)

Find the distance from ship A to the lighthouse L after 2.5 hours. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 148° − 35°	✓ 1
A1 113°	✓ 1
Part (b)
M1 AP = 40 km and BP = 30 km	✓ 1
M1 cosine rule: d² = 40² + 30² − 2 × 40 × 30 × cos 113°	✓ 1
A1 58.6 km	✓ 1
Part (c)
M1 sine rule: sin(PAB)/30 = sin 113°/58.63…	✓ 1
A1 angle PAB = 28.1°	✓ 1
A1 bearing = (035° + 180°) − 28.1° = 187°	✓ 1
Part (d)
M1 distance apart is proportional to time: rate = 58.63…/2.5 = 23.45… km h⁻¹	✓ 1
A1 100/23.45… = 4.26 hours	✓ 1
Part (e)
M1 angle APL = 100° − 35° = 65°	✓ 1
M1 cosine rule: AL² = 40² + 20² − 2 × 40 × 20 × cos 65°	✓ 1
A1 36.4 km	✓ 1

Exam tip: When two objects move at constant velocities from the same point, the separation grows in proportion to time, so one cosine-rule calculation gives every later separation.

+ Add to test✓ In your test323IBMI3S-3-03r44-14Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules Paper 2Hard12 marks

A volcanic island occupies the square region 0 ≤ x ≤ 10, 0 ≤ y ≤ 10 on a map in which one unit represents 1 km. Three monitoring stations are at A(1, 2), B(7, 2) and C(4, 8). A Voronoi diagram is drawn for the three stations. The perpendicular bisector of BC has equation y = 0.5x + 2.25.

(a)

Write down the equation of the boundary between the regions of A and B.

(2)

(b)

Find the equation of the perpendicular bisector of AC, giving your answer in the form y = mx + c.

(3)

(c)

Find the coordinates of the vertex where the three regions meet.

(2)

(d)

The region of C is bounded by the edges of the island and by parts of the two bisectors. Find the area of this region.

(4)

(e)

The 1800 residents of the island are spread uniformly over its area. Find the number of residents whose nearest station is C.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of AB is (4, 2) and AB is horizontal, so the bisector is vertical	✓ 1
A1 x = 4	✓ 1
Part (b)
M1 midpoint of AC = (2.5, 5) and gradient of AC = 2	✓ 1
M1 perpendicular gradient −0.5 and substitution of midpoint	✓ 1
A1 y = −0.5x + 6.25	✓ 1
Part (c)
M1 substitute x = 4 into one bisector	✓ 1
A1 (4, 4.25)	✓ 1
Part (d)
M1 bisector of AC meets x = 0 at (0, 6.25) and bisector of BC meets x = 10 at (10, 7.25)	✓ 1
A1 both boundary points correct	✓ 1
M1 valid method for the area, e.g. area of the square above y = 4.25 minus the two triangles below the bisectors, or the shoelace formula on (0, 6.25), (4, 4.25), (10, 7.25), (10, 10), (0, 10)	✓ 1
A1 44.5 km²	✓ 1
Part (e)
A1 44.5/100 × 1800 = 801	✓ 1

Exam tip: To find the area of a Voronoi cell on a bounded map, locate where each edge meets the boundary first, then split the cell into simple shapes.

+ Add to test✓ In your test324IBMI3S-3-01r46-08Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard14 marks

Three tyre-supply points for a Formula 1 circuit are located at A(0, 0), B(10, 0) and C(4, 8) on a plan where one unit represents 100 m. A mechanic always walks to the nearest supply point, so the regions served form a Voronoi diagram.

(a)

Write down the equation of the perpendicular bisector of AB.

(1)

(b)

Find the equation of the perpendicular bisector of AC, giving your answer in the form y = mx + c.

(3)

(c)

Hence find the coordinates of the Voronoi vertex where the edges from parts (a) and (b) meet.

(2)

(d)

Show that the perpendicular bisector of BC passes through the vertex found in part (c).

(4)

(e)

A mechanic stands at Z(3, 5). Determine which supply point is nearest, showing the three distances (to 3 significant figures) in plan units.

(2)

(f)

Calculate the real distance, in metres, from the vertex to supply point A. Give your answer to the nearest metre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x = 5	✓ 1
Part (b)
A1 midpoint of AC = (2, 4)	✓ 1
M1 gradient of AC = 2, so the perpendicular gradient is −1/2	✓ 1
A1 y − 4 = −½(x − 2), so y = −0.5x + 5	✓ 1
Part (c)
M1 substitutes x = 5 into y = −0.5x + 5	✓ 1
A1 vertex = (5, 2.5)	✓ 1
Part (d)
A1 midpoint of BC = (7, 4)	✓ 1
M1 gradient of BC = 8 / (−6) = −4/3, so the perpendicular gradient is 3/4	✓ 1
A1 equation y = 0.75x − 1.25	✓ 1
R1 when x = 5, y = 3.75 − 1.25 = 2.5, so the line passes through (5, 2.5)	✓ 1
Part (e)
M1 calculates distances ZA = √34 = 5.83, ZB = √74 = 8.60 and ZC = √10 = 3.16	✓ 1
A1 C is nearest	✓ 1
Part (f)
M1 distance = √(5² + 2.5²) = √31.25 = 5.59… units	✓ 1
A1 distance = 559 m	✓ 1

Exam tip: Check each Voronoi vertex by testing that it is the same distance from all three sites; the check takes seconds and catches arithmetic slips.

+ Add to test✓ In your test325IBMI3S-3-01r46-13Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard12 marks

A hospital oxygen storage tank is made from a cylinder with a hemisphere attached to each end. The radius of the cylinder and of each hemisphere is r metres. The overall length of the tank, from the far point of one hemisphere to the far point of the other, is 3.40 m.

(a)

Show that the volume of the tank is V = πr²(3.4 − (2/3)r) m³.

(3)

(b)

The tank must hold exactly 1.20 m³. Find the value of r, to 3 significant figures, with 0 < r < 1.7.

(3)

(c)

Find the length of the cylindrical part of the tank, to 3 significant figures. Use the unrounded value of r.

(1)

(d)

Calculate the total outer surface area of the tank, in m², to 3 significant figures. Use the unrounded values of r and the cylinder length.

(3)

(e)

The outside of the tank is coated at a cost of €4.80 per m². Using the 3 significant figure surface area from part (d), find the cost of coating the tank, to the nearest cent.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 length of cylindrical part = 3.4 − 2r	✓ 1
M1 V = πr²(3.4 − 2r) + (4/3)πr³ (two hemispheres make one sphere)	✓ 1
A1 simplifies: πr²(3.4 − 2r + (4/3)r) = πr²(3.4 − (2/3)r) (AG)	✓ 1
Part (b)
M1 equates πr²(3.4 − (2/3)r) = 1.2	✓ 1
M1 solves the equation using a graphing or equation-solving tool	✓ 1
A1 r = 0.347 m	✓ 1
Part (c)
A1 length = 3.4 − 2 × 0.3472… = 2.71 m	✓ 1
Part (d)
M1 curved area of cylinder = 2πr × (cylinder length) and area of two hemispheres = 4πr²	✓ 1
M1 total area = 2π × 0.3472… × 2.7056 + 4π × 0.3472…² (or equivalent)	✓ 1
A1 surface area = 7.42 m²	✓ 1
Part (e)
M1 7.42 × 4.80	✓ 1
A1 cost = €35.62	✓ 1

Exam tip: When a 'show that' result is given, write every line of the derivation; you cannot work backwards from the printed answer.

+ Add to test✓ In your test326IBMI3S-3-03r46-13Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard13 marks

Three mobile-phone masts are at A(2, 2), B(10, 2) and C(6, 10), where one unit represents 1 km. The Voronoi diagram for these sites has a vertex at V(6, 5).

(a)

Show that V is equidistant from A, B and C.

(2)

(b)

A new mast D is built at V. Find the equation of the perpendicular bisector of AD, in the form $y = mx + c$.

(3)

(c)

The perpendicular bisector of CD is $y = 7.5$. Find the coordinates of the vertex where the edges AD and CD meet.

(2)

(d)

The cell of D is a triangle with vertices (1, 7.5), (11, 7.5) and a third vertex T where the edges AD and BD meet, on the line $x = 6$. Find the coordinates of T.

(2)

(e)

Find the area of the cell of D, in km², to 3 significant figures.

(2)

(f)

There are on average 150 customers per km². Using the exact area, find the expected number of customers served by D.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 distance VA = $\sqrt{4^2+3^2}$ = 5 (and VB = 5 by the same working)	✓ 1
A1 VC = 5, so all three distances are equal	✓ 1
Part (b)
M1 midpoint of AD = (4, 3.5) and gradient of AD = 3/4	✓ 1
M1 perpendicular gradient = −4/3 and substitution	✓ 1
A1 $y = -\tfrac43 x + \tfrac{53}{6}$ (equivalent forms accepted)	✓ 1
Part (c)
M1 substitutes $y = 7.5$ into the equation from (b)	✓ 1
A1 (1, 7.5)	✓ 1
Part (d)
M1 substitutes $x = 6$ into the equation from (b)	✓ 1
A1 $\left(6, \tfrac{5}{6}\right)$	✓ 1
Part (e)
M1 area = ½ × base 10 × height (7.5 − 5/6)	✓ 1
A1 33.3 km² (exactly 100/3)	✓ 1
Part (f)
M1 150 × 100/3	✓ 1
A1 5000 customers	✓ 1

Exam tip: A Voronoi vertex is the circumcentre of the three sites around it; checking equal distances is a quick test of your coordinates.

+ Add to test✓ In your test327IBMI3S-3-03r46-14Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Right-angled trigonometry, bearings and the sine and cosine rules; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams; Voronoi diagrams; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; Volume and surface area in context; Right-angled trigonometry, bearings and the sine and cosine rules; volume and surface area in context; Voronoi diagrams Paper 2Hard12 marks

Three mountain huts H, K and L are connected by straight paths. Hut K is 6.0 km from H on a bearing of 035°. Hut L is 9.1 km from K on a bearing of 128°. Hut L is 1150 m higher than hut H.

(a)

Show that angle HKL = 87°.

(2)

(b)

Find the distance HL, in km, to 3 significant figures.

(3)

(c)

Find the bearing of L from H, to the nearest degree.

(3)

(d)

Find the shortest distance from K to the straight line HL, to 3 significant figures.

(2)

(e)

Find the angle of elevation of L from H, to 3 significant figures. Use the unrounded value of HL.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of H from K = 035° + 180° = 215°	✓ 1
A1 215° − 128° = 87° (AG)	✓ 1
Part (b)
M1 cosine rule $HL^2 = 6.0^2 + 9.1^2 - 2\times6.0\times9.1\cos 87°$	✓ 1
A1 correct substitution, $HL^2 = 113.1…$	✓ 1
A1 10.6 km	✓ 1
Part (c)
M1 sine rule $\sin H = \dfrac{9.1\sin 87°}{HL}$	✓ 1
A1 angle KHL = 58.7°	✓ 1
A1 bearing = 035° + 58.7° = 094°	✓ 1
Part (d)
M1 distance = HK sin(angle KHL) = 6.0 × sin 58.7…°	✓ 1
A1 5.13 km	✓ 1
Part (e)
M1 $\tan\theta = \dfrac{1.15}{10.63…}$	✓ 1
A1 6.17°	✓ 1

Common error: Keep unrounded values from earlier parts in your calculator; rounding HL to 10.6 km before part (e) gives 6.19°.

Marking point	Mark	Notes
Part (a)
M1 for substituting into \(V=\pi r^2 h\): \(\pi\times1.2^2\times2.5\)	✓ 1
A1 for \(V=11.3\) m³ (11.3097…)	✓ 1
Part (b)
A1 for 11 300 litres, using 1 m³ = 1000 litres; accept 11 310 or 11 309.7; follow through from their answer to part (a)	✓ 1
Part (c)
M1 for dividing their capacity by the rate, e.g. \(11309.7\div15\)	✓ 1
A1 for 754 minutes (753.98…), which is about 12.6 hours; accept 753 minutes as follow-through from 11 300 litres	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for Pythagoras with radius 18: \(h=\sqrt{70^2-18^2}\)	✓ 1
A1 for \(h=67.6\) cm (67.646…)	✓ 1
Part (b)
M1 for using \(\pi r l\): \(\pi\times18\times70\)	✓ 1
A1 for 3960 cm² (3958.4…)	✓ 1

Marking point	Mark	Notes
Part (a)
\(l=\sqrt{9^2+20^2}\) M1 ; l = 21.9 cm (21.932…) A1	✓ 2
Part (b)
\(V=\frac{1}{3}\pi(9^2)(20)\) M1 ; V = 1700 cm³ (1696.5…) A1	✓ 2
Part (c)
Curved surface area \(=\pi r l=\pi(9)(21.932…)\) M1 ; 620 cm² (620.10…) A1	✓ 2

Marking point	Mark	Notes
Part (a)
\(V=\frac{4}{3}\pi(0.325)^3\) M1 ; 0.144 m³ (0.14379…) A1	✓ 2
Part (b)
\(A=4\pi(0.325)^2\) M1 ; 1.33 m² (1.3273…) A1	✓ 2
Part (c)
1.3273… × 8.40 × 50 M1 ; 557.48 USD A1 (accept 557 or 557.5 USD; accept 558.60 USD from using 1.33)	✓ 2

Marking point	Mark	Notes
Part (a)
\(x = 3\) A1	✓ 1
\(y = 2\) A1	✓ 1
Part (b)
V(3, 2) A1	✓ 1
Part (c)
V is equidistant from P and Q and from P and R, so it is equidistant from Q and R and lies on the bisector of QR (or: the midpoint of QR is (3, 2)) R1	✓ 1
Part (d)
\(\sqrt{3^2 + 2^2}\) M1	✓ 1
3.61 km A1	✓ 1

Marking point	Mark	Notes
Part (a)
(i) \(x = 5\) A1	✓ 1
(ii) \(y = 3\) A1	✓ 1
Part (b)
(5, 3) A1	✓ 1
Part (c)
Distance = \(\sqrt{5^2 + 3^2}\) M1	✓ 1
= 5.83 km (equidistant from all three stations) A1	✓ 1

Marking point	Mark	Notes
Part (a)
Use of distance formula, e.g. XA = \(\sqrt{4^2 + 1^2}\) M1	✓ 1
XA = \(\sqrt{17}\) = 4.12 km, XB = \(\sqrt{13}\) = 3.61 km A1	✓ 1
XC = \(\sqrt{32}\) = 5.66 km A1	✓ 1
Part (b)
X is closest to B, so X lies in the cell of B R1	✓ 1
Estimated rainfall = 35 mm A1	✓ 1
Part (c)
\(\frac{\|35 - 39\|}{39} \times 100\) M1	✓ 1
= 10.3% A1	✓ 1

Marking point	Mark	Notes
Part (a)
M1 uses \(\sin 4.8^\circ=\dfrac{0.75}{L}\) or equivalent	✓ 1
A1 \(L=8.96\) m	✓ 1
Part (b)
M1 uses \(\tan 4.8^\circ=\dfrac{0.75}{x}\) (or Pythagoras with the answer to (a)) – ECF from (a) if Pythagoras is used	✓ 1
A1 \(x=8.93\) m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(V=\pi\times1.4^2\times2.2\)	✓ 1
A1 \(V=13.5\) m³	✓ 1
Part (b)
M1 converts to litres and divides: \(\dfrac{13547}{40}\) (using unrounded volume)	✓ 1
A1 339 minutes	✓ 1

Marking point	Mark	Notes
Part (a)
M1 attempt to use \(\tan 38° = \frac{h}{85}\)	✓ 1
A1 \(h = 66.4\) m	✓ 1
Part (b)
M1 attempt to use \(\cos 38° = \frac{85}{d}\) or Pythagoras with the unrounded height	✓ 1
A1 \(108\) m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 bearing of H from T is \(070° + 180° = 250°\) (or equivalent angle reasoning)	✓ 1
A1 \(250° - 160° = 90°\)	✓ 1
Part (b)
M1 attempt to use Pythagoras, \(\sqrt{6.4^2 + 4.8^2}\)	✓ 1
A1 HU = 8.0 km	✓ 1
Part (c)
M1 attempt to find angle UHT using \(\tan(\text{UHT}) = \frac{4.8}{6.4}\)	✓ 1
A1 angle UHT = 36.9°	✓ 1
A1 bearing = \(70° + 36.9° = 106.9°\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 attempt to use \(V = \pi r^2 h\)	✓ 1
A1 \(567\) m³	✓ 1
Part (b)
M1 multiply volume by 116	✓ 1
A1 £65,824	✓ 1

Marking point	Mark	Notes
Part (a)
M1 use of \(V=\pi r^2h\) with \(r=1.2\)	✓ 1
A1 6.786 (accept 6.79), so 6.79 m³ to 3 s.f.	✓ 1
Part (b)
M1 multiply volume by 18, i.e. 6.7858 × 18 = 122.15	✓ 1
A1 122 kg (122 kg)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 use of \(A=\pi rl\) with \(r=9\) and \(l=24\)	✓ 1
A1 678.6 so 679 cm²	✓ 1
Part (b)
M1 multiply by 1.15, 678.58 × 1.15 = 780.37	✓ 1
A1 780 cm²	✓ 1

Marking point	Mark	Notes
Part (a)
M1 use of \(V=\pi r^2h\) with \(r=18\), \(h=45\)	✓ 1
A1 \(V=45\,800\) cm³ (45 804.4…)	✓ 1
Part (b)
M1 multiply volume by density (ECF from (a)) and convert from g to kg (÷1000)	✓ 1
A1 35.7 kg (35.73…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\tan\theta=\dfrac{185}{2600}\)	✓ 1
A1 \(\theta=4.07^\circ\) (4.0696…)	✓ 1
Part (b)
M1 Pythagoras \(\sqrt{2600^2+185^2}\), or \(\dfrac{185}{\sin\theta}\) with the unrounded angle	✓ 1
A1 2607 m (2606.6…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\tan32^\circ=\dfrac{48}{DT}\) (angle of depression equals the angle of elevation of the drone from T)	✓ 1
A1 \(DT=76.8\) m (76.82…)	✓ 1
Part (b)
M1 \(\sin32^\circ=\dfrac{48}{\text{distance}}\), or Pythagoras with the unrounded value from (a)	✓ 1
A1 90.6 m (90.58…; accept ECF from (a))	✓ 1

Marking point	Mark	Notes
Part (a)
M1 use of the distance formula \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)	✓ 1
A1 two distances correct	✓ 1
A1 PA = 5.00 km, PB = 3.16 km, PC = 4.12 km	✓ 1
Part (b)
A1 station B; P lies in the Voronoi cell of site B	✓ 1

Marking point	Mark	Notes
Part (a)
M1 uses \(V=\frac{1}{3}\pi r^2h\)	✓ 1
A1 correct substitution \(\frac{1}{3}\pi(4.5)^2(1.8)\)	✓ 1
A1 38.2 km³	✓ 1
Part (b)
M1 \(\tan\theta=\frac{1.8}{4.5}\)	✓ 1
A1 21.8°	✓ 1

Marking point	Mark	Notes
Part (a)
M1 uses \(\pi r^2h\) for the cylinder (0.603 m³) and \(\frac{2}{3}\pi r^3\) for the hemisphere (0.134 m³)	✓ 1
M1 adds the two volumes	✓ 1
A1 0.737 m³	✓ 1
Part (b)
M1 curved area of cylinder \(2\pi rh\) (3.02) plus hemisphere \(2\pi r^2\) (1.01)	✓ 1
A1 4.02 m²	✓ 1
Part (c)
M1 area for two coats \(2\times4.0212=8.04\) m², so \(8.04\div6=1.34\) tins, rounded up to 2 tins	✓ 1
A1 cost = 2 × 14.50 = €29.00	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\tan3.5^{\circ}=\frac{32}{d}\)	✓ 1
A1 \(d=523\) m	✓ 1
Part (b)
M1 second distance \(\frac{32}{\tan6^{\circ}}\)	✓ 1
A1 304 m	✓ 1
A1 distance sailed = 219 m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(V = \pi \times 0.80^2 \times 1.20\)	✓ 1
A1 2.41 m³	✓ 1
Part (b)
A1 volume is about 2410 litres (2413 litres unrounded)	✓ 1
M1 time = volume (in litres) ÷ 15	✓ 1
A1 161 minutes	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(V = \pi \times 4.2^2 \times 11.5\)	✓ 1
A1 637 cm³	✓ 1
Part (b)
M1 mass = 637.30… × 1.42	✓ 1
A1 905 g	✓ 1
Part (c)
M1 \(A = 2\pi \times 4.2 \times 11.5\)	✓ 1
A1 303 cm²	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for using \(\frac{1}{2}ab\sin C\): \(\frac{1}{2}\times2.1^2\times\sin70°\)	✓ 1
A1 for 2.07 m² (2.07202…)	✓ 1
Part (b)
A1 for \(2.07202\times3=6.22\) m³	✓ 1
Part (c)
M1 for the cosine rule, \(\sqrt{2.1^2+2.1^2-2(2.1)(2.1)\cos70°}\), or for \(2\times2.1\sin35°\)	✓ 1
A1 for 2.41 m (2.40902…)	✓ 1
Part (d)
M1 for finding the three rectangular face areas: two sloping rectangles \(2\times2.1\times3=12.6\) and the groundsheet \(2.40902\times3=7.227\)	✓ 1
M1 for adding the two triangular ends, \(2\times2.07202=4.144\), to the rectangular areas, so that all five faces are summed	✓ 1
A1 for 24.0 m² (23.971…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for the cylinder volume, \(\pi\times3.5^2\times12=461.8\)	✓ 1
M1 for the hemisphere volume, \(\frac{2}{3}\pi\times3.5^3=89.80\)	✓ 1
A1 for 552 m³ (551.61…)	✓ 1
Part (b)
M1 for \(0.8\times551.61\times750\)	✓ 1
A1 for 331 000 kg (331 tonnes)	✓ 1
Part (c)
M1 for adding the curved cylinder area and half a sphere: \(2\pi(3.5)(12)+2\pi(3.5)^2\)	✓ 1
A1 for 341 m² (340.86…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{1}{2}(0.9+0.5)\times0.4\)	✓ 1
A1 for 0.28 m²	✓ 1
Part (b)
M1 for \(0.28\times4=1.12\) m³	✓ 1
A1 for 1120 litres	✓ 1
Part (c)
M1 for a horizontal offset of 0.2 m on each side, with \(\tan\theta=\frac{0.4}{0.2}\)	✓ 1
A1 for 63.4°	✓ 1

Marking point	Mark	Notes
Part (a)
\(y = -\frac{1}{2}x + 4.5\), so gradient of edge = \(-\frac{1}{2}\) A1	✓ 1
The edge is the perpendicular bisector of AB, so gradient of AB = 2 R1	✓ 1
Part (b)
\(y = 2x - 3\) A1	✓ 1
Part (c)
Substituting: \(x + 2(2x - 3) = 9\) M1	✓ 1
\(x = 3\), \(y = 3\), so M(3, 3) A1	✓ 1
Part (d)
M is the midpoint of AB, so B = (2×3 − 2, 2×3 − 1) = (4, 5) A1	✓ 1
Part (e)
AB = \(\sqrt{2^2 + 4^2} = \sqrt{20}\) = 4.47 km A1	✓ 1

Marking point	Mark	Notes
Part (a)
\(x = 4\) A1	✓ 1
Part (b)
Midpoint (1, 3) and gradient of AC = 3, so perpendicular gradient \(-\frac{1}{3}\) M1	✓ 1
\(y = -\frac{1}{3}x + \frac{10}{3}\) (or equivalent, e.g. \(x + 3y = 10\)) A1	✓ 1
Part (c)
At \(x = 4\): \(y = -\frac{4}{3} + \frac{10}{3} = 2\), so V(4, 2) A1	✓ 1
Part (d)
VA = \(\sqrt{4^2 + 2^2}\), VB = \(\sqrt{4^2 + 2^2}\), VC = \(\sqrt{2^2 + 4^2}\) M1	✓ 1
Each distance = \(\sqrt{20}\) ≈ 4.47 km A1	✓ 1
Part (e)
The plant is 4.47 km from all three villages. Moving it anywhere else in the triangle brings it closer to at least one village, so 4.47 km is the greatest possible distance to the nearest village R1	✓ 1

Marking point	Mark	Notes
Part (a)
A1 \(180^\circ-72^\circ-55^\circ=53^\circ\)	✓ 1
Part (b)
M1 \(\dfrac{AT}{\sin55^\circ}=\dfrac{60}{\sin53^\circ}\)	✓ 1
A1 correct substitution	✓ 1
A1 \(AT=61.5\) m	✓ 1
Part (c)
M1 \(AT\times\sin72^\circ\) (ECF from (b))	✓ 1
A1 \(58.5\) m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(h=\sqrt{15^2-6^2}\)	✓ 1
A1 \(h=13.7\) cm	✓ 1
Part (b)
M1 \(V=\frac13\pi\times6^2\times13.75\) (ECF from (a))	✓ 1
A1 \(V=518\) cm³	✓ 1
Part (c)
M1 \(\pi rl=\pi\times6\times15\)	✓ 1
A1 \(283\) cm²	✓ 1