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International GCSE Maths · Pearson Edexcel 4MA1 · Unit 4

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 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →

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1IGME-061-02-02Area, surface area and volume Paper FEasy4 marks

A circular coaster has radius 4.8 cm.

(a)

Work out the area of the coaster. Give your answer correct to 3 significant figures.

(2)
(b)

Work out the circumference of the coaster. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
π × 4.82 (M1)✓ 1
= 72.4 cm2 (A1)✓ 1
Part (b)
2 × π × 4.8 or π × 9.6 (M1)✓ 1
= 30.2 cm (A1)✓ 1

Exam tip: Area uses r²; circumference uses the diameter. Write the formula before substituting so a slip in the calculator still earns the method mark.

2IGME-061-02-03Area, surface area and volume Paper FEasy5 marks

A gift box is a cuboid measuring 8 cm by 5 cm by 3.5 cm.

(a)

Work out the volume of the box.

(2)
(b)

Work out the total surface area of the box.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
8 × 5 × 3.5 (M1)✓ 1
= 140 cm3 (A1)✓ 1
Part (b)
Finds at least two different face areas from 40, 28, 17.5 (M1)✓ 1
2 × (40 + 28 + 17.5) (M1)✓ 1
= 171 cm2 (A1)✓ 1

Common error: A cuboid has six faces in three matching pairs. Adding only three face areas gives half the correct surface area.

3IGME-061-02-04Area, surface area and volume Paper FEasy4 marks

A cylindrical tin is completely filled with soup. The tin has radius 3.2 cm and height 11 cm.

(a)

Work out the volume of soup in the tin. Give your answer correct to 3 significant figures.

(2)
(b)

The soup has density 1.05 g/cm3.
Work out the mass of soup in the tin. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
π × 3.22 × 11 (M1)✓ 1
= 354 cm3 (A1)✓ 1
Part (b)
353.86... × 1.05 (M1)✓ 1
= 372 g (A1)✓ 1

Exam tip: Keep the unrounded volume in your calculator for part (b). Using 354 gives 371.7, which still rounds to 372, but early rounding can change the final figure in other questions.

4IGME-061-01-02Pythagoras' theorem and trigonometry Paper FEasy3 marks

ABC is a right-angled triangle with angle ABC = 90°.
AB = 7.4 cm and BC = 5.2 cm.
Work out the size of angle BAC.
Give your answer correct to 1 decimal place.

(a)

Calculate angle BAC.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for recognising tan: tan BAC = \(\frac{5.2}{7.4}\)✓ 1
M1 for BAC = tan−1\(\left(\frac{5.2}{7.4}\right)\)✓ 1
A1 for 35.1°✓ 1

Exam tip: Label the sides from the angle you want: BC is opposite angle A and AB is adjacent to it. Writing O, A and H on the sides first helps you choose the right ratio.

5IGME-061-01-03Pythagoras' theorem and trigonometry Paper FEasy5 marks

PQR is a right-angled triangle with angle PQR = 90°.
PR = 12 cm and angle QPR = 28°.

(a)

Calculate the length of QR.
Give your answer correct to 3 significant figures.

(3)
(b)

Use Pythagoras' theorem and your answer to part (a) to calculate the length of PQ.
Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for sin 28° = \(\frac{QR}{12}\)✓ 1
M1 for QR = 12 × sin 28°✓ 1
A1 for 5.63 cm✓ 1
Part (b)
M1 for \(\sqrt{12^2 - 5.63...^2}\)✓ 1
A1 for 10.6 cm✓ 1

Exam tip: Carry the unrounded value of QR from your calculator into part (b). Using a value rounded too early can change the final digit.

6IGME-061-01-04Pythagoras' theorem and trigonometry Paper FEasy3 marks

Point A has coordinates (−3, 4).
Point B has coordinates (5, −2).
Work out the length of the line segment AB.

(a)

Calculate the length of AB.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for horizontal distance 8 and vertical distance 6✓ 1
M1 for \(\sqrt{8^2 + 6^2}\)✓ 1
A1 for 10✓ 1

Common error: Subtracting a negative coordinate is easy to get wrong: from −3 to 5 is a distance of 8, not 2. Sketch the points to check the horizontal and vertical distances.

7IGME-061-00-03Angles, polygons and circle properties Paper FEasy4 marks

The straight lines AB and CD are parallel, with AB above CD. A is to the left of B and C is to the left of D. A straight line EF crosses AB at G and crosses CD at H, with G above H.
Angle AGH = 68°.

(a)

Find the size of angle GHD. Give a reason for your answer.

(2)
(b)

Find the size of angle GHC. Give a reason for your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Angle GHD = 68°✓ 1
Reason: alternate angles are equal✓ 1
Part (b)
Angle GHC = 180° − 68° = 112°✓ 1
Reason: either angles on a straight line add up to 180° (using angle GHD), or co-interior angles add up to 180° (using angle AGH)✓ 1

Exam tip: Write reasons using the correct geometric words, such as 'alternate angles are equal' or 'co-interior angles add up to 180°'. Informal names like 'Z angles' or 'C angles' do not earn the reason mark.

8IGME-061-01-05Pythagoras' theorem and trigonometry Paper FEasy3 marks

A surveyor stands on horizontal ground 40 m from the base of a vertical tower.
She uses a measuring instrument whose eyepiece is 1.5 m above the ground.
From the eyepiece, the angle of elevation of the top of the tower is 32°.

(a)

Work out the height of the tower.
Give your answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for tan 32° = \(\frac{h}{40}\), where h is the height of the top above the eyepiece, or 40 × tan 32° (= 24.99...)✓ 1
M1 for 40 × tan 32° + 1.5 (= 26.49...)✓ 1
A1 for 26.5 m✓ 1
(An answer of 25.0 m, with the 1.5 m not added, scores M1 M0 A0.)—

Common error: The angle is measured from the eyepiece, not from the ground. Add the 1.5 m height of the instrument after using tan to find the height above the eyepiece.

9IGME-061-01-01Pythagoras' theorem and trigonometry Paper FEasy3 marks

A ladder of length 6.5 m leans against a vertical wall. The foot of the ladder is on horizontal ground, 1.8 m from the bottom of the wall.
Work out how far up the wall the ladder reaches.
Give your answer correct to 3 significant figures.

(a)

Work out the height reached by the top of the ladder.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for 6.52 − 1.82 (= 42.25 − 3.24 = 39.01)✓ 1
M1 for \(\sqrt{6.5^2 - 1.8^2}\) or \(\sqrt{39.01}\)✓ 1
A1 for 6.25 m (accept 6.245… seen before rounding)✓ 1

Common error: The ladder is the hypotenuse, so you subtract the squares. Adding them gives 6.74 m, which is longer than the ladder and cannot be right.

10IGME-061-00-06Angles, polygons and circle properties Paper FEasy4 marks

The four angles of a quadrilateral are \(x\)°, \(2x\)°, \((3x − 20)\)° and 110°.

(a)

Find the value of \(x\).

(3)
(b)

Find the size of the largest angle in the quadrilateral.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses angle sum 360°: \(x + 2x + 3x − 20 + 110 = 360\)✓ 1
Simplifies to \(6x + 90 = 360\), so \(6x = 270\)✓ 1
\(x = 45\)✓ 1
Part (b)
The angles are 45°, 90°, 115° and 110°, so the largest is \(3(45) − 20 = 115\)°✓ 1

Common error: The angles of a quadrilateral add up to 360°, not 180°. Using 180° gives \(x = 15\), and then the angle \((3x − 20)\)° would be 25°. This looks reasonable but is wrong, so always check that your four angles add up to 360°.

11IGME-061-02-05Area, surface area and volume Paper FEasy5 marks

A lawn is made from a rectangle measuring 12 m by 9 m with a semicircle joined along one of the 9 m sides. The diameter of the semicircle is 9 m.

(a)

Work out the total area of the lawn. Give your answer correct to 3 significant figures.

(3)
(b)

Turf costs £6.40 per m2.
Work out the cost of turf for the whole lawn, using your unrounded area. Give your answer to the nearest penny.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Rectangle: 12 × 9 = 108 (B1)✓ 1
Semicircle: \(\frac{1}{2}\) × π × 4.52 (= 31.8...) (M1)✓ 1
Total = 140 m2 (from 139.8...) (A1)✓ 1
Part (b)
139.808... × 6.40 (M1)✓ 1
= £894.78 (from 894.775...) (A1)✓ 1

Common error: Using a radius of 9 m instead of 4.5 m, or forgetting to halve the circle, is the most frequent slip with semicircles.

12IGME-061-00-02Angles, polygons and circle properties Paper FEasy5 marks

Each interior angle of a regular polygon is 156°.

(a)

Write down the size of each exterior angle.

(1)
(b)

Work out the number of sides of the polygon.

(2)
(c)

Work out the sum of the interior angles of the polygon.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
180 − 156 = 24°✓ 1
Part (b)
360 ÷ 24✓ 1
15 sides✓ 1
Part (c)
(15 − 2) × 180 or 15 × 156✓ 1
2340°✓ 1

Common error: Do not divide 360 by the interior angle. The exterior angles of any polygon add up to 360°, so 360 must be divided by the exterior angle.

13IGME-061-00-04Angles, polygons and circle properties Paper FEasy3 marks

In triangle PQR, PQ = PR and angle QPR = 38°. The side QR is extended beyond R to the point S.

(a)

Work out the size of angle PQR.

(2)
(b)

Work out the size of angle PRS.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses the two equal base angles: (180 − 38) ÷ 2✓ 1
Angle PQR = 71°✓ 1
Part (b)
Angle PRS = 180 − 71 = 109° (or 38 + 71 = 109°), follow through from their answer to the first part✓ 1

Common error: The two equal angles are opposite the two equal sides. Because PQ = PR, the equal angles are at Q and R, not at P.

14IGME-061-02-01Area, surface area and volume Paper FEasy4 marks

A flower bed is in the shape of a trapezium. The parallel sides have lengths 7.5 m and 12.3 m. The perpendicular distance between the parallel sides is 6 m.

(a)

Work out the area of the flower bed.

(2)
(b)

Fertiliser is spread over the whole flower bed at a rate of 45 g per m2.
Work out the total mass of fertiliser needed. Give your answer in kilograms.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(\frac{1}{2}(7.5+12.3)\times 6\) (M1)✓ 1
= 59.4 m2 (A1)✓ 1
Part (b)
59.4 × 45 (= 2673 g), or 59.4 × 0.045; follow through from their area in part (a) (M1)✓ 1
= 2.673 kg (accept 2.67 kg) (A1)✓ 1

Common error: Multiplying 7.5 × 12.3 × 6, or forgetting the ½, gives the wrong area. Always add the parallel sides first. Then divide grams by 1000 to get kilograms, not by 100.

15IGME-061-00-01Angles, polygons and circle properties Paper FEasy3 marks

Three angles lie together on a straight line. Their sizes are \(2x\)°, \((3x + 15)\)° and 45°.

(a)

Find the value of \(x\).

(2)
(b)

Find the size of the largest of the three angles.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(2x + 3x + 15 + 45 = 180\) or \(5x + 60 = 180\)✓ 1
\(x = 24\)✓ 1
Part (b)
\(3(24) + 15 = 87\)° (the other angles are 48° and 45°)✓ 1

Exam tip: Angles on a straight line add up to 180°. Write the equation first, then collect like terms. That way the method mark is earned even if the algebra slips.

16IGME-061-02-06Area, surface area and volume Paper FMedium6 marks

A wooden doorstop is a triangular prism of length 15 cm. Its cross-section is a right-angled triangle in which the two shorter sides are 6 cm and 8 cm.

(a)

Work out the volume of the doorstop.

(2)
(b)

Work out the total surface area of the doorstop.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(\frac{1}{2}\) × 6 × 8 × 15 (M1)✓ 1
= 360 cm3 (A1)✓ 1
Part (b)
Hypotenuse = √(62 + 82) = 10 cm (M1)✓ 1
Two triangles: 2 × 24 = 48 (M1)✓ 1
Rectangles: 6 × 15 + 8 × 15 + 10 × 15 = 360 (M1)✓ 1
Total = 408 cm2 (A1)✓ 1

Exam tip: A triangular prism has five faces: two triangles and three rectangles. Use Pythagoras to find the width of the sloping rectangle first.

17IGME-061-01-06Pythagoras' theorem and trigonometry Paper FMedium5 marks

An isosceles triangle has two sides of length 9 cm and a base of length 11 cm.

(a)

Work out the perpendicular height of the triangle.
Give your answer correct to 3 significant figures.

(3)
(b)

Work out the area of the triangle.
Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 for using half the base, 5.5 cm✓ 1
M1 for \(\sqrt{9^2 - 5.5^2}\)✓ 1
A1 for 7.12 cm✓ 1
Part (b)
M1 for \(\frac{1}{2}\) × 11 × 7.12...✓ 1
A1 for 39.2 cm2✓ 1

Exam tip: The height of an isosceles triangle cuts the base in half and makes two right-angled triangles. Use 5.5 cm, not 11 cm, in Pythagoras' theorem.

18IGME-061-01-07Pythagoras' theorem and trigonometry Paper FMedium7 marks

A ship sails from port P for 35 km due north to point Q.
It then sails 48 km due east to point R.

(a)

Calculate the direct distance PR.
Give your answer correct to 3 significant figures.

(3)
(b)

Calculate the bearing of R from P.
Give your answer correct to 1 decimal place.

(3)
(c)

Write down the bearing of P from R.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for 352 + 482 (= 3529)✓ 1
M1 for \(\sqrt{3529}\)✓ 1
A1 for 59.4 km✓ 1
Part (b)
M1 for tan θ = \(\frac{48}{35}\) with θ the angle at P✓ 1
M1 for θ = tan−1\(\left(\frac{48}{35}\right)\)✓ 1
A1 for 053.9°✓ 1
Part (c)
B1 for 233.9° (follow through 180° + their answer to (b))✓ 1

Exam tip: Bearings are measured clockwise from north and written with three figures. The back bearing differs from the original bearing by 180°.

19IGME-061-00-07Angles, polygons and circle properties Paper FMedium7 marks

ABCDE is a regular pentagon. ABFG is a square drawn on the outside of the pentagon, so that the square and the pentagon share the side AB. The points E, A and G are joined to form triangle EAG.

(a)

Work out the size of each interior angle of the pentagon.

(2)
(b)

Work out the size of angle EAG.

(2)
(c)

Work out the size of angle AGE. Give a reason for your answer.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
(5 − 2) × 180 ÷ 5 or 180 − 360 ÷ 5✓ 1
108°✓ 1
Part (b)
Angles around point A sum to 360°: 360 − 108 − 90✓ 1
162°✓ 1
Part (c)
Reason: AE = AB (sides of the regular pentagon) and AB = AG (sides of the square), so triangle EAG is isosceles and its base angles are equal✓ 1
(180 − 162) ÷ 2✓ 1
9°✓ 1

Exam tip: The pentagon and the square have the same side length, so AE = AG. Spotting this hidden isosceles triangle is the key step, and stating it earns the reason mark.

20IGME-061-02-11Area, surface area and volume Paper FMedium6 marks

A water tank is a cuboid with internal dimensions 1.2 m by 0.8 m by 0.5 m. The tank is empty.
1 m3 = 1000 litres.

(a)

The base of the tank has area 0.96 m2. Change 0.96 m2 into cm2.

(1)
(b)

Work out the capacity of the tank in litres.

(2)
(c)

Water flows into the tank at a rate of 0.2 litres per second. Work out how long it takes to fill the tank completely. Give your answer in minutes.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
9600 cm2 (B1)✓ 1
Part (b)
1.2 × 0.8 × 0.5 = 0.48 m3 (M1)✓ 1
= 480 litres (A1)✓ 1
Part (c)
Their capacity ÷ 0.2, e.g. 480 ÷ 0.2, or a rate of 0.2 × 60 = 12 litres per minute (M1)✓ 1
2400 seconds, or their capacity ÷ 12 (M1)✓ 1
= 40 minutes (A1)✓ 1
Follow through from their capacity in the previous part for both M marks.—

Common error: 1 m² = 10 000 cm², not 100 cm². Square the length conversion factor for areas and cube it for volumes. Check the time unit too: a rate per second gives an answer in seconds, so divide by 60 to get minutes.

21IGME-061-01-08Pythagoras' theorem and trigonometry Paper FMedium4 marks

Points B, C and D lie in that order on a straight horizontal line.
Point A is vertically above B, so that AB is perpendicular to BD.
AB = 8 cm, angle ACB = 40° and angle ADB = 25°.
Work out the length of CD.
Give your answer correct to 3 significant figures.

(a)

Work out the length of CD, giving your answer correct to 3 significant figures.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for BC = \(\frac{8}{\tan 40°}\) (= 9.53...) or an equivalent first step, e.g. AC = \(\frac{8}{\sin 40°}\) (= 12.4...)✓ 1
M1 for BD = \(\frac{8}{\tan 25°}\) (= 17.15...) or an equivalent, e.g. angle CAD = 15° used with the sine rule in triangle ACD✓ 1
M1 for a complete method for CD, e.g. BD − BC or CD = \(\frac{AC \sin 15°}{\sin 25°}\)✓ 1
A1 for 7.62 cm (accept 7.62 to 7.63 from premature rounding)✓ 1

Common error: AB is the opposite side in both right-angled triangles and BC, BD are adjacent, so divide 8 by tan. Multiplying by tan gives lengths that are far too short. Keep full calculator values until the final subtraction.

22IGME-061-00-05Angles, polygons and circle properties Paper FMedium5 marks

T is a point on a circle with centre O. The radius of the circle is 5 cm.
The line TP is the tangent to the circle at T.
Angle TOP = 52°.

(a)

Write down the size of angle OTP. Give a reason for your answer.

(2)
(b)

Work out the size of angle OPT.

(1)
(c)

Calculate the length of TP. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Angle OTP = 90°✓ 1
Reason: a tangent is perpendicular to the radius at the point of contact✓ 1
Part (b)
Angle OPT = 180° − 90° − 52° = 38°✓ 1
Part (c)
TP = 5 × tan 52° (or an equivalent, e.g. TP = 5 ÷ tan 38°)✓ 1
TP = 6.40 cm✓ 1

Exam tip: Whenever a radius meets a tangent, mark the right angle straight away. This turns the problem into right-angled trigonometry.

23IGME-061-00-14Angles, polygons and circle properties Paper FHard7 marks

Regular polygons can sometimes be placed around a single point so that they fit exactly, with no gaps and no overlaps.

(a)

A regular pentagon and two identical regular polygons are placed around a point. Show that they cannot fit exactly.

(3)
(b)

A square, a regular hexagon and a regular polygon with \(n\) sides fit exactly around a point. Find the value of \(n\).

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
Interior angle of a regular pentagon = 180 − 360 ÷ 5 = 108°✓ 1
Each of the two polygons would need an interior angle of (360 − 108) ÷ 2 = 126°✓ 1
Exterior angle = 180 − 126 = 54°, and 360 ÷ 54 = 6.67 (3 s.f.), which is not a whole number, so the polygons cannot fit✓ 1
Part (b)
Uses square angle 90° and hexagon angle 120°✓ 1
Interior angle of third polygon = 360 − 90 − 120 = 150°✓ 1
Exterior angle = 180 − 150 = 30°✓ 1
\(n\) = 360 ÷ 30 = 12✓ 1

Exam tip: When polygons fit exactly around a point, their interior angles add up to 360°. A regular polygon exists only if 360 divided by its exterior angle is a whole number.

24IGME-061-00-12Angles, polygons and circle properties Paper FHard7 marks

A boat sails from port P to a buoy Q on a bearing of 065°. It then sails from Q to a lighthouse R on a bearing of 160°.

(a)

Work out the bearing of P from Q.

(2)
(b)

Work out the size of angle PQR.

(2)
(c)

The distance QP is equal to the distance QR. Work out the bearing of R from P.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
65 + 180✓ 1
245°✓ 1
Part (b)
245 − 160✓ 1
85°✓ 1
Part (c)
Angle QPR = (180 − 85) ÷ 2 = 47.5°✓ 1
65 + 47.5✓ 1
112.5°✓ 1

Exam tip: Draw a north line at every point and measure bearings clockwise from north. A back bearing differs from the original bearing by exactly 180°.

25IGME-061-03-01Similar shapes and circle theorems Paper HEasy4 marks

Triangle ABC is similar to triangle PQR. Angle A corresponds to angle P, angle B to angle Q and angle C to angle R.
AB = 6 cm, BC = 8 cm, PQ = 9 cm and PR = 10.5 cm.

(a)

Work out the length of QR.

(2)
(b)

Work out the length of AC.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Scale factor = 9 ÷ 6 = 1.5 (M1)—
QR = 8 × 1.5 = 12 cm (A1)—
✓ 2
Part (b)
AC = 10.5 ÷ 1.5 (M1)—
AC = 7 cm (A1)—
✓ 2

Exam tip: Match corresponding vertices first. Going from the small triangle to the large one, multiply by the scale factor. Going back, divide by it.

26IGME-061-03-04Similar shapes and circle theorems Paper HEasy5 marks

T is a point outside a circle with centre O. The line TP is a tangent to the circle at the point P.
OP = 5 cm and OT = 13 cm.

(a)

Work out the length of PT.

(3)
(b)

Work out the size of angle POT. Give your answer correct to 1 decimal place.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Angle OPT = 90° because a tangent is perpendicular to the radius (B1)—
PT2 = 132 − 52 (M1)—
PT = 12 cm (A1)—
✓ 3
Part (b)
cos POT = 5/13 (M1)—
Angle POT = 67.4° (A1)—
✓ 2

Exam tip: A tangent meeting a radius gives a right angle. Once you see it, Pythagoras and SOHCAHTOA can both be used.

27IGME-061-03-05Similar shapes and circle theorems Paper HEasy3 marks

Two photographs are mathematically similar rectangles.
The smaller photograph is 12 cm long and has an area of 96 cm2.
The larger photograph is 18 cm long.

(a)

Work out the area of the larger photograph.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Length scale factor = 18 ÷ 12 = 1.5 (M1)—
Area scale factor = 1.52 = 2.25 (M1)—
Area = 96 × 2.25 = 216 cm2 (A1)—
✓ 3

Common error: Multiplying the area by the length scale factor gives 144 cm², which is wrong. Areas scale by the square of the length scale factor.

28IGME-061-03-02Similar shapes and circle theorems Paper HEasy4 marks

A, B and C are points on a circle with centre O. C lies on the major arc AB.
Angle AOB = 124°.

(a)

Find the size of angle ACB. Give a reason for your answer.

(2)
(b)

Work out the size of angle OAB.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Angle ACB = 124 ÷ 2 = 62°✓ 1
Reason: the angle at the centre is twice the angle at the circumference, both subtended by the same arc AB✓ 1
Part (b)
OA = OB (radii), so triangle AOB is isosceles and angle OAB = (180 − 124) ÷ 2✓ 1
Angle OAB = 28°✓ 1

Exam tip: Two radii always form an isosceles triangle. Look for this whenever the centre is joined to two points on the circle, and state 'radii are equal' as your reason.

29IGME-061-03-03Similar shapes and circle theorems Paper HEasy4 marks

ABCD is a cyclic quadrilateral.
Angle ABC = (3x + 10)° and angle ADC = (2x + 20)°.

(a)

Work out the value of x.

(3)
(b)

Find the size of angle ADC.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses the fact that opposite angles of a cyclic quadrilateral add up to 180°: 3x + 10 + 2x + 20 = 180✓ 1
Simplifies to 5x + 30 = 180, so 5x = 150✓ 1
x = 30✓ 1
Part (b)
Angle ADC = 2 × 30 + 20 = 80°✓ 1

Common error: Do not set the two opposite angles equal to each other. In a cyclic quadrilateral, opposite angles add up to 180°. Check your answer: here 100° + 80° = 180°.

30IGME-061-03-07Similar shapes and circle theorems Paper HMedium5 marks

Two solid cones, P and Q, are mathematically similar.
The surface area of P is 50 cm2 and the surface area of Q is 162 cm2.
Cone P has a height of 10 cm and a volume of 250 cm3.

(a)

Work out the height of cone Q.

(3)
(b)

Work out the volume of cone Q.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Area scale factor = 162 ÷ 50 = 3.24 (M1)—
Length scale factor = √3.24 = 1.8 (M1)—
Height = 10 × 1.8 = 18 cm (A1)—
✓ 3
Part (b)
250 × 1.83 (M1)—
Volume = 1458 cm3 (A1)—
✓ 2

Common error: Using 3.24 as the length scale factor is wrong. Take the square root of the area ratio to get the length ratio, then cube that for volume.

31IGME-061-03-08Similar shapes and circle theorems Paper HMedium4 marks

A, B and C are points on a circle. The straight line DAE is the tangent to the circle at A. D and B are on the same side of the line AC.
Angle DAB = 58° and angle BAC = 47°.

(a)

Find the size of angle ACB. Give a reason for your answer.

(2)
(b)

Work out the size of angle ABC.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Angle ACB = 58° (B1)—
Reason: alternate segment theorem (the angle between a tangent and a chord equals the angle in the alternate segment) (B1)—
✓ 2
Part (b)
180 − 58 − 47, using the angle sum of triangle ABC (or angle CAE = 75° and the alternate segment theorem) (M1)—
Angle ABC = 75° (A1)—
✓ 2

Exam tip: For the alternate segment theorem, the angle between the tangent and chord AB equals the angle subtended by AB on the opposite side of the chord.

32IGME-061-02-07Area, surface area and volume Paper HMedium5 marks

A slice of cake is modelled as a sector of a circle with radius 9 cm and angle 140° at the centre.

(a)

Work out the area of the sector. Give your answer correct to 3 significant figures.

(2)
(b)

Work out the perimeter of the sector. Give your answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(\frac{140}{360}\) × π × 92 (M1)✓ 1
= 99.0 cm2 (98.96...) (A1)✓ 1
Part (b)
Arc = \(\frac{140}{360}\) × 2 × π × 9 (= 7π = 21.99...) (M1)✓ 1
Adds two radii: 21.99... + 9 + 9 (M1)✓ 1
= 40.0 cm (A1)✓ 1

Common error: The perimeter of a sector includes the two straight radii as well as the arc. Giving only the arc length loses the last two marks.

33IGME-061-02-09Area, surface area and volume Paper HMedium5 marks

A solid cone has base radius 6 cm and perpendicular height 8 cm.
[Curved surface area of a cone = \(\pi r l\), where \(l\) is the slant height]

(a)

Work out the slant height of the cone.

(2)
(b)

Work out the total surface area of the cone. Give your answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
√(62 + 82) (M1)✓ 1
= 10 cm (A1)✓ 1
Part (b)
Curved surface: π × 6 × 10 (= 60π) (M1)✓ 1
Base: π × 62 (= 36π) added (M1)✓ 1
= 302 cm2 (96π) (A1)✓ 1

Common error: Using the perpendicular height 8 cm in πrl instead of the slant height gives 48π for the curved surface, which is wrong.

34IGME-061-02-10Area, surface area and volume Paper HMedium5 marks

Triangle PQR has PQ = 8.4 cm, PR = 11.2 cm and angle QPR = 52°.

(a)

Work out the area of triangle PQR. Give your answer correct to 3 significant figures.

(2)
(b)

Work out the length of QR. Give your answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(\frac{1}{2}\) × 8.4 × 11.2 × sin 52° (M1)✓ 1
= 37.1 cm2 (A1)✓ 1
Part (b)
QR2 = 8.42 + 11.22 − 2 × 8.4 × 11.2 × cos 52° (M1)✓ 1
QR2 = 80.15... (M1)✓ 1
QR = 8.95 cm (A1)✓ 1

Exam tip: ½ab sin C needs the angle between the two given sides. Check that 52° is the included angle before substituting.

35IGME-061-01-09Pythagoras' theorem and trigonometry Paper HMedium6 marks

A closed box is in the shape of a cuboid.
The base of the box is a rectangle 12 cm long and 5 cm wide. The box is 7 cm high.
A thin straight rod fits exactly from a bottom corner of the box to the opposite top corner.

(a)

Calculate the length of the rod.
Give your answer correct to 3 significant figures.

(3)
(b)

Calculate the angle between the rod and the base of the box.
Give your answer correct to 1 decimal place.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for base diagonal \(\sqrt{12^2 + 5^2}\) = 13✓ 1
M1 for \(\sqrt{13^2 + 7^2}\) or \(\sqrt{12^2 + 5^2 + 7^2}\)✓ 1
A1 for 14.8 cm✓ 1
Part (b)
M1 for identifying the right-angled triangle with sides 13 cm and 7 cm✓ 1
M1 for tan−1\(\left(\frac{7}{13}\right)\) or equivalent using the rod length✓ 1
A1 for 28.3°✓ 1

Exam tip: The angle between a line and a plane is measured to the line's projection on that plane. Here the projection is the base diagonal, not an edge of the base.

36IGME-061-01-11Pythagoras' theorem and trigonometry Paper HMedium4 marks

Two boats leave a harbour H at the same time.
Boat A sails 14 km on a bearing of 040°.
Boat B sails 9 km on a bearing of 125°.
Calculate the distance between the two boats.
Give your answer correct to 3 significant figures.

(a)

Calculate the distance AB.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 for angle AHB = 85°✓ 1
M1 for AB2 = 142 + 92 − 2 × 14 × 9 × cos 85°✓ 1
M1 for AB = \(\sqrt{255.03...}\)✓ 1
A1 for 16.0 km✓ 1

Exam tip: Find the angle between the two directions first by subtracting the bearings: 125° − 40° = 85°. The cosine rule uses the angle between the two known sides.

37IGME-061-03-06Similar shapes and circle theorems Paper HMedium5 marks

Two bottles are mathematically similar.
The smaller bottle is 15 cm tall and the larger bottle is 24 cm tall.

(a)

The smaller bottle has a volume of 500 cm3. Work out the volume of the larger bottle.

(3)
(b)

The larger bottle has a surface area of 960 cm2. Work out the surface area of the smaller bottle.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Length scale factor = 24 ÷ 15 = 1.6✓ 1
Volume scale factor = 1.63 = 4.096✓ 1
Volume = 500 × 4.096 = 2048 cm3✓ 1
Part (b)
Area scale factor = 1.62 = 2.56, so the method is 960 ÷ 2.56✓ 1
Surface area = 375 cm2✓ 1

Exam tip: Write down k, k² and k³ at the start. Then use the right one for length, area or volume.

38IGME-061-03-09Similar shapes and circle theorems Paper HMedium7 marks

A, B, C and D are points on a circle. The chords AC and BD cross at the point X.
Angle BAC = 35° and angle ABD = 50°.
AX = 4 cm, XC = 6 cm and BX = 3 cm.

(a)

Find the size of angle BDC. Give a reason for your answer.

(2)
(b)

Work out the size of angle AXB.

(2)
(c)

Triangles AXB and DXC are similar, with A corresponding to D and B corresponding to C. Use this to work out the length of XD.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Angle BDC = 35°✓ 1
Reason: angles in the same segment are equal, since angles BAC and BDC both stand on arc BC✓ 1
Part (b)
Uses the angle sum of triangle AXB: 180 − 35 − 50✓ 1
Angle AXB = 95°✓ 1
Part (c)
Writes a correct ratio of corresponding sides: AX/DX = BX/CX✓ 1
4/XD = 3/6, so XD = (4 × 6)/3✓ 1
XD = 8 cm✓ 1
Also accept the intersecting chords result 4 × 6 = 3 × XD for full marks.—

Common error: Pairing the wrong sides gives the wrong answer. Match sides that lie opposite equal angles. AX is opposite angle ABX (50°), and DX is opposite the equal angle DCX (50°).

39IGME-061-01-10Pythagoras' theorem and trigonometry Paper HMedium6 marks

In triangle ABC, angle BAC = 48°, angle ABC = 67° and BC = 9.5 cm.

(a)

Calculate the length of AC.
Give your answer correct to 3 significant figures.

(3)
(b)

Calculate the area of triangle ABC.
Give your answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for \(\frac{AC}{\sin 67°} = \frac{9.5}{\sin 48°}\)✓ 1
M1 for AC = \(\frac{9.5 \sin 67°}{\sin 48°}\)✓ 1
A1 for 11.8 cm (accept 11.76 to 11.77)✓ 1
Part (b)
B1 for angle ACB = 65°✓ 1
M1 for a correct ½ab sin C using two known sides and the included angle, e.g. \(\frac{1}{2}\) × 9.5 × 11.767... × sin 65°✓ 1
A1 for 50.7 cm2 (accept answers from 50.6 to 50.8)✓ 1

Common error: For the area formula ½ab sin C, the angle must be the one between the two sides you use. Here BC and AC meet at C, so you need angle ACB. Keep the unrounded value of AC in your calculator: using 11.8 gives 50.8 rather than 50.7.

40IGME-061-00-08Angles, polygons and circle properties Paper HMedium4 marks

A, B, C and D are points, in that order, on a circle with centre O. B lies on the major arc AC and D lies on the minor arc AC.
Angle AOC = 130°, where 130° is the angle subtended at O by the minor arc AC.

(a)

Find the size of angle ABC. Give a reason for your answer.

(2)
(b)

Find the size of angle ADC. Give a reason for your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Angle ABC = 130 ÷ 2 = 65°✓ 1
Reason: the angle at the centre is twice the angle at the circumference subtended by the same arc✓ 1
Part (b)
Angle ADC = 180 − 65 = 115°✓ 1
Reason: opposite angles of a cyclic quadrilateral add up to 180°✓ 1
(Alternative: reflex angle AOC = 360 − 130 = 230°, so ADC = 230 ÷ 2 = 115°, because the angle at the centre is twice the angle at the circumference; this earns full marks.)—

Common error: Do not double the angle at the centre. The angle at the circumference is half the angle at the centre, not twice it. Always give the full circle theorem as your reason, not just the numbers.

41IGME-061-00-09Angles, polygons and circle properties Paper HMedium5 marks

A, B and C are points on a circle. The straight line TAS is the tangent to the circle at A. The angle between the tangent and the chord AB, on the side of AB away from C, is angle TAB = 58°. Angle BAC = 47°.

(a)

Write down the size of angle ACB. Give a reason for your answer.

(2)
(b)

Work out the size of angle ABC.

(1)
(c)

Find the size of angle SAC, the angle between the tangent and the chord AC on the side of AC away from B. Give a reason for your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Angle ACB = 58°✓ 1
Reason: alternate segment theorem, since the angle between a tangent and a chord equals the angle in the alternate segment✓ 1
Part (b)
Angle ABC = 180 − 58 − 47 = 75°✓ 1
Part (c)
Angle SAC = 75°✓ 1
Reason: alternate segment theorem (angle SAC = angle ABC). Also accept angles on a straight line sum to 180°, so 180 − 58 − 47✓ 1

Exam tip: In the alternate segment theorem, the matching angle is in the segment on the other side of the chord. Trace the chord, then look across it.

42IGME-061-00-11Angles, polygons and circle properties Paper HMedium6 marks

ABCD is a cyclic quadrilateral.
Angle DAB = \((3x + 10)\)°, angle BCD = \((2x + 30)\)°, angle ABC = \((2y + 25)\)° and angle CDA = \((3y - 5)\)°.

(a)

Find the value of \(x\).

(3)
(b)

Find the size of angle ABC.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses opposite angles of a cyclic quadrilateral sum to 180°: \(3x + 10 + 2x + 30 = 180\)✓ 1
\(5x + 40 = 180\), so \(5x = 140\)✓ 1
\(x = 28\)✓ 1
Part (b)
\(2y + 25 + 3y - 5 = 180\)✓ 1
\(5y + 20 = 180\), so \(5y = 160\) and \(y = 32\)✓ 1
Angle ABC = \(2(32) + 25 = 89\)°✓ 1

Common error: Only opposite angles of a cyclic quadrilateral add up to 180°. Adding neighbouring angles, such as angle A and angle B, gives the wrong equation.

43IGME-061-02-08Area, surface area and volume Paper HMedium5 marks

A bowl is in the shape of a hemisphere with internal radius 7.5 cm. The bowl is completely filled with water.
[Volume of a sphere = \(\frac{4}{3}\pi r^3\)]

(a)

Work out the volume of water in the bowl. Give your answer correct to 3 significant figures.

(2)
(b)

All the water is poured into an empty cylindrical jar with internal radius 5 cm. Work out the depth of water in the jar.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(\frac{1}{2}\) × \(\frac{4}{3}\) × π × 7.53 (= 281.25π)✓ 1
= 884 cm3 (883.57…)✓ 1
Part (b)
π × 52 × h = their volume from the first part (e.g. 281.25π or 883.57… or 884)✓ 1
h = their volume ÷ (25π)✓ 1
= 11.25 cm (accept answers in the range 11.2 to 11.3 cm, e.g. 11.3 cm from using 884)✓ 1

Exam tip: Leaving the volume as 281.25π lets the π cancel when you divide by 25π, giving an exact answer with no rounding error.

44IGME-061-03-10Similar shapes and circle theorems Paper HMedium7 marks

In triangle ABC, D is a point on AB and E is a point on AC so that DE is parallel to BC.
AD = 4 cm, DB = 6 cm, DE = 5 cm and AE = 3 cm.
The area of triangle ADE is 6 cm2.

(a)

Work out the length of BC.

(2)
(b)

Work out the length of EC.

(2)
(c)

Work out the area of the trapezium DBCE.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Scale factor = AB ÷ AD = 10 ÷ 4 = 2.5✓ 1
BC = 5 × 2.5 = 12.5 cm✓ 1
Part (b)
AC = 3 × 2.5 = 7.5 cm, or EC ÷ AE = DB ÷ AD = 6 ÷ 4✓ 1
EC = 7.5 − 3 = 4.5 cm✓ 1
Part (c)
Area scale factor = 2.52 = 6.25✓ 1
Area of ABC = 6 × 6.25 = 37.5 cm2✓ 1
Area of DBCE = 37.5 − 6 = 31.5 cm2✓ 1

Common error: Using DB ÷ AD = 1.5 as the scale factor is wrong. The similar triangles are ADE and ABC, so compare AB with AD. For areas, square the length scale factor.

45IGME-061-02-13Area, surface area and volume Paper HHard5 marks

Two perfume bottles, A and B, are mathematically similar.
Bottle A has volume 400 cm3 and surface area 240 cm2.
Bottle B has volume 1350 cm3.

(a)

Bottle A has height 12 cm. Work out the height of bottle B.

(2)
(b)

Work out the surface area of bottle B.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Length scale factor = ∛(1350 ÷ 400) = ∛3.375 = 1.5 (M1)✓ 1
12 × 1.5 = 18 cm (A1)✓ 1
Part (b)
Area scale factor = 1.52 (M1)✓ 1
= 2.25 (A1)✓ 1
240 × 2.25 = 540 cm2 (A1)✓ 1

Common error: Multiplying the surface area by the volume ratio 3.375 is wrong. Find the length scale factor first, then square it for areas.

46IGME-061-02-14Area, surface area and volume Paper HHard6 marks

A solid metal sphere of radius 6 cm is melted down and all the metal is recast into a solid cone with base radius 9 cm.
[Volume of a sphere = \(\frac{4}{3}\pi r^3\), volume of a cone = \(\frac{1}{3}\pi r^2 h\), curved surface area of a cone = \(\pi r l\)]

(a)

Show that the perpendicular height of the cone is \(\frac{32}{3}\) cm.

(3)
(b)

Work out the curved surface area of the cone. Give your answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Sphere volume: \(\frac{4}{3}\) × π × 63 = 288π (M1)✓ 1
\(\frac{1}{3}\) × π × 92 × h = 288π, i.e. 27πh = 288π (M1)✓ 1
h = \(\frac{288}{27} = \frac{32}{3}\) (A1)✓ 1
Part (b)
l = √(92 + (\(\frac{32}{3}\))2) (M1)✓ 1
l = 13.95... ; π × 9 × 13.95... (M1)✓ 1
= 395 cm2 (A1)✓ 1

Exam tip: In a 'show that' part, write every step including the equation with π on both sides; just stating 10.7 earns no marks.

47IGME-061-01-13Pythagoras' theorem and trigonometry Paper HHard7 marks

A solid pyramid has a square base ABCD with sides of length 10 cm.
The vertex V is vertically above the centre M of the base.
Each sloping edge VA, VB, VC and VD has length 13 cm.

(a)

Calculate the height VM of the pyramid.
Give your answer correct to 3 significant figures.

(3)
(b)

Calculate the angle between the edge VA and the base ABCD.
Give your answer correct to 1 decimal place.

(2)
(c)

Calculate the angle between the triangular face VAB and the base ABCD.
Give your answer correct to 1 decimal place.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for AM = \(\frac{1}{2}\sqrt{10^2 + 10^2}\) = \(5\sqrt{2}\) (= 7.07...)✓ 1
M1 for \(\sqrt{13^2 - (5\sqrt{2})^2}\) = \(\sqrt{119}\)✓ 1
A1 for 10.9 cm✓ 1
Part (b)
M1 for cos−1\(\left(\frac{5\sqrt{2}}{13}\right)\) or equivalent✓ 1
A1 for 57.0°✓ 1
Part (c)
M1 for tan−1\(\left(\frac{10.9...}{5}\right)\), using the midpoint of AB✓ 1
A1 for 65.4°✓ 1

Common error: For the angle between a face and the base, use the line from M to the midpoint of AB (length 5 cm), not the line from M to a corner. The two angles in (b) and (c) are different.

48IGME-061-01-14Pythagoras' theorem and trigonometry Paper HHard7 marks

A right-angled triangle has sides of length \(x\) cm, \((x + 7)\) cm and \((x + 8)\) cm, where \((x + 8)\) cm is the hypotenuse.

(a)

Show that \(x^2 - 2x - 15 = 0\)

(3)
(b)

Solve \(x^2 - 2x - 15 = 0\) and hence find the perimeter of the triangle.

(2)
(c)

Calculate the size of the smallest angle in the triangle.
Give your answer correct to 1 decimal place.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for \(x^2 + (x + 7)^2 = (x + 8)^2\)✓ 1
M1 for correct expansion \(x^2 + x^2 + 14x + 49 = x^2 + 16x + 64\)✓ 1
A1 for simplifying correctly to \(x^2 - 2x - 15 = 0\) with no errors✓ 1
Part (b)
M1 for \((x - 5)(x + 3) = 0\), giving \(x = 5\) (rejecting \(x = -3\))✓ 1
A1 for perimeter 30 cm✓ 1
Part (c)
M1 for tan−1\(\left(\frac{5}{12}\right)\) or sin−1\(\left(\frac{5}{13}\right)\)✓ 1
A1 for 22.6°✓ 1

Common error: \((x + 7)^2\) is not \(x^2 + 49\); you must include the middle term \(14x\). Also reject \(x = -3\), because a length cannot be negative.

49IGME-061-03-13Similar shapes and circle theorems Paper HHard5 marks

Two solid statues are mathematically similar and are made from the same metal.
The larger statue has a mass of 2.7 kg and the smaller statue has a mass of 0.8 kg.
The smaller statue is 16 cm tall. The larger statue has a surface area of 540 cm2.

(a)

Work out the height of the larger statue.

(3)
(b)

Work out the surface area of the smaller statue.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Volume scale factor = 2.7 ÷ 0.8 = 3.375✓ 1
Length scale factor = \(\sqrt[3]{3.375}\) = 1.5✓ 1
Height = 16 × 1.5 = 24 cm✓ 1
Part (b)
Area scale factor = 1.52 = 2.25, so use 540 ÷ 2.25✓ 1
Surface area = 240 cm2✓ 1

Common error: Both statues are made of the same metal, so the ratio of their masses equals the ratio of their volumes. Take the cube root to get the length scale factor, not the square root. Then square the length scale factor to get the area scale factor.

50IGME-061-03-12Similar shapes and circle theorems Paper HHard7 marks

A and B are points on a circle with centre O and radius 7 cm. TA and TB are tangents to the circle from the point T.
Angle ATB = 48°. C is a point on the major arc AB.

(a)

Work out the size of angle AOB.

(2)
(b)

Find the size of angle ACB. Give a reason for your answer.

(2)
(c)

Work out the length of TA. Give your answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses angle OAT = angle OBT = 90° (tangent meets radius at 90°), e.g. 360 − 90 − 90 − 48✓ 1
Angle AOB = 132°✓ 1
Part (b)
Angle ACB = 66°✓ 1
Reason: the angle at the centre is twice the angle at the circumference✓ 1
Part (c)
Angle ATO = 48 ÷ 2 = 24° (OT bisects angle ATB), or angle AOT = 132 ÷ 2 = 66°✓ 1
TA = 7 ÷ tan 24°, or TA = 7 × tan 66°✓ 1
TA = 15.7 cm✓ 1

Exam tip: OT bisects angle ATB because the two tangents from T are equal in length and the figure is symmetrical. Triangle OAT is right-angled at A, so you can use either the 24° angle at T or the 66° angle at O.

51IGME-061-01-12Pythagoras' theorem and trigonometry Paper HHard7 marks

A triangular plot of land has sides of length 7 m, 10 m and 12 m.

(a)

Calculate the size of the largest angle of the triangle.
Give your answer correct to 1 decimal place.

(3)
(b)

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

(2)
(c)

A straight path runs from the corner opposite the 12 m side to meet the 12 m side at right angles.
Calculate the length of the path.
Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for correct substitution into the cosine rule for the angle opposite the 12 m side, e.g. 122 = 72 + 102 − 2 × 7 × 10 × cos θ✓ 1
M1 for cos θ = \(\frac{7^2 + 10^2 - 12^2}{2 \times 7 \times 10}\) (= \(\frac{5}{140}\) = 0.0357...)✓ 1
A1 for 88.0°✓ 1
Part (b)
M1 for \(\frac{1}{2}\) × 7 × 10 × sin 87.95...° or any equivalent method, e.g. Heron's formula \(\sqrt{14.5 \times 7.5 \times 4.5 \times 2.5}\)✓ 1
A1 for 35.0 m2✓ 1
Part (c)
M1 for \(\frac{1}{2}\) × 12 × h = 34.97... or any equivalent method, e.g. h = 7 × sin(angle between the 7 m and 12 m sides)✓ 1
A1 for 5.83 m✓ 1

Exam tip: The largest angle is always opposite the longest side. Using the area from part (b) with the 12 m side as the base avoids a second long trigonometry calculation.

52IGME-061-02-12Area, surface area and volume Paper HHard5 marks

A solid cone has base radius 12 cm and perpendicular height 20 cm. A smaller cone of height 5 cm is cut from the top, parallel to the base, leaving a frustum.
[Volume of a cone = \(\frac{1}{3}\pi r^2 h\)]

(a)

Work out the radius of the circular top of the frustum.

(2)
(b)

Work out the volume of the frustum. Give your answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses similarity: scale factor 5 ÷ 20 = \(\frac{1}{4}\) or 12 × \(\frac{5}{20}\)✓ 1
= 3 cm✓ 1
Part (b)
Volume of large cone: \(\frac{1}{3}\) × π × 122 × 20 (= 960π)✓ 1
Volume of small cone: \(\frac{1}{3}\) × π × 32 × 5 (= 15π), or 960π × \((\frac{1}{4})^3\); subtracted from the large cone, or frustum = 960π × \(\frac{63}{64}\)✓ 1
= 2970 cm3 (945π)✓ 1

Exam tip: A frustum is a large cone minus a similar small cone. Find the missing radius with similar triangles before using the volume formula. Alternatively, cube the length scale factor to find the small cone's volume.

53IGME-061-03-11Similar shapes and circle theorems Paper HHard6 marks

A, B, C and D are points, in that order, on a circle. AC is a diameter of the circle.
Angle BAC = 2x° and angle ADB = (3x − 5)°.

(a)

Explain why angle ABC = 90°.

(1)
(b)

Work out the value of x. Give reasons for your working.

(4)
(c)

Find the size of angle ACB.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
AC is a diameter, so angle ABC is an angle in a semicircle, which is a right angle✓ 1
Part (b)
Angle ACB = 180 − 90 − 2x = (90 − 2x)°, because angles in a triangle add up to 180°✓ 1
Angle ADB = angle ACB, because angles in the same segment are equal✓ 1
3x − 5 = 90 − 2x✓ 1
x = 19✓ 1
Part (c)
Angle ACB = 90 − 2 × 19 = 52°✓ 1

Exam tip: Angles ADB and ACB both stand on the chord AB, on the same side of it. Spotting this pair is the key step. When a question asks for reasons, name each circle theorem in full.