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International GCSE Maths · Pearson Edexcel 4MA1 · Unit 4

Geometry and trigonometry

385 free, high-quality past-paper-style questions on Geometry and trigonometry for Edexcel IGCSE Maths, each with a full mark scheme. Covers: Area, surface area and volume; Pythagoras' theorem and trigonometry; Angles, polygons and circle properties; and more. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.2,202 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

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+ Add to test✓ In your test1IGME-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 point	Mark	Notes
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.

+ Add to test✓ In your test2IGME-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.

Diagram not to scale

(a)

Work out the volume of the box.

(2)

(b)

Work out the total surface area of the box.

(3)

Show mark scheme

Marking point	Mark	Notes
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.

+ Add to test✓ In your test3IGME-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 point	Mark	Notes
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.

+ Add to test✓ In your test4IGME-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.

Diagram not to scale

(a)

Calculate angle BAC.

(3)

Show mark scheme

Marking point	Mark	Notes
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.

+ Add to test✓ In your test5IGME-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°.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test6IGME-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 point	Mark	Notes
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.

+ Add to test✓ In your test7IGME-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°.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test8IGME-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°.

Diagram not to scale

(a)

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

(3)

Show mark scheme

Marking point	Mark	Notes
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.

+ Add to test✓ In your test9IGME-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.

Diagram not to scale

(a)

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

(3)

Show mark scheme

Marking point	Mark	Notes
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.

+ Add to test✓ In your test10IGME-061-00-06Angles, polygons and circle properties Paper FEasy4 marks

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

Diagram not to scale

(a)

Find the value of $x$.

(3)

(b)

Find the size of the largest angle in the quadrilateral.

(1)

Show mark scheme

Marking point	Mark	Notes
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°.

+ Add to test✓ In your test11IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test12IGME-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 point	Mark	Notes
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.

+ Add to test✓ In your test13IGME-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.

Diagram not to scale

(a)

Work out the size of angle PQR.

(2)

(b)

Work out the size of angle PRS.

(1)

Show mark scheme

Marking point	Mark	Notes
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.

+ Add to test✓ In your test14IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test15IGME-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°.

Diagram not to scale

(a)

Find the value of $x$.

(2)

(b)

Find the size of the largest of the three angles.

(1)

Show mark scheme

Marking point	Mark	Notes
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.

+ Add to test✓ In your test16IGME3-4-10d-01Angles, polygons and circle properties Paper FEasy5 marks

A garden gazebo has a floor in the shape of a regular octagon.

(a)

Work out the size of one interior angle of the octagon.

(2)

(b)

A different regular polygon has interior angles of 156°. How many sides does it have?

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 exterior angle 360 ÷ 8 = 45°	✓ 1
A1 135°	✓ 1
Part (b)
M1 exterior angle 180 − 156 = 24°	✓ 1
M1 360 ÷ 24	✓ 1
A1 15	✓ 1

Exam tip: The exterior angles of any polygon add up to 360°, so the number of sides of a regular polygon is 360 divided by one exterior angle.

+ Add to test✓ In your test17IGME3-4-10d-02Pythagoras' theorem and trigonometry Paper FEasy5 marks

A ladder of length 6.5 m leans against a vertical wall on level ground. The foot of the ladder is 2.5 m from the wall.

Diagram not to scale

(a)

Work out how far up the wall the ladder reaches.

(2)

(b)

The foot of the ladder is moved so that the top of the ladder is 5.2 m up the wall. How far is the foot of the ladder now from the wall?

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6.5² − 2.5²	✓ 1
A1 6 m	✓ 1
Part (b)
M1 6.5² − 5.2² = 42.25 − 27.04	✓ 1
M1 √15.21	✓ 1
A1 3.9 m	✓ 1

Common error: The ladder is the hypotenuse. Subtract the squares of the shorter sides from it; do not add them.

+ Add to test✓ In your test18IGME3-4-10d-03Pythagoras' theorem and trigonometry Paper FEasy5 marks

A wheelchair ramp is a straight slope 4.2 m long. It rises 0.7 m vertically from the ground to a doorway.

Diagram not to scale

(a)

Work out the angle the ramp makes with the horizontal. Give your answer to 1 decimal place.

(3)

(b)

Work out the horizontal distance covered by the ramp. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin x = 0.7 ÷ 4.2	✓ 1
M1 x = sin⁻¹(0.1667)	✓ 1
A1 9.6°	✓ 1
Part (b)
M1 √(4.2² − 0.7²)	✓ 1
A1 4.14 m	✓ 1

Exam tip: Label the sides opposite, adjacent and hypotenuse before choosing sin, cos or tan.

+ Add to test✓ In your test19IGME3-4-10d-04Area, surface area and volume Paper FEasy5 marks

A cylindrical water tank has radius 0.8 m and height 1.5 m. (1 m³ = 1000 litres)

(a)

Work out the volume of the tank in m³. Give your answer to 3 significant figures.

(2)

(b)

The tank is filled from empty at 40 litres per minute. Work out how many minutes it takes to fill. Use the exact volume and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 0.8² × 1.5	✓ 1
A1 3.02 m³	✓ 1
Part (b)
M1 volume in litres = 3015.9...	✓ 1
M1 3015.9... ÷ 40	✓ 1
A1 75.4 minutes	✓ 1

Common error: Convert m³ to litres by multiplying by 1000, not 100. Keep the full calculator value until the final step.

+ Add to test✓ In your test20IGME3-4-10d-05Angles, polygons and circle properties Paper FEasy5 marks

The roof truss of a footbridge contains an isosceles triangle. The angle at the top is 4x degrees and each of the two equal base angles is (3x + 10) degrees.

Diagram not to scale

(a)

Write an equation in x and solve it.

(3)

(b)

The base of the triangle is extended in a straight line past one corner. Work out the exterior angle formed at that corner.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4x + 2(3x + 10) = 180	✓ 1
M1 10x + 20 = 180	✓ 1
A1 x = 16	✓ 1
Part (b)
M1 base angle 3 × 16 + 10 = 58° (ECF from (a))	✓ 1
A1 180 − 58 = 122°	✓ 1

Exam tip: Angles in a triangle add up to 180°, and angles on a straight line add up to 180°. Say which fact you are using.

+ Add to test✓ In your test21IGME3-4-11r5-01Area, surface area and volume Paper FEasy3 marks

A runway at a regional airport is a rectangle 2.4 km long and 45 m wide.

(a)

Calculate the area of the runway, in m².

(2)

(b)

1 hectare = 10 000 m². Work out the area of the runway in hectares.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 convert 2.4 km to 2400 m and attempt 2400 × 45	✓ 1
A1 108 000 m²	✓ 1
Part (b)
B1 10.8 hectares (ECF from (a))	✓ 1

Common error: Convert every length to the same unit before multiplying; mixing km and m gives an area that is wrong by a factor of 1000.

+ Add to test✓ In your test22IGME3-4-11r5-02Pythagoras' theorem and trigonometry Paper FEasy4 marks

A rectangular field is 84 m long and 135 m wide. A farmer walks from one corner to the opposite corner along the diagonal.

Diagram not to scale

(a)

Calculate the length of the diagonal.

(3)

(b)

The farmer could instead walk along two sides of the field. How many metres shorter is the diagonal route?

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 84² + 135² (= 25 281)	✓ 1
M1 square root of their sum	✓ 1
A1 159 m	✓ 1
Part (b)
B1 84 + 135 − 159 = 60 m (ECF from (a))	✓ 1

Exam tip: When using Pythagoras to find the longest side, add the squares; to find a shorter side, subtract.

+ Add to test✓ In your test23IGME3-4-11r5-03Angles, polygons and circle properties Paper FEasy5 marks

A drummer's practice pad has the shape of a regular decagon (a polygon with 10 equal sides and 10 equal angles).

(a)

Work out the size of one interior angle of the decagon.

(3)

(b)

A different regular polygon has an exterior angle of 15°. Work out the number of sides of this polygon.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 exterior angle = 360 ÷ 10 (= 36°)	✓ 1
M1 180 − 36	✓ 1
A1 144°	✓ 1
Part (b)
M1 360 ÷ 15	✓ 1
A1 24	✓ 1

Exam tip: Exterior angles of any polygon add up to 360°, so for a regular polygon each exterior angle is 360 divided by the number of sides.

+ Add to test✓ In your test24IGME3-4-11r5-14Similar shapes and circle theorems Paper FEasy4 marks

Triangle ABC is isosceles with AB = AC. Angle BAC = 48°. The line DE is a straight line through A, parallel to BC, with D on the same side of A as B.

Diagram not to scale

(a)

Work out the size of angle ABC.

(2)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (180 − 48) ÷ 2	✓ 1
A1 66°	✓ 1
Part (b)
B1 66°	✓ 1
B1 alternate angles are equal (DE is parallel to BC)	✓ 1

Common error: Name the angle rule in full, for example 'alternate angles are equal'; informal labels such as 'Z angles' do not explain why the angles are equal.

+ Add to test✓ In your test25IGME3-4-13r5-02Pythagoras' theorem and trigonometry Paper FEasy4 marks

Calculator

In a video game, a character starts at the origin of a flat grid and walks 24 units east and then 7 units north, stopping at point P.

Diagram not to scale

(a)

Calculate the straight-line distance from the origin to P.

(2)

(b)

Calculate the angle between the straight line from the origin to P and the east direction. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 24² + 7² (= 625)	✓ 1
A1 25 (units)	✓ 1
Part (b)
M1 tan(angle) = 7/24 (or tan⁻¹(7/24))	✓ 1
A1 16.3°	✓ 1

Exam tip: Sketch a right-angled triangle and label the sides before choosing sin, cos or tan.

+ Add to test✓ In your test26IGME3-4-13r5-03Area, surface area and volume Paper FEasy5 marks

Calculator

A water storage tank is a cuboid with internal length 2.5 m, width 1.6 m and height 1.2 m. 1 m³ = 1000 litres.

Diagram not to scale

(a)

Calculate the volume of the tank in litres.

(3)

(b)

The empty tank is filled at a constant rate of 25 litres per minute. Work out how long it takes to fill the tank completely. Give your answer in hours and minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2.5 × 1.6 × 1.2	✓ 1
A1 4.8 (m³)	✓ 1
B1 4800 litres	✓ 1
Part (b)
M1 4800 ÷ 25 (= 192 minutes)	✓ 1
A1 3 hours 12 minutes	✓ 1

Common error: Convert between m³ and litres before dividing by a rate given in litres.

+ Add to test✓ In your test27IGME3-4-13r5-04Angles, polygons and circle properties Paper FEasy4 marks

Calculator

A game map is tiled with regular polygons. One tile has an exterior angle of 24°.

(a)

Work out the number of sides of the polygon.

(2)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 24	✓ 1
A1 15	✓ 1
Part (b)
M1 (15 − 2) × 180	✓ 1
A1 2340°	✓ 1

Exam tip: The exterior angles of any polygon add up to 360°, so a regular polygon has 360 ÷ (exterior angle) sides.

+ Add to test✓ In your test28IGME3-4-11r6-01Angles, polygons and circle properties Paper FEasy3 marks

A floor mosaic found at an excavation is in the shape of a regular octagon.

(a)

Work out the size of one exterior angle of the octagon.

(2)

(b)

Work out the size of one interior angle of the octagon.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 8	✓ 1
A1 45°	✓ 1
Part (b)
B1 135° (or 180 − 45)	✓ 1

Exam tip: Interior and exterior angles at a vertex add up to 180°, which gives a quick check.

+ Add to test✓ In your test29IGME3-4-11r6-02Pythagoras' theorem and trigonometry Paper FEasy5 marks

A cyclist rides 12 km due east and then 5 km due north. She then rides straight back to her starting point in a straight line.

Diagram not to scale

(a)

Calculate the length of the straight ride back to the start.

(3)

(b)

The whole journey takes 1.5 hours. Calculate her average speed for the whole journey, in km/h.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12² + 5² (= 169)	✓ 1
M1 √169	✓ 1
A1 13 km	✓ 1
Part (b)
M1 total distance 12 + 5 + 13 = 30 km (ECF from (a))	✓ 1
A1 30 ÷ 1.5 = 20 km/h	✓ 1

Common error: Average speed uses the total distance of every leg, not just the first two.

+ Add to test✓ In your test30IGME3-4-11r6-03Area, surface area and volume Paper FEasy5 marks

A data centre has a server room in the shape of a cuboid, 15 m long, 9 m wide and 3.5 m high.

(a)

Work out the volume of the server room.

(2)

(b)

The floor is to be covered with square tiles of side 0.6 m. Work out the number of tiles needed to cover the floor exactly, with no cutting.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 15 × 9 × 3.5	✓ 1
A1 472.5 m³	✓ 1
Part (b)
M1 15 ÷ 0.6 = 25 or 9 ÷ 0.6 = 15 (or floor area 135 and tile area 0.36)	✓ 1
M1 25 × 15 (or 135 ÷ 0.36)	✓ 1
A1 375 tiles	✓ 1

Exam tip: Check that the units of length are the same before dividing one by the other.

+ Add to test✓ In your test31IGME3-4-11r6-14Area, surface area and volume Paper FEasy5 marks

An archaeological trench has the shape of a rectangle measuring 8 m by 5 m with a semicircle added on one of the 5 m sides. The diameter of the semicircle is 5 m.

Diagram not to scale

(a)

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

(2)

(b)

Calculate the perimeter of the trench. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rectangle 8 × 5 = 40 and semicircle ½ × π × 2.5²	✓ 1
A1 49.8 m²	✓ 1
Part (b)
M1 straight edges 8 + 8 + 5 = 21 (the 5 m side with the semicircle is not part of the perimeter)	✓ 1
M1 semicircle arc ½ × π × 5	✓ 1
A1 28.9 m	✓ 1

Common error: The side where the semicircle is attached is inside the shape, so leave it out of the perimeter.

+ Add to test✓ In your test32IGME3-4-11r9-01Pythagoras' theorem and trigonometry Paper FEasy3 marks

A railway line climbs a steady straight slope. Over a horizontal distance of 640 m the track gains a height of 48 m.

Diagram not to scale (vertical scale exaggerated)

(a)

Calculate the length of track along the slope. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 640² + 48² (= 411 904)	✓ 1
M1 for square root of their sum	✓ 1
A1 for 642 m (641.8…)	✓ 1

Common error: Adding the two shorter sides and forgetting to square root, or subtracting the squares when the unknown side is the longest side.

+ Add to test✓ In your test33IGME3-4-11r9-02Area, surface area and volume Paper FEasy5 marks

A rectangular training pool is 25 m long, 12 m wide and has a constant depth of 1.8 m.

Diagram not to scale (depth exaggerated)

(a)

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

(2)

(b)

1 m³ = 1000 litres. The empty pool is filled at a rate of 450 litres per minute. Calculate the time taken to fill the pool completely. Give your answer in hours.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 25 × 12 × 1.8	✓ 1
A1 for 540 m³	✓ 1
Part (b)
M1 for 540 × 1000 (= 540 000 litres)	✓ 1
M1 for 540 000 ÷ 450 ÷ 60	✓ 1
A1 for 20 hours (ECF from (a))	✓ 1

Exam tip: Convert units before dividing: m³ to litres first, then minutes to hours at the end.

+ Add to test✓ In your test34IGME3-4-11r9-03Angles, polygons and circle properties Paper FEasy4 marks

The floor of a small observatory is a regular polygon. Each exterior angle of the polygon is 24°.

(a)

Work out the number of sides of the polygon.

(2)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 360 ÷ 24	✓ 1
A1 for 15	✓ 1
Part (b)
M1 for (15 − 2) × 180 (or 15 × 180 − 360)	✓ 1
A1 for 2340° (ECF from (a))	✓ 1

Common error: The exterior angles of any polygon add up to 360°, but the interior angles do not; use (n − 2) × 180° for those.

+ Add to test✓ In your test35IGME3-4-11r9-04Pythagoras' theorem and trigonometry Paper FEasy4 marks

A signal mast stands on level ground. From a point on the ground 18 m from the foot of the mast, the angle of elevation of the top of the mast is 32°.

Diagram not to scale

(a)

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

(2)

(b)

A cable runs in a straight line from the top of the mast to the point on the ground. Calculate the length of the cable. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for tan 32° = h ÷ 18 or h = 18 × tan 32°	✓ 1
A1 for 11.2 m	✓ 1
Part (b)
M1 for cos 32° = 18 ÷ cable length (or a correct Pythagoras method using unrounded h)	✓ 1
A1 for 21.2 m	✓ 1

Exam tip: Label the opposite, adjacent and hypotenuse relative to the given angle before choosing sin, cos or tan.

+ Add to test✓ In your test36IGME3-4-12r9-01Pythagoras' theorem and trigonometry Paper FEasy4 marks

A support cable for a wind turbine mast is fixed to the mast at a point 24 m above level ground. The other end is anchored to the ground 7 m from the foot of the mast.

Diagram not to scale

(a)

Calculate the length of the cable.

(3)

(b)

The cable costs €18.50 per metre. Work out the cost of one cable.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 7² + 24² (= 625)	✓ 1
M1 for square root of 625	✓ 1
A1 for 25 m	✓ 1
Part (b)
B1 for €462.50 (25 × 18.50, ECF from (a))	✓ 1

Exam tip: The cable is the longest side of a right-angled triangle, so add the squares of the two shorter sides.

+ Add to test✓ In your test37IGME3-4-12r9-02Area, surface area and volume Paper FEasy6 marks

A rectangular shop window is 3.6 m wide and 2.5 m high. A circular logo of radius 0.4 m is already printed on the glass. The rest of the window is to be covered with coloured film.

Diagram not to scale

(a)

Work out the area of the window.

(1)

(b)

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

(2)

(c)

The film costs £14 per square metre. Calculate the cost of the film needed. Use the unrounded logo area and give your answer to the nearest penny.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 9 m² (3.6 × 2.5)	✓ 1
Part (b)
M1 for π × 0.4²	✓ 1
A1 for 0.503 m² (0.50265…)	✓ 1
Part (c)
M1 for 9 − 0.50265… (= 8.497…)	✓ 1
M1 for × 14	✓ 1
A1 for £118.96	✓ 1

Common error: Subtract the area of the logo from the window before multiplying by the price; pricing the whole window overcharges.

+ Add to test✓ In your test38IGME3-4-12r9-03Angles, polygons and circle properties Paper FEasy4 marks

A pie chart shows how 480 patients in a study were treated. The angles in the pie chart add up to 360°. The sector for “recovered at home” represents 156 patients. The sector for “treated in hospital” has an angle of 84°.

(a)

Work out the angle of the sector for “recovered at home”.

(2)

(b)

Work out the number of patients treated in hospital.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 156 ÷ 480 × 360	✓ 1
A1 for 117°	✓ 1
Part (b)
M1 for 84 ÷ 360 × 480	✓ 1
A1 for 112	✓ 1

Exam tip: Each patient is worth 360 ÷ 480 = 0.75° of the pie chart, which gives a quick check on both answers.

+ Add to test✓ In your test39IGME3-4-12r9-04Pythagoras' theorem and trigonometry Paper FEasy4 marks

Solar panels are fixed on a flat roof using supports so that each panel is a flat rectangle tilted upwards at an angle to the horizontal. The lower edge of every panel rests on the roof.

Side view of each panel. Diagram not to scale

(a)

One panel is 2.0 m long (measured up the slope) and is tilted at 35°. Calculate the height of its upper edge above the roof. Give your answer correct to 3 significant figures.

(2)

(b)

A second panel is 2.4 m long and its upper edge is 0.9 m above the roof. Calculate the angle at which this panel is tilted. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 2.0 × sin 35°	✓ 1
A1 for 1.15 m	✓ 1
Part (b)
M1 for sin x = 0.9 ÷ 2.4	✓ 1
A1 for 22.0°	✓ 1

Common error: Using tan instead of sin: the panel length is the hypotenuse, because it lies along the slope.

+ Add to test✓ In your test40IGME3-4-12r10-01Angles, polygons and circle properties Paper FEasy3 marks

The rim of a space probe's antenna dish is a regular polygon. Each exterior angle of the polygon is 18°.

(a)

Work out the number of sides of the polygon.

(2)

(b)

Work out the size of one interior angle.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 18	✓ 1
A1 20	✓ 1
Part (b)
B1 162°	✓ 1

Exam tip: Exterior angles of any polygon add up to 360°, and each interior angle is 180° minus its exterior angle.

+ Add to test✓ In your test41IGME3-4-12r10-03Area, surface area and volume Paper FEasy3 marks

A coffee-roasting drum is a cylinder with radius 30 cm and length 80 cm.

(a)

Work out the volume of the drum in cm³. Give your answer to 3 significant figures.

(2)

(b)

Convert this volume to litres (1 litre = 1000 cm³). Use the unrounded volume and give your answer to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 30² × 80	✓ 1
A1 226 000 (cm³)	✓ 1
Part (b)
B1 226 (litres)	✓ 1

Common error: Use the radius, not the diameter, in the cylinder volume formula πr²h.

+ Add to test✓ In your test42IGME3-4-12r10-06Angles, polygons and circle properties Paper FEasy4 marks

The three angles of a triangular bridge truss are in the ratio 3 : 4 : 5.

(a)

Work out the size of the largest angle.

(3)

(b)

Is the truss triangle right-angled? Give a reason for your answer.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3 + 4 + 5 = 12	✓ 1
M1 180 ÷ 12 × 5	✓ 1
A1 75°	✓ 1
Part (b)
B1 No, because the largest angle is 75°, not 90°	✓ 1

Exam tip: In a ratio question, add the parts first, then find the size of one part.

+ Add to test✓ In your test43IGME3-4-12r10-14Pythagoras' theorem and trigonometry Paper FEasy4 marks

The drum of a coffee roaster has a diameter of 0.6 m.

(a)

A point on the rim of the drum travels round during 50 complete turns. Work out the distance travelled by the point. Give your answer to 3 significant figures.

(2)

(b)

The drum turns at 55 revolutions per minute. Work out the distance travelled by a point on the rim in 1 hour. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 0.6 × 50	✓ 1
A1 94.2 (m)	✓ 1
Part (b)
M1 55 × 60 × π × 0.6	✓ 1
A1 6220 (m)	✓ 1

Common error: Circumference uses the diameter as πd; if you are given the radius, use 2πr.

+ Add to test✓ In your test44IGME3-4-13r10-01Area, surface area and volume Paper FEasy5 marks

A rectangular tank at a marine research centre has internal length 2.4 m, width 1.5 m and depth 0.8 m.

Diagram not to scale

(a)

Calculate the volume of the tank in m³.

(2)

(b)

1 m³ = 1000 litres. Write the volume of the tank in litres.

(1)

(c)

The tank is filled completely from empty by a pump delivering 36 litres every minute. Work out how many minutes the pump takes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2.4 × 1.5 × 0.8	✓ 1
A1 2.88 (m³)	✓ 1
Part (b)
B1 2880 (litres)	✓ 1
Part (c)
M1 2880 ÷ 36 (oe)	✓ 1
A1 80 (minutes), ECF from (b)	✓ 1

Exam tip: Convert units before dividing: a volume in m³ and a rate in litres per minute cannot be combined directly.

+ Add to test✓ In your test45IGME3-4-13r10-02Pythagoras' theorem and trigonometry Paper FEasy6 marks

A wheelchair ramp at the entrance to a museum rises 0.45 m vertically over a horizontal distance of 5.4 m. The ramp is a straight slope.

Diagram not to scale (vertical scale exaggerated)

(a)

Calculate the length of the sloping surface of the ramp. Give your answer in metres correct to 3 significant figures.

(3)

(b)

Calculate the angle the ramp makes with the horizontal. Give your answer in degrees correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.45² + 5.4² (= 29.3625)	✓ 1
M1 √(29.3625)	✓ 1
A1 5.42 (m)	✓ 1
Part (b)
M1 tan x = 0.45 ÷ 5.4 (or sin x = 0.45 ÷ 5.42 etc.)	✓ 1
M1 x = tan⁻¹(0.45 ÷ 5.4)	✓ 1
A1 4.8 (°)	✓ 1

Common error: Using the hypotenuse as one of the shorter sides: check that the longest side of a right-angled triangle is opposite the right angle.

+ Add to test✓ In your test46IGME3-4-13r10-03Angles, polygons and circle properties Paper FEasy4 marks

A logistics company designs a protective cap for the corner of a pallet in the shape of a regular decagon (a regular polygon with 10 sides).

(a)

Calculate the size of one interior angle of a regular decagon.

(2)

(b)

A different regular polygon has each exterior angle equal to 15°. Work out the number of sides of this polygon.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 10 = 36 (exterior angle) or (10 − 2) × 180 ÷ 10	✓ 1
A1 144 (°)	✓ 1
Part (b)
M1 360 ÷ 15	✓ 1
A1 24	✓ 1

Exam tip: The exterior angles of any polygon add up to 360°, so for a regular polygon the number of sides is 360 divided by one exterior angle.

+ Add to test✓ In your test47IGME3-4-13r10-10Similar shapes and circle theorems Paper FEasy4 marks

A roof support is an isosceles triangle ABC with AB = AC. Angle BAC = 48°. The side BC is extended to a point D beyond C.

Diagram not to scale

(a)

Work out the size of angle ABC.

(2)

(b)

Work out the size of angle ACD.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (180 − 48) ÷ 2	✓ 1
A1 66 (°)	✓ 1
Part (b)
M1 180 − 66 (angles on a straight line) or 48 + 66 (exterior angle)	✓ 1
A1 114 (°)	✓ 1

Exam tip: Base angles of an isosceles triangle are equal, so subtract the apex angle from 180° and halve the result.

+ Add to test✓ In your test48IGME3-4-13r12-01Pythagoras' theorem and trigonometry Paper FEasy5 marks

A solar panel of length 2.4 m rests on a flat roof. Its lower edge touches the roof and its upper edge is held up by a vertical strut of height 0.9 m. The panel and strut lie in the same vertical plane and the roof is horizontal.

Side view. Diagram not to scale

(a)

Calculate the horizontal distance from the foot of the strut to the lower edge of the panel. Give your answer in metres to 3 significant figures.

(3)

(b)

Calculate the angle between the panel and the roof. Give your answer in degrees to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $2.4^2 - 0.9^2$ or $\sqrt{2.4^2-0.9^2}$	✓ 1
M1 for square root of their value, e.g. $\sqrt{4.95}$	✓ 1
A1 for 2.22 m (2.2248...)	✓ 1
Part (b)
M1 for a correct trigonometric ratio, e.g. $\sin x = \frac{0.9}{2.4}$ (or $\cos x = \frac{2.2249}{2.4}$ or $\tan x = \frac{0.9}{2.2249}$)	✓ 1
A1 for 22.0° (22.02...)	✓ 1

Exam tip: Choose the ratio that uses the two sides you were given in the question; using given data avoids carrying a rounding error forward.

+ Add to test✓ In your test49IGME3-4-13r12-02Area, surface area and volume Paper FEasy4 marks

A food company sells soup in cylindrical tins. Each tin has radius 3.5 cm and height 11 cm.

(a)

Calculate the volume of a tin. Give your answer in cm³ to 3 significant figures.

(2)

(b)

A paper label covers the whole curved surface of the tin exactly once, with no overlap. Calculate the area of the label. Give your answer in cm² to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\pi \times 3.5^2 \times 11$	✓ 1
A1 for 423 cm³ (423.3...)	✓ 1
Part (b)
M1 for $2 \times \pi \times 3.5 \times 11$	✓ 1
A1 for 242 cm² (241.9...)	✓ 1

Common error: Do not include the circular ends when a question asks only for the curved surface; the label here is a rectangle of width $2\pi r$ and height 11 cm.

+ Add to test✓ In your test50IGME3-4-13r12-04Pythagoras' theorem and trigonometry Paper FEasy5 marks

A mobile testing unit leaves a depot and drives 12 km due north and then 5 km due east, ending at point P. The roads are straight.

Diagram not to scale

(a)

Calculate the straight-line distance from the depot to P.

(2)

(b)

Calculate the bearing of P from the depot. Give your answer in degrees to 1 decimal place, as a three-figure bearing.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $12^2 + 5^2$ (= 169)	✓ 1
A1 for 13 km	✓ 1
Part (b)
M1 for a correct ratio for the angle east of north, e.g. $\tan x = \frac{5}{12}$	✓ 1
A1 for 22.6° (22.62...)	✓ 1
B1 for three-figure bearing written as 022.6° (or 022.6)	✓ 1

Common error: A bearing is measured clockwise from north. Sketch the two legs first so you can see that the angle east of north uses 5 (opposite) and 12 (adjacent).

+ Add to test✓ In your test51IGME3-4-12r14-02Pythagoras' theorem and trigonometry Paper FEasy5 marks

A support cable runs from the top of a vertical signal mast to a point on level ground. The mast is 8.4 m high and the point is 3.5 m from the base of the mast.

Diagram not to scale

(a)

Calculate the length of the cable.

(3)

(b)

Calculate the angle the cable makes with the ground. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
8.4² + 3.5² = 82.81	✓ 1
√82.81	✓ 1
9.1 m	✓ 1
Part (b)
tan x = 8.4 ÷ 3.5	✓ 1
67.4°	✓ 1

Common error: To find an angle from a ratio, use the inverse function (tan⁻¹) on your calculator; do not stop at the value of tan x.

+ Add to test✓ In your test52IGME3-4-12r14-03Similar shapes and circle theorems Paper FEasy6 marks

An athletics infield is a rectangle 90 m long and 50 m wide, with a semicircle of diameter 50 m added at each of the two shorter ends. Give each answer to 3 significant figures.

Diagram not to scale

(a)

Calculate the perimeter of the infield.

(3)

(b)

Calculate the area of the infield.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Two semicircles make a circle: π × 50 = 157.08…	✓ 1
Add 2 × 90 = 180	✓ 1
337 m	✓ 1
Part (b)
Rectangle 90 × 50 = 4500	✓ 1
Circle π × 25² = 1963.5…	✓ 1
6460 m²	✓ 1

Exam tip: Two semicircles of the same size make one full circle, which saves calculating them separately.

+ Add to test✓ In your test53IGME3-4-12r14-04Pythagoras' theorem and trigonometry Paper FEasy4 marks

A railway embankment slopes at 28° to the horizontal. The sloping side is 14.5 m long. Give each answer to 3 significant figures.

Diagram not to scale

(a)

Calculate the vertical height of the embankment.

(2)

(b)

Calculate the horizontal width of the embankment.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
sin 28° = height ÷ 14.5 or 14.5 × sin 28°	✓ 1
6.81 m	✓ 1
Part (b)
14.5 × cos 28°	✓ 1
12.8 m	✓ 1

Common error: Check the calculator is in degree mode; in radian mode sin 28 gives a completely different value.

+ Add to test✓ In your test54IGME3-4-12r17-01Angles, polygons and circle properties Paper FEasy4 marks

A window in a new building is the shape of a regular polygon. Each exterior angle of the polygon is 24°.

(a)

Work out the number of sides of the polygon.

(2)

(b)

Work out the size of one interior angle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 24	✓ 1
A1 15	✓ 1
Part (b)
M1 180 − 24 (or 180 − 360 ÷ 15)	✓ 1
A1 156°	✓ 1

Exam tip: The exterior angles of any polygon add up to 360°, and an interior angle and its exterior angle add up to 180°.

+ Add to test✓ In your test55IGME3-4-12r17-02Pythagoras' theorem and trigonometry Paper FEasy4 marks

A loading ramp for a delivery van rises 0.9 m vertically over a horizontal distance of 2.4 m. The ramp is a straight, flat slope.

Diagram not to scale

(a)

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

(2)

(b)

Work out the angle the ramp makes with the horizontal. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.9² + 2.4² (= 6.57)	✓ 1
A1 2.56 m	✓ 1
Part (b)
M1 tan θ = 0.9 ÷ 2.4 (or sin θ = 0.9 ÷ 2.5632 or cos θ = 2.4 ÷ 2.5632)	✓ 1
A1 20.6°	✓ 1

Common error: Check that the hypotenuse is the longest side and that the calculator is in degree mode before using inverse trigonometric functions.

+ Add to test✓ In your test56IGME3-4-12r17-03Area, surface area and volume Paper FEasy4 marks

A marine biology laboratory has a tank in the shape of a cylinder. The tank has radius 0.8 m and height 1.5 m.

(a)

Work out the volume of the tank in m³. Give your answer correct to 3 significant figures.

(2)

(b)

1 m³ = 1000 litres. Work out the capacity of the tank in litres, to the nearest litre. Use the unrounded value of the volume.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 0.8² × 1.5	✓ 1
A1 3.02 m³	✓ 1
Part (b)
M1 π × 0.8² × 1.5 × 1000 (= 3015.9...)	✓ 1
A1 3016 litres	✓ 1

Exam tip: Keep the unrounded value on your calculator between parts so that a rounded answer does not change the final result.

+ Add to test✓ In your test57IGME3-4-12r17-08Similar shapes and circle theorems Paper FEasy4 marks

A model of a whale is mathematically similar to the real whale. The model is 45 cm long and the real whale is 12.6 m long. The length of a fin on the model is 8.5 cm.

(a)

Work out the scale factor from the model to the real whale.

(2)

(b)

Work out the length of the real fin in metres.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12.6 m = 1260 cm, 1260 ÷ 45 (or 12.6 ÷ 0.45)	✓ 1
A1 28	✓ 1
Part (b)
M1 8.5 × 28 (= 238 cm)	✓ 1
A1 2.38 m	✓ 1

Exam tip: Convert both lengths to the same unit before dividing to find the scale factor.

+ Add to test✓ In your test58IGME3-4-13r18-01Pythagoras' theorem and trigonometry Paper FEasy5 marks

A rectangular reservoir plot measures 48 m by 20 m. A water pipe is laid in a straight line across the plot from one corner to the opposite corner.

Diagram not to scale

(a)

Calculate the length of the pipe.

(3)

(b)

The pipe costs £14.50 per metre. Work out the total cost of the pipe.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of Pythagoras: length² = 48² + 20²	✓ 1
M1 length² = 2704, or √2704	✓ 1
A1 52 m	✓ 1
Part (b)
M1 52 × 14.50 (ECF from (a))	✓ 1
A1 £754	✓ 1

Exam tip: Write the Pythagoras statement before substituting; it earns the method mark even if the arithmetic slips.

+ Add to test✓ In your test59IGME3-4-13r18-02Angles, polygons and circle properties Paper FEasy5 marks

A board game uses tiles shaped as regular polygons.

(a)

A tile is a regular octagon. Work out the size of one exterior angle.

(2)

(b)

Another tile is a regular polygon with each interior angle equal to 156°. Work out the number of sides of this polygon.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 8	✓ 1
A1 45°	✓ 1
Part (b)
M1 exterior angle = 180 − 156 = 24°	✓ 1
M1 360 ÷ 24	✓ 1
A1 15	✓ 1

Common error: Dividing 360 by the interior angle instead of the exterior angle gives a non-whole number of sides; check your answer is an integer.

+ Add to test✓ In your test60IGME3-4-13r18-03Area, surface area and volume Paper FEasy5 marks

A cylindrical medicine bottle has an internal radius of 2.5 cm and an internal height of 11 cm.

(a)

Show that the volume of the bottle is 216 cm³, correct to 3 significant figures.

(3)

(b)

The bottle is filled to 80% of its volume with medicine. Work out the volume of medicine, in cm³, correct to 3 significant figures. Use the unrounded volume from (a).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of V = πr²h	✓ 1
M1 π × 2.5² × 11	✓ 1
A1 215.98… (or 68.75π) leading to 216	✓ 1
Part (b)
M1 0.8 × 215.98…	✓ 1
A1 173	✓ 1

Exam tip: In a 'show that' question, write the unrounded value before the rounded one.

+ Add to test✓ In your test61IGME3-4-13r18-04Pythagoras' theorem and trigonometry Paper FEasy6 marks

In a video game, a ramp is a straight slope of length 12.5 m. The top of the ramp is 3.0 m above the ground.

Diagram not to scale

(a)

Work out the angle the ramp makes with the horizontal. Give your answer correct to 1 decimal place.

(3)

(b)

Work out the horizontal distance covered by the ramp. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin x = 3.0 ÷ 12.5	✓ 1
M1 x = sin⁻¹(0.24)	✓ 1
A1 13.9°	✓ 1
Part (b)
M1 12.5² − 3.0²	✓ 1
M1 √147.25	✓ 1
A1 12.1 m	✓ 1

Common error: Add the squares only when finding the hypotenuse; when a shorter side is needed, subtract.

+ Add to test✓ In your test62IGME3-4-13r19-01Angles, polygons and circle properties Paper FEasy4 marks

Archaeologists uncover the floor of a Roman bath house. The floor is a regular octagon (8 equal sides and 8 equal angles).

(a)

Work out the size of one exterior angle of the octagon.

(2)

(b)

Work out the size of one interior angle of the octagon.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 8	✓ 1
A1 45°	✓ 1
Part (b)
M1 180 − 45	✓ 1
A1 135°	✓ 1

Common error: Dividing the interior angle sum by 8 without first finding it; use exterior angle = 360° ÷ number of sides, then subtract from 180°.

+ Add to test✓ In your test63IGME3-4-13r19-04Area, surface area and volume Paper FEasy4 marks

Archaeologists dig a rectangular trench 3.5 m long, 2 m wide and 1.2 m deep.

(a)

Calculate the volume of soil removed from the trench.

(2)

(b)

Each wheelbarrow carries 0.07 m³ of soil. Work out the number of full wheelbarrow loads needed to remove all the soil.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3.5 × 2 × 1.2	✓ 1
A1 8.4 m³	✓ 1
Part (b)
M1 8.4 ÷ 0.07	✓ 1
A1 120	✓ 1

Exam tip: Check that all lengths are in the same unit before multiplying; here every length is already in metres, so the volume is in m³.

+ Add to test✓ In your test64IGME3-4-13r19-05Area, surface area and volume Paper FEasy6 marks

A circular mosaic has radius 1.8 m.

(a)

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

(2)

(b)

Calculate the circumference of the mosaic. Give your answer correct to 3 significant figures.

(2)

(c)

A metal edging strip for the mosaic costs £6.50 per metre. Using the unrounded circumference from your calculator, work out the cost of edging all the way round. Give your answer to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 1.8²	✓ 1
A1 10.2 m²	✓ 1
Part (b)
M1 2 × π × 1.8	✓ 1
A1 11.3 m	✓ 1
Part (c)
M1 11.3097... × 6.50	✓ 1
A1 £73.51	✓ 1

Common error: Mixing up the area and circumference formulae; area has r squared, circumference does not.

+ Add to test✓ In your test65IGME3-4-13r19-06Angles, polygons and circle properties Paper FEasy5 marks

A bicycle frame includes an isosceles triangle with two equal sides. The angle between the two equal sides is 48°.

(a)

Work out the size of each of the other two angles in the triangle.

(2)

(b)

A tile in a floor pattern is a regular polygon with each interior angle equal to 156°. Work out the number of sides of the polygon.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (180 − 48) ÷ 2	✓ 1
A1 66°	✓ 1
Part (b)
M1 exterior angle = 180 − 156 = 24°	✓ 1
M1 360 ÷ 24	✓ 1
A1 15	✓ 1

Exam tip: For regular polygons, the exterior angles always total 360°, which is often quicker than using the interior angle sum.

+ Add to test✓ In your test66IGME3-4-11r20-01Angles, polygons and circle properties Paper FEasy3 marks

A farmer fences a field in the shape of a regular pentagon.

(a)

Calculate the size of one exterior angle of the pentagon.

(2)

(b)

Calculate the size of one interior angle of the pentagon.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $360 \div 5$	✓ 1
A1 $72^\circ$	✓ 1
Part (b)
B1 $180 - 72 = 108^\circ$ (ECF from the exterior angle in the previous part)	✓ 1

Exam tip: Interior and exterior angles at a vertex of a polygon add up to 180°, which is often quicker than using the angle-sum formula.

+ Add to test✓ In your test67IGME3-4-11r20-02Pythagoras' theorem and trigonometry Paper FEasy5 marks

A light aircraft flies 120 km due east from airport A to a landing strip B. It then flies 50 km due north from B to a second strip C.

Diagram not to scale

(a)

Calculate the direct distance from A to C.

(3)

(b)

The aircraft travels at a constant speed of 600 km/h. Calculate the time taken, in minutes, to fly the whole route from A to B to C and then straight back to A.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $120^2 + 50^2$	✓ 1
M1 $\sqrt{14400+2500}=\sqrt{16900}$	✓ 1
A1 130 km	✓ 1
Part (b)
M1 total distance $120+50+130=300$ km, time $=300\div600$ hours (ECF from the previous part)	✓ 1
A1 30 minutes	✓ 1

Common error: Add the squares of the two shorter sides, then take the square root; do not forget the square root at the end.

+ Add to test✓ In your test68IGME3-4-11r20-03Area, surface area and volume Paper FEasy5 marks

A farm has a water tank in the shape of a cylinder. The inside radius is 1.5 m and the inside height is 2.4 m.

(a)

Calculate the volume of the tank. Give your answer in m³ correct to 3 significant figures.

(2)

(b)

The tank is filled from empty at a rate of 40 litres per minute. Calculate the time taken to fill the tank completely. Use the unrounded volume from the previous part, and give your answer in minutes correct to 3 significant figures. (1 m³ = 1000 litres)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\pi\times1.5^2\times2.4$	✓ 1
A1 17.0 m³	✓ 1
Part (b)
M1 convert volume to litres: $16.96\ldots\times1000$	✓ 1
M1 divide by 40	✓ 1
A1 424 minutes	✓ 1

Common error: Convert between m³ and litres carefully: 1 m³ is 1000 litres, not 100.

+ Add to test✓ In your test69IGME3-4-11r20-04Area, surface area and volume Paper FEasy4 marks

The circular skin of a drum has a diameter of 36 cm.

(a)

Calculate the circumference of the drum skin. Give your answer correct to 1 decimal place.

(2)

(b)

Calculate the area of the drum skin. Give your answer correct to the nearest cm².

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\pi\times36$	✓ 1
A1 113.1 cm	✓ 1
Part (b)
M1 $\pi\times18^2$	✓ 1
A1 1018 cm²	✓ 1

Exam tip: Check whether you are given the diameter or the radius before using the area formula; the radius here is 18 cm.

+ Add to test✓ In your test70IGME3-4-12r21-01Angles, polygons and circle properties Paper FEasy6 marks

In a board game, the playing tokens are regular polygons.

(a)

Work out the size of one interior angle of a regular octagon.

(2)

(b)

Each exterior angle of a different regular polygon is 24°. Work out the number of sides of this polygon.

(2)

(c)

Work out the sum of the interior angles of the polygon in part (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 8 = 45 (exterior angle) or (8 − 2) × 180 ÷ 8	✓ 1
A1 135°	✓ 1
Part (b)
M1 360 ÷ 24	✓ 1
A1 15	✓ 1
Part (c)
M1 (15 − 2) × 180	✓ 1
A1 2340°	✓ 1

Exam tip: Exterior angles of any polygon add up to 360°, so for a regular polygon the exterior angle is 360 divided by the number of sides.

+ Add to test✓ In your test71IGME3-4-12r21-02Pythagoras' theorem and trigonometry Paper FEasy5 marks

A rectangular field is 126 m long and 84 m wide. A water pipe is laid in a straight line from one corner of the field to the opposite corner.

Diagram not to scale

(a)

Calculate the length of the pipe. Give your answer correct to 1 decimal place.

(3)

(b)

The pipe costs £18.50 per metre. Using 151.4 m as the length, work out the cost of the pipe.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 126² + 84² (= 22 932)	✓ 1
M1 square root of the sum	✓ 1
A1 151.4 m	✓ 1
Part (b)
M1 151.4 × 18.5	✓ 1
A1 £2800.90	✓ 1

Common error: To find the longest side of a right-angled triangle add the squares of the other two sides; subtracting them is only for finding a shorter side.

+ Add to test✓ In your test72IGME3-4-12r21-04Area, surface area and volume Paper FEasy6 marks

A pharmacist stores syrup in a cylindrical bottle of radius 3.0 cm and height 10 cm. 1 cm³ = 1 ml.

(a)

Calculate the volume of the bottle. Give your answer correct to 3 significant figures.

(2)

(b)

The bottle is 80% full of syrup. Work out the volume of syrup, in ml, correct to 3 significant figures.

(2)

(c)

A paper label covers the whole curved surface of the bottle. Work out the area of the label, correct to the nearest cm².

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 3.0² × 10	✓ 1
A1 283 cm³	✓ 1
Part (b)
M1 0.8 × 282.7... (use the unrounded volume)	✓ 1
A1 226 ml	✓ 1
Part (c)
M1 2 × π × 3.0 × 10	✓ 1
A1 188 cm²	✓ 1

Common error: The curved surface of a cylinder is 2πrh; the formula πr²h gives the volume, so check which one the question needs.

+ Add to test✓ In your test73IGME3-4-13r22-01Angles, polygons and circle properties Paper FEasy5 marks

Calculator

A food-science lab designs a biscuit cutter in the shape of a regular polygon. Each exterior angle of the polygon is 18°.

(a)

Work out the number of sides of the polygon.

(2)

(b)

Work out the size of one interior angle.

(1)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 18	✓ 1
A1 20	✓ 1
Part (b)
B1 162°	✓ 1
Part (c)
M1 (20 − 2) × 180 or 20 × 162	✓ 1
A1 3240°	✓ 1

Common error: Using 180 ÷ n for the exterior angle instead of 360 ÷ n.

+ Add to test✓ In your test74IGME3-4-13r22-03Pythagoras' theorem and trigonometry Paper FEasy5 marks

Calculator

A mountaineer leaves base camp B and walks 1.2 km due east to point P. She then walks 0.9 km due north to a hut H. The ground is level and the paths are straight.

Diagram not to scale

(a)

Work out the direct distance from B to H.

(2)

(b)

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

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1.2² + 0.9² or √(1.44 + 0.81)	✓ 1
A1 1.5 km	✓ 1
Part (b)
M1 tan (PBH) = 0.9 ÷ 1.2	✓ 1
M1 tan⁻¹(0.75)	✓ 1
A1 36.9° (36.87...)	✓ 1

Exam tip: Label the opposite, adjacent and hypotenuse sides before choosing sin, cos or tan.

+ Add to test✓ In your test75IGME3-4-13r22-05Similar shapes and circle theorems Paper FEasy6 marks

Points A and B lie on a circle with centre O. Angle AOB = 124°. The tangents to the circle at A and B meet at the point T, so OATB is a quadrilateral.

Diagram not to scale

(a)

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

(2)

(b)

Work out the size of angle ATB.

(2)

(c)

Work out the size of angle OAB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 90°	✓ 1
B1 the tangent is perpendicular to the radius at the point of contact	✓ 1
Part (b)
M1 360 − 90 − 90 − 124	✓ 1
A1 56°	✓ 1
Part (c)
M1 triangle OAB is isosceles (OA = OB), so angle OAB = (180 − 124) ÷ 2	✓ 1
A1 28°	✓ 1

Exam tip: When asked for a reason in circle questions, name the property in words, for example: tangent perpendicular to radius.

+ Add to test✓ In your test76IGME3-4-13r22-10Area, surface area and volume Paper FEasy8 marks

Calculator

A telecoms company fences a rectangular enclosure around a relay tower. The enclosure is a rectangle 14 m long and 9 m wide with a semicircle added on one of the 9 m sides, so the semicircle has diameter 9 m. The fence goes all the way round the outside of the combined shape.

Diagram not to scale

(a)

Work out the area of the enclosure. Give your answer to the nearest m².

(3)

(b)

Work out the total length of the fence. Give your answer to 3 significant figures.

(3)

(c)

The fence costs £8.50 per metre. Use the unrounded length from (b) and work out the total cost, to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rectangle 14 × 9 = 126	✓ 1
M1 semicircle ½ × π × 4.5² (= 31.8...)	✓ 1
A1 158 m² (157.8...)	✓ 1
Part (b)
M1 arc length ½ × π × 9 (= 14.137...)	✓ 1
M1 14 + 14 + 9 + arc	✓ 1
A1 51.1 m (51.137...)	✓ 1
Part (c)
M1 51.137... × 8.5	✓ 1
A1 £434.67	✓ 1

Exam tip: For a composite shape, mark which edges are inside the shape: they are not part of the perimeter.

+ Add to test✓ In your test77IGME3-4-11r24-01Angles, polygons and circle properties Paper FEasy5 marks

A solar panel on a space probe is shaped like a regular polygon. Each interior angle of the panel is 150°.

(a)

Work out the size of one exterior angle.

(1)

(b)

Work out the number of sides of the panel.

(2)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 30°	✓ 1
Part (b)
M1 360 ÷ 30	✓ 1
A1 12	✓ 1
Part (c)
M1 (12 − 2) × 180 or 12 × 150	✓ 1
A1 1800°	✓ 1

Exam tip: Exterior angles of any polygon add up to 360°, and an interior angle and its exterior angle add up to 180°.

+ Add to test✓ In your test78IGME3-4-11r24-03Area, surface area and volume Paper FEasy7 marks

A coffee roasting drum is a cylinder of radius 0.3 m and length 0.9 m.

(a)

Work out the volume of the drum. Give your answer in m³ correct to 3 significant figures.

(3)

(b)

Green coffee beans fill 60% of the volume of the drum. Work out the volume of beans. Use the unrounded volume from part (a). Give your answer correct to 3 significant figures.

(2)

(c)

The density of the green beans is 650 kg/m³. Work out the mass of beans in the drum. Use the unrounded volume of beans. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 0.3²	✓ 1
M1 × 0.9	✓ 1
A1 0.254 m³	✓ 1
Part (b)
M1 0.6 × 0.25446...	✓ 1
A1 0.153 m³	✓ 1
Part (c)
M1 0.15268... × 650	✓ 1
A1 99.2 kg	✓ 1

Exam tip: Keep the full calculator value between parts and round only the final answer to the accuracy asked for.

+ Add to test✓ In your test79IGME3-4-11r24-04Pythagoras' theorem and trigonometry Paper FEasy6 marks

A ramp on a footbridge is a straight slope 12 m long. The top of the ramp is 1.5 m above the bottom.

Diagram not to scale

(a)

Work out the angle the ramp makes with the horizontal. Give your answer correct to 1 decimal place.

(3)

(b)

A second ramp rises 1.5 m at an angle of 5° to the horizontal. Work out the length of this ramp. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin x = 1.5 ÷ 12	✓ 1
M1 x = sin⁻¹(0.125)	✓ 1
A1 7.2°	✓ 1
Part (b)
M1 sin 5° = 1.5 ÷ L	✓ 1
M1 L = 1.5 ÷ sin 5°	✓ 1
A1 17.2 m	✓ 1

Common error: Check that your calculator is in degree mode before using sin, cos or tan.

+ Add to test✓ In your test80IGME3-4-14r29-01Area, surface area and volume Paper FEasy5 marks

The stage of a theatre is a rectangle measuring 12 m by 8.5 m.

(a)

Work out the area of the stage.

(2)

(b)

Work out the perimeter of the stage.

(1)

(c)

New carpet for the stage costs £14.50 per square metre. Work out the cost of carpeting the whole stage.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $12\times8.5$	✓ 1
A1 102 m$^{2}$	✓ 1
Part (b)
B1 41 m	✓ 1
Part (c)
M1 $102\times14.50$ (ECF from (a))	✓ 1
A1 £1479	✓ 1

Common error: Perimeter is a length in metres and area is in square metres; do not mix up the two formulae.

+ Add to test✓ In your test81IGME3-4-14r29-02Pythagoras' theorem and trigonometry Paper FEasy4 marks

A ship sails 9 km due east from a harbour and then 12 km due north.

Diagram not to scale

(a)

Work out the straight-line distance of the ship from the harbour.

(2)

(b)

The ship returns directly to the harbour at a constant speed of 20 km/h. Work out the time taken, in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\sqrt{9^{2}+12^{2}}$	✓ 1
A1 15 km	✓ 1
Part (b)
M1 $15\div20$ (ECF from (a))	✓ 1
A1 45 minutes	✓ 1

Exam tip: The east and north legs meet at a right angle, so the straight-line distance is the hypotenuse.

+ Add to test✓ In your test82IGME3-4-14r29-03Angles, polygons and circle properties Paper FEasy4 marks

The stage in a small theatre is shaped like a regular octagon.

(a)

Work out the size of one exterior angle of the octagon.

(1)

(b)

Work out the size of one interior angle of the octagon.

(1)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $360\div8=45^{\circ}$	✓ 1
Part (b)
B1 $180-45=135^{\circ}$	✓ 1
Part (c)
M1 $(8-2)\times180$ or $8\times135$	✓ 1
A1 $1080^{\circ}$	✓ 1

Exam tip: Exterior angles of any polygon add up to $360^{\circ}$, so a regular polygon's exterior angle is $360^{\circ}$ divided by the number of sides.

+ Add to test✓ In your test83IGME3-4-14r29-04Area, surface area and volume Paper FEasy4 marks

A volcanologist stores a rock sample in a cylindrical container of radius 4 cm and height 10 cm. The container is completely full of rock.

(a)

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

(2)

(b)

The density of the rock is 2.7 g/cm$^{3}$. Work out the mass of the rock. Give your answer in grams, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\pi\times4^{2}\times10$	✓ 1
A1 503 cm$^{3}$	✓ 1
Part (b)
M1 volume $\times2.7$	✓ 1
A1 1360 g	✓ 1

Common error: Use the radius, not the diameter, in $\pi r^{2}h$.

+ Add to test✓ In your test84IGME3-4-13r32-01Area, surface area and volume Paper FEasy4 marks

A shop sells paint in cylindrical tins. Each tin has radius 7 cm and height 12 cm.

(a)

Work out the volume of a tin. Give your answer to 3 significant figures.

(2)

(b)

The label covers the whole curved surface of the tin. Work out the area of the label. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 7² × 12	✓ 1
A1 1850 cm³ (accept 1847.3 seen)	✓ 1
Part (b)
M1 2 × π × 7 × 12	✓ 1
A1 528 cm² (527.79 seen)	✓ 1

Common error: Using the diameter in place of the radius in the circle formulas makes the volume four times too big.

+ Add to test✓ In your test85IGME3-4-13r32-02Pythagoras' theorem and trigonometry Paper FEasy6 marks

A support wire for a wind-turbine mast is fixed to level ground at a point 14 m from the foot of the mast. The wire is attached to the mast at a point 30 m above the ground. The mast is vertical.

Diagram not to scale

(a)

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

(3)

(b)

Work out the angle between the wire and the ground. Give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 14² + 30² (= 1096)	✓ 1
M1 square root of their sum	✓ 1
A1 33.1 m (33.11 seen)	✓ 1
Part (b)
M1 tan x = 30 ÷ 14 (or equivalent trigonometric ratio)	✓ 1
M1 inverse tangent of their ratio	✓ 1
A1 65.0° (64.98 seen)	✓ 1

Exam tip: Label the sides opposite, adjacent and hypotenuse relative to the angle before choosing sin, cos or tan.

+ Add to test✓ In your test86IGME3-4-13r32-03Angles, polygons and circle properties Paper FEasy6 marks

The floor of a hospital atrium is covered by a mosaic in the shape of a regular polygon. Each exterior angle of the polygon is 24°.

(a)

Work out the number of sides of the polygon.

(2)

(b)

Work out the size of one interior angle.

(2)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 24	✓ 1
A1 15	✓ 1
Part (b)
M1 180 − 24	✓ 1
A1 156°	✓ 1
Part (c)
M1 (15 − 2) × 180 (or 15 × 156)	✓ 1
A1 2340°	✓ 1

Exam tip: The exterior angles of any polygon add up to 360°, so an exterior angle of a regular polygon gives its number of sides directly.

+ Add to test✓ In your test87IGME3-4-13r32-04Area, surface area and volume Paper FEasy6 marks

A solar farm is built on a field made from a rectangle 80 m long and 50 m wide with a semicircle added on one of the 50 m sides. The diameter of the semicircle is 50 m.

Diagram not to scale

(a)

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

(4)

(b)

Solar panels cover 35% of the area of the field. Work out the area covered by the panels. Use your answer to part (a) and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 80 × 50 (= 4000)	✓ 1
M1 ½ × π × 25²	✓ 1
M1 4000 + their semicircle area	✓ 1
A1 4980 m² (4981.7 seen)	✓ 1
Part (b)
M1 0.35 × their area from (a)	✓ 1
A1 1740 m² (accept 1743 or 1743.6 from unrounded working) (ECF from (a))	✓ 1

Common error: The semicircle has radius 25 m, not 50 m; halve the diameter before using the circle area formula.

+ Add to test✓ In your test88IGME3-4-12r36-02Pythagoras' theorem and trigonometry Paper FEasy5 marks

A theatre stage is made from a rectangle 12 m long and 8 m wide. A rectangular corner measuring 5 m by 3 m is then removed to leave an L-shaped stage.

Diagram not to scale

(a)

Work out the area of the L-shaped stage.

(3)

(b)

Work out the perimeter of the L-shaped stage.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 × 8 = 96	✓ 1
M1 5 × 3 = 15	✓ 1
A1 81 m²	✓ 1
Part (b)
M1 12 + 8 + 12 + 8 or recognising the perimeter is the same as the original rectangle	✓ 1
A1 40 m	✓ 1

Common error: Removing a corner changes the area but not the perimeter of the shape.

+ Add to test✓ In your test89IGME3-4-12r36-05Pythagoras' theorem and trigonometry Paper FEasy5 marks

A theatre has a stage in the shape of a semicircle with diameter 10 m.

Diagram not to scale

(a)

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

(2)

(b)

Calculate the perimeter of the stage. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × π × 5²	✓ 1
A1 39.3 m²	✓ 1
Part (b)
M1 ½ × π × 10 (curved edge, 15.7...)	✓ 1
M1 add the straight edge of 10 m	✓ 1
A1 25.7 m	✓ 1

Common error: The perimeter of a semicircle includes the straight diameter as well as the curved edge.

+ Add to test✓ In your test90IGME3-4-13r36-01Angles, polygons and circle properties Paper FEasy4 marks

An archaeologist is reconstructing a mosaic border in the shape of a regular polygon. Each exterior angle of the polygon is 15°.

(a)

Work out the number of sides of the polygon.

(2)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 15	✓ 1
A1 24	✓ 1
Part (b)
M1 (24 − 2) × 180 (ECF from (a))	✓ 1
A1 3960°	✓ 1

Exam tip: The exterior angles of any polygon add up to 360°, so divide 360 by the exterior angle to find the number of sides.

+ Add to test✓ In your test91IGME3-4-13r36-02Pythagoras' theorem and trigonometry Paper FEasy5 marks

A cyclist rides 9 km due west and then 4.8 km due north. She then returns directly to her start in a straight line.

(a)

Calculate the length of the straight-line return journey.

(3)

(b)

Work out how much shorter the straight-line return is than riding back along the two original sections of the route.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 9² + 4.8² (= 104.04)	✓ 1
M1 square root of 104.04	✓ 1
A1 10.2 km	✓ 1
Part (b)
M1 (9 + 4.8) − 10.2	✓ 1
A1 3.6 km	✓ 1

Common error: When finding the hypotenuse, add the squares of the two shorter sides before taking the square root; do not forget the final square-root step.

+ Add to test✓ In your test92IGME3-4-13r36-08Similar shapes and circle theorems Paper FEasy4 marks

Triangle ABC is a roof truss. AB = AC and angle BAC = 48°. Point D is on AB and point E is on AC so that DE is parallel to BC.

Diagram not to scale

(a)

Work out the size of angle ABC.

(2)

(b)

Write down the size of angle ADE and give a reason.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (180 − 48) ÷ 2	✓ 1
A1 66°	✓ 1
Part (b)
B1 66° (ECF from (a))	✓ 1
B1 Corresponding angles on parallel lines are equal	✓ 1

Exam tip: In an isosceles triangle the two base angles are equal, so subtract the apex angle from 180° and halve.

+ Add to test✓ In your test93IGME3-4-13r36-12Pythagoras' theorem and trigonometry Paper FEasy5 marks

A ramp in a data centre rises 1.2 m vertically over a horizontal distance of 8 m.

Diagram not to scale

(a)

Calculate the angle the ramp makes with the horizontal. Give your answer correct to 1 decimal place.

(2)

(b)

Calculate the length of the sloping surface of the ramp. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan⁻¹(1.2 ÷ 8)	✓ 1
A1 8.5°	✓ 1
Part (b)
M1 8² + 1.2² (= 65.44)	✓ 1
M1 square root of 65.44	✓ 1
A1 8.09 m	✓ 1

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

+ Add to test✓ In your test94IGME3-4-14r37-01Angles, polygons and circle properties Paper FEasy4 marks

A library has a reading room in the shape of a regular octagon.

(a)

Work out the size of one interior angle of the regular octagon.

(2)

(b)

Another regular polygon has an exterior angle of 40°. Work out the number of sides of this polygon.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 8 = 45 or (8 − 2) × 180 ÷ 8	✓ 1
A1 135°	✓ 1
Part (b)
M1 360 ÷ 40	✓ 1
A1 9	✓ 1

Exam tip: Exterior angles of any polygon add up to 360°, and interior + exterior = 180° at each vertex.

+ Add to test✓ In your test95IGME3-4-14r37-02Pythagoras' theorem and trigonometry Paper FEasy5 marks

A wind turbine tower is 45 m tall and stands on level ground. A support cable runs in a straight line from the top of the tower to an anchor point on the ground 28 m from the base of the tower.

Diagram not to scale

(a)

Work out the length of the cable.

(3)

(b)

The cable costs £12.40 per metre. Work out the cost of the cable.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 45² + 28² (= 2809)	✓ 1
M1 √(their 2809)	✓ 1
A1 53 m	✓ 1
Part (b)
M1 53 × 12.40	✓ 1
A1 £657.20 (ECF from (a))	✓ 1

Common error: When finding the longest side add the squares; to find a shorter side subtract them.

+ Add to test✓ In your test96IGME3-4-14r37-11Angles, polygons and circle properties Paper FEasy4 marks

A triangular warning sign at a wind farm has angles 3x°, (2x + 15)° and (x + 45)°.

Diagram not to scale

(a)

Work out the value of x.

(3)

(b)

Work out the size of the largest angle of the sign.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3x + 2x + 15 + x + 45 = 180	✓ 1
M1 6x + 60 = 180	✓ 1
A1 x = 20	✓ 1
Part (b)
B1 65° (ECF from (a))	✓ 1

Exam tip: Angles in a triangle sum to 180°; after finding x, substitute it into each angle and check they total 180°.

+ Add to test✓ In your test97IGME3-4-14r37-12Pythagoras' theorem and trigonometry Paper FEasy6 marks

A maintenance boat leaves harbour H and sails 12 km due east to wind turbine T. It then sails 9 km due north to turbine U.

Diagram not to scale

(a)

Work out the distance HU.

(3)

(b)

Work out the bearing of U from H. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12² + 9² (= 225)	✓ 1
M1 √225	✓ 1
A1 15 km	✓ 1
Part (b)
M1 tan (angle THU) = 9/12	✓ 1
A1 36.9° (angle THU)	✓ 1
A1 053.1° (90 − 36.87)	✓ 1

Common error: Bearings are measured clockwise from north, so the angle found inside the triangle from east must be subtracted from 90°.

+ Add to test✓ In your test98IGME3-4-14r44-01Area, surface area and volume Paper FEasy5 marks

A window in a museum is made from a rectangle with a semicircle on top. The rectangle is 2.4 m wide and 3.0 m high. The diameter of the semicircle is the 2.4 m width of the rectangle.

Diagram not to scale

(a)

Work out the total area of the window. Give your answer in m² correct to 3 significant figures.

(3)

(b)

The glass costs £45 for each m². Work out the cost of the glass for the window. Use the unrounded area from (a) and give your answer to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rectangle area 2.4 × 3.0 = 7.2	✓ 1
M1 semicircle area ½ × π × 1.2²	✓ 1
A1 9.46 m²	✓ 1
Part (b)
M1 area × 45 (9.46195… × 45)	✓ 1
A1 £425.79	✓ 1

Exam tip: Split a compound shape into simple shapes, find each area separately and add. The radius is half the diameter.

+ Add to test✓ In your test99IGME3-4-14r44-02Pythagoras' theorem and trigonometry Paper FEasy6 marks

A loading ramp for a lorry is a straight plank of length 4.8 m. The top of the plank rests on the lorry bed, which is 1.4 m above the horizontal ground.

Diagram not to scale

(a)

Work out the horizontal distance from the foot of the ramp to the point on the ground directly below the top of the ramp. Give your answer in metres to 3 significant figures.

(3)

(b)

Work out the angle between the ramp and the ground. Give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4.8² − 1.4² (= 21.08)	✓ 1
M1 √21.08	✓ 1
A1 4.59 m	✓ 1
Part (b)
M1 sin x = 1.4 ÷ 4.8 (or cos x = 4.59 ÷ 4.8, tan x = 1.4 ÷ 4.59)	✓ 1
M1 sin⁻¹(1.4 ÷ 4.8)	✓ 1
A1 17.0°	✓ 1

Exam tip: The ramp is the longest side, the hypotenuse. If you are given the hypotenuse and the side opposite the angle, use sine.

+ Add to test✓ In your test100IGME3-4-14r44-03Angles, polygons and circle properties Paper FEasy4 marks

A marine research platform is in the shape of a regular octagon.

(a)

Work out the size of one exterior angle of the octagon.

(1)

(b)

Work out the size of one interior angle of the octagon.

(1)

(c)

Another regular polygon has exterior angles of 18°. Work out the number of sides of this polygon.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 45°	✓ 1
Part (b)
B1 135° (ECF from (a))	✓ 1
Part (c)
M1 360 ÷ 18	✓ 1
A1 20	✓ 1

Exam tip: The exterior angles of any polygon add up to 360°. For a regular polygon, divide 360° by the number of sides to find each one.

+ Add to test✓ In your test101IGME3-4-14r44-04Area, surface area and volume Paper FEasy5 marks

A cylindrical tank at a marine biology centre has a diameter of 2.6 m and a depth of 1.5 m. 1 m³ = 1000 litres.

(a)

Work out the volume of the tank in litres. Give your answer to the nearest litre.

(3)

(b)

The tank is filled from empty at a rate of 40 litres per minute. Use the value 7964 litres to work out how long it takes to fill the tank. Give your answer in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 radius 1.3 m and π × 1.3² × 1.5	✓ 1
M1 × 1000 to convert m³ to litres	✓ 1
A1 7964 litres	✓ 1
Part (b)
M1 7964 ÷ 40	✓ 1
A1 199.1 minutes	✓ 1

Common error: Use the radius, not the diameter, in the formula for the volume of a cylinder: V = πr²h.

+ Add to test✓ In your test102IGME3-4-14r45-01Pythagoras' theorem and trigonometry Paper FEasy4 marks

A warehouse robot drives 36 cm east and then 77 cm north. It then returns to its starting point along the straight line joining its finishing point to its starting point.

Diagram not to scale

(a)

Work out the length of the straight return path.

(2)

(b)

Work out the total distance travelled by the robot.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\sqrt{36^2+77^2}$	✓ 1
A1 85 cm	✓ 1
Part (b)
M1 36 + 77 + 85 (ECF from (a))	✓ 1
A1 198 cm	✓ 1

Exam tip: The east and north directions are at right angles, so the return path is the hypotenuse of a right-angled triangle.

+ Add to test✓ In your test103IGME3-4-14r45-02Angles, polygons and circle properties Paper FEasy5 marks

A fashion designer cuts regular octagons from fabric to make a patchwork.

(a)

Work out the size of one exterior angle of a regular octagon.

(2)

(b)

Work out the size of one interior angle of a regular octagon.

(1)

(c)

Two regular octagons and one other regular polygon fit together at a point with no gaps. Work out the name of the other polygon.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 8	✓ 1
A1 45°	✓ 1
Part (b)
B1 135° (ECF from (a))	✓ 1
Part (c)
M1 360 − 2 × 135 = 90	✓ 1
A1 square (angle 90° means 4 sides)	✓ 1

Common error: Do not confuse the interior angle with the exterior angle. They add up to 180° on a straight line at each vertex.

+ Add to test✓ In your test104IGME3-4-14r45-03Pythagoras' theorem and trigonometry Paper FEasy4 marks

A fishing boat lowers a rope of length 24 m from the surface to an anchor on the sea bed. The rope is straight and makes an angle of 35° below the horizontal.

Diagram not to scale

(a)

Calculate the depth of the sea bed below the boat. Give your answer to 3 significant figures.

(2)

(b)

At another site the sea is 12.0 m deep at a point that is 18.0 m horizontally from the boat. Calculate the angle below the horizontal of the straight line from the boat to this point. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 24 × sin 35°	✓ 1
A1 13.8 m	✓ 1
Part (b)
M1 tan⁻¹(12 ÷ 18)	✓ 1
A1 33.7°	✓ 1

Exam tip: Label the sides opposite, adjacent and hypotenuse relative to the angle before choosing sin, cos or tan.

+ Add to test✓ In your test105IGME3-4-13r46-01Angles, polygons and circle properties Paper FEasy4 marks

A beekeeper studies the cells of a honeycomb frame.

(a)

Each honeycomb cell is a regular hexagon. Work out the size of one interior angle of a regular hexagon.

(2)

(b)

The beekeeper cuts a plastic label in the shape of a regular polygon. Each exterior angle of the label is 18°. Work out the number of sides of the polygon.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 6 = 60 for the exterior angle (or (6 − 2) × 180 ÷ 6)	✓ 1
A1 120°	✓ 1
Part (b)
M1 360 ÷ 18	✓ 1
A1 20	✓ 1

Exam tip: Exterior angles of any polygon add up to 360°, so for a regular polygon divide 360° by the number of sides.

+ Add to test✓ In your test106IGME3-4-13r46-02Pythagoras' theorem and trigonometry Paper FEasy5 marks

The rectangular roof of a beehive measures 54 cm by 42 cm.

(a)

Calculate the length of a diagonal of the roof. Give your answer correct to 3 significant figures.

(3)

(b)

Hasan has a straight rod of length 75 cm. He says it can lie flat on the roof. Is he correct? Give a reason for your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 54² + 42² (= 4680)	✓ 1
M1 √4680	✓ 1
A1 68.4 cm	✓ 1
Part (b)
B1 No (or Hasan is not correct)	✓ 1
B1 The longest straight line on the rectangle is the diagonal, 68.4 cm, which is less than 75 cm	✓ 1

Common error: Adding the two sides instead of using Pythagoras' theorem gives 96 cm, which is far too long for a diagonal.

+ Add to test✓ In your test107IGME3-4-13r46-03Area, surface area and volume Paper FEasy5 marks

A hospital stores water in a tank shaped like a cuboid. The tank is 2.5 m long, 1.2 m wide and 0.8 m high.

(a)

Work out the volume of the tank in m³.

(2)

(b)

Write the volume in litres. 1 m³ = 1000 litres.

(1)

(c)

The full tank is emptied at a constant rate of 40 litres per minute. Work out the time taken to empty the tank.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2.5 × 1.2 × 0.8	✓ 1
A1 2.4 m³	✓ 1
Part (b)
B1 2400 litres	✓ 1
Part (c)
M1 2400 ÷ 40	✓ 1
A1 60 minutes	✓ 1

Exam tip: Convert to the unit used in the rate before dividing, so that litres are divided by litres per minute.

+ Add to test✓ In your test108IGME3-4-13r46-04Pythagoras' theorem and trigonometry Paper FEasy6 marks

In a Formula 1 garage, cars are wheeled up a straight service ramp on to a platform.

Diagram not to scale

(a)

The first ramp is 14 m long and is inclined at 8° to the horizontal. Calculate the height of the top of the ramp above the ground. Give your answer correct to 3 significant figures.

(3)

(b)

A second ramp rises 1.2 m over a horizontal distance of 9 m. Calculate the angle this ramp makes with the horizontal. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height = 14 × sin 8°	✓ 1
M1 correct evaluation 1.948…	✓ 1
A1 1.95 m	✓ 1
Part (b)
M1 tan x = 1.2 ÷ 9	✓ 1
M1 x = tan⁻¹(1.2 ÷ 9)	✓ 1
A1 7.6°	✓ 1

Exam tip: Label the sides opposite, adjacent and hypotenuse before choosing sin, cos or tan.

+ Add to test✓ In your test109IGME3-4-14r46-01Angles, polygons and circle properties Paper FEasy4 marks

A wind turbine has three identical blades of length 45 m, equally spaced around the hub.

Diagram not to scale

(a)

Work out the angle between two neighbouring blades.

(2)

(b)

Calculate the distance travelled by the tip of a blade in one complete turn. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 3	✓ 1
A1 120°	✓ 1
Part (b)
M1 2 × π × 45	✓ 1
A1 283 m	✓ 1

Exam tip: The tip of the blade moves round a circle whose radius is the blade length, so use circumference = 2πr.

+ Add to test✓ In your test110IGME3-4-14r46-02Pythagoras' theorem and trigonometry Paper FEasy6 marks

In a library, a ladder of length 3.9 m leans against a straight bookshelf. The ladder reaches a height h up the shelf. The ground is horizontal and the shelf is vertical.

Diagram not to scale

(a)

The foot of the ladder is 1.5 m from the bottom of the shelf. Calculate the height h.

(3)

(b)

The foot of the ladder is moved to 2.1 m from the shelf. Calculate the new height reached by the ladder. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3.9² − 1.5² (= 12.96)	✓ 1
M1 √12.96	✓ 1
A1 3.6 m	✓ 1
Part (b)
M1 3.9² − 2.1² (= 10.8)	✓ 1
M1 √10.8	✓ 1
A1 3.29 m	✓ 1

Common error: When the unknown side is a shorter side, subtract the squares; adding them gives the hypotenuse.

+ Add to test✓ In your test111IGME3-4-14r46-03Area, surface area and volume Paper FEasy5 marks

A swimming pool has a rectangular floor 12 m by 8 m. The floor is to be covered with square tiles of side 25 cm.

(a)

Work out the area of the floor of the pool in m².

(2)

(b)

Work out the number of tiles needed to cover the floor completely.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 × 8	✓ 1
A1 96 m²	✓ 1
Part (b)
M1 area of one tile 0.25 × 0.25 = 0.0625 m² (or 625 cm²)	✓ 1
M1 96 ÷ 0.0625	✓ 1
A1 1536	✓ 1

Exam tip: Make sure both areas are in the same units before dividing: either m² for both or cm² for both.

+ Add to test✓ In your test112IGME3-4-14r46-04Pythagoras' theorem and trigonometry Paper FEasy6 marks

Wind turbine towers stand on level ground on a wind farm.

Diagram not to scale

(a)

From a point 80 m from the base of the first tower, the angle of elevation of the top of the tower is 52°. Calculate the height of the tower. Give your answer correct to 3 significant figures.

(3)

(b)

A second tower is 95 m tall. From a point P on the ground, the angle of elevation of the top of this tower is 35°. Calculate the distance of P from the base of the tower. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height = 80 × tan 52°	✓ 1
M1 correct evaluation 102.39…	✓ 1
A1 102 m	✓ 1
Part (b)
M1 tan 35° = 95 ÷ distance	✓ 1
M1 distance = 95 ÷ tan 35°	✓ 1
A1 136 m	✓ 1

Exam tip: If the unknown is in the denominator, rearrange before using the calculator: distance = height ÷ tan(angle).

+ Add to test✓ In your test113IGME-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.

Diagram not to scale

(a)

Work out the volume of the doorstop.

(2)

(b)

Work out the total surface area of the doorstop.

(4)

Show mark scheme

Marking point	Mark	Notes
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.

+ Add to test✓ In your test114IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test115IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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°.

+ Add to test✓ In your test116IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test117IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test118IGME-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.

Diagram not to scale

(a)

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

(4)

Show mark scheme

Marking point	Mark	Notes
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.

+ Add to test✓ In your test119IGME-061-00-05Similar shapes and circle theorems 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°.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test120IGME3-4-10d-06Area, surface area and volume Paper FMedium7 marks

A camping tent is a triangular prism lying on one of its rectangular faces. Each triangular end is isosceles with base 4 m and perpendicular height 1.5 m. The tent is 3.2 m long.

Diagram not to scale

(a)

Work out the length of each sloping side of a triangular end.

(2)

(b)

Work out the volume of the tent.

(2)

(c)

The two sloping rectangular sides and the two triangular ends are made from fabric. The floor is not. Work out the total area of fabric.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(2² + 1.5²)	✓ 1
A1 2.5 m	✓ 1
Part (b)
M1 cross-section area ½ × 4 × 1.5 = 3 m²	✓ 1
A1 3 × 3.2 = 9.6 m³	✓ 1
Part (c)
M1 2 × 2.5 × 3.2 = 16 m² (ECF from (a))	✓ 1
M1 2 × ½ × 4 × 1.5 = 6 m²	✓ 1
A1 22 m²	✓ 1

Common error: The sloping length is needed for the rectangular sides, but the perpendicular height is needed for the triangle area. Do not mix them up.

+ Add to test✓ In your test121IGME3-4-10d-07Pythagoras' theorem and trigonometry Paper FMedium6 marks

A lighthouse is 60 m high. From the top of the lighthouse, the angle of depression of a boat at sea is 24°. The boat then sails directly towards the base of the lighthouse until the angle of depression is 35°. The sea is level and the base of the lighthouse is at sea level.

Diagram not to scale

(a)

Work out the original distance of the boat from the base of the lighthouse. Give your answer to 3 significant figures.

(3)

(b)

Work out how far the boat sailed. Do not round until your final answer, which should be to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 24° = 60 ÷ d (angle of depression equals the angle of elevation from the boat)	✓ 1
M1 d = 60 ÷ tan 24°	✓ 1
A1 135 m	✓ 1
Part (b)
M1 new distance 60 ÷ tan 35° = 85.69...	✓ 1
M1 134.76... − 85.68...	✓ 1
A1 49.1 m	✓ 1

Exam tip: The angle of depression from the top equals the angle of elevation from the boat, because they are alternate angles.

+ Add to test✓ In your test122IGME3-4-11r5-05Area, surface area and volume Paper FMedium5 marks

A grain store is a cylinder of radius 2.5 m and height 6 m with a hemisphere of radius 2.5 m fixed on top of it. Use the formulae: volume of a cylinder = πr²h; volume of a sphere = (4/3)πr³.

Cross-section. Diagram not to scale

(a)

Calculate the volume of the cylinder. Give your answer to 1 decimal place.

(2)

(b)

Calculate the total volume of the grain store. Give your answer to 1 decimal place. Use unrounded values in your working.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 2.5² × 6	✓ 1
A1 117.8 m³	✓ 1
Part (b)
M1 hemisphere volume = ½ × (4/3) × π × 2.5³ (= 32.7…)	✓ 1
M1 add to cylinder volume	✓ 1
A1 150.5 m³	✓ 1

Exam tip: A hemisphere is half a sphere: halve the sphere formula before adding it to the volume of the other solid.

+ Add to test✓ In your test123IGME3-4-11r5-06Area, surface area and volume Paper FMedium7 marks

A drum is a closed cylinder of diameter 40 cm and height 25 cm. Give money answers to the nearest penny and use unrounded values in your working.

(a)

Calculate the area of one circular end of the drum. Give your answer to the nearest cm².

(2)

(b)

Calculate the area of the curved surface. Give your answer to the nearest cm².

(2)

(c)

The whole outside of the drum is covered with a material that costs £0.0065 per cm². Calculate the cost of the material.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 20²	✓ 1
A1 1257 cm²	✓ 1
Part (b)
M1 π × 40 × 25 (or 2π × 20 × 25)	✓ 1
A1 3142 cm²	✓ 1
Part (c)
M1 total area = 2 × π × 20² + π × 40 × 25 (= 5654.8…)	✓ 1
M1 multiply total area by 0.0065	✓ 1
A1 £36.76	✓ 1

Common error: Remember a closed cylinder has two circular ends as well as the curved surface; leaving one out is a frequent slip.

+ Add to test✓ In your test124IGME3-4-11r5-08Area, surface area and volume Paper FMedium6 marks

A rotating sprinkler waters a sector of a circular field. The sector has radius 60 m and angle 140°.

Diagram not to scale

(a)

Calculate the area watered. Give your answer to the nearest m².

(2)

(b)

Calculate the length of the curved edge of the sector. Give your answer to 1 decimal place.

(2)

(c)

A fence is built along the whole boundary of the sector (the curved edge and the two straight edges). Calculate the length of the fence. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (140 ÷ 360) × π × 60²	✓ 1
A1 4398 m²	✓ 1
Part (b)
M1 (140 ÷ 360) × 2 × π × 60	✓ 1
A1 146.6 m	✓ 1
Part (c)
M1 2 × 60 + 146.6… (the two radii plus the arc)	✓ 1
A1 266.6 m	✓ 1

Common error: The boundary of a sector includes the two straight radii as well as the arc, so add 2r to the arc length.

+ Add to test✓ In your test125IGME3-4-13r5-05Pythagoras' theorem and trigonometry Paper FMedium5 marks

Calculator

A straight water pipe runs down a hillside. The pipe is 85 m long and makes an angle of 12° with the horizontal.

Diagram not to scale (vertical scale exaggerated)

(a)

Calculate the vertical drop of the pipe. Give your answer correct to 3 significant figures.

(2)

(b)

A second pipe falls 11 m vertically over a horizontal distance of 70 m. Calculate the angle this pipe makes with the horizontal. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 85 × sin 12°	✓ 1
A1 17.7 m	✓ 1
Part (b)
M1 tan(angle) = 11/70	✓ 1
M1 tan⁻¹(11/70)	✓ 1
A1 8.9°	✓ 1

Common error: Check the calculator is in degree mode; a radian setting gives a completely different value.

+ Add to test✓ In your test126IGME3-4-13r5-06Area, surface area and volume Paper FMedium4 marks

Calculator

A circular lawn sprinkler waters a sector of a circle of radius 18 m. The angle of the sector is 140°. Use the π button on your calculator.

Diagram not to scale

(a)

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

(2)

(b)

Calculate the length of the curved edge of the watered region. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (140/360) × π × 18²	✓ 1
A1 396 m²	✓ 1
Part (b)
M1 (140/360) × 2 × π × 18	✓ 1
A1 44.0 m	✓ 1

Common error: Area uses r² and arc length uses 2πr; mixing the two formulae is a frequent slip.

+ Add to test✓ In your test127IGME3-4-13r5-07Angles, polygons and circle properties Paper FMedium5 marks

Calculator

On a game screen, a triangular badge has angles of (2x + 10)°, (3x − 5)° and (x + 25)°.

Diagram not to scale

(a)

Show that 6x + 30 = 180.

(2)

(b)

Solve the equation to find the value of x.

(2)

(c)

Work out the size of the smallest angle in the badge.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sum of the three angles = 180	✓ 1
A1 2x + 10 + 3x − 5 + x + 25 collected correctly to 6x + 30 = 180	✓ 1
Part (b)
M1 6x = 150	✓ 1
A1 x = 25	✓ 1
Part (c)
A1 50°	✓ 1

Exam tip: In a 'show that' question, write every step; the final line must match the statement given.

+ Add to test✓ In your test128IGME3-4-13r5-11Area, surface area and volume Paper FMedium5 marks

Calculator

A drinking-water trough is a prism. Its cross-section is a trapezium with parallel sides 0.6 m and 0.9 m, and the distance between them is 0.4 m. The trough is 5 m long. 1 m³ = 1000 litres.

Cross-section of the trough. Diagram not to scale

(a)

Calculate the area of the cross-section.

(2)

(b)

Calculate the volume of the trough in m³.

(1)

(c)

Water flows into the empty trough at 12 litres per minute. Calculate the time taken to fill it.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (0.6 + 0.9) × 0.4	✓ 1
A1 0.3 m²	✓ 1
Part (b)
B1 0.3 × 5 = 1.5 m³	✓ 1
Part (c)
M1 1.5 × 1000 ÷ 12 (1500 litres)	✓ 1
A1 125 minutes	✓ 1

Exam tip: Write the prism rule as volume = area of cross-section × length so each step is visible.

+ Add to test✓ In your test129IGME3-4-11r6-04Area, surface area and volume Paper FMedium4 marks

A circular stone disc found at a dig has a diameter of 64 cm. A decorated sector of the disc has an angle of 72° at the centre. Give answers to 3 significant figures.

Diagram not to scale

(a)

Calculate the length of the curved edge (arc) of the decorated sector.

(2)

(b)

Calculate the area of the decorated sector.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 72/360 × π × 64	✓ 1
A1 40.2 cm	✓ 1
Part (b)
M1 72/360 × π × 32²	✓ 1
A1 643 cm²	✓ 1

Common error: The formula for area needs the radius, so halve the diameter first.

+ Add to test✓ In your test130IGME3-4-11r6-05Pythagoras' theorem and trigonometry Paper FMedium6 marks

A cycling club builds two ramps, each in the shape of a right-angled triangle when viewed from the side. The first ramp has a sloping surface 4.5 m long and rises 0.6 m vertically. The second ramp makes an angle of 9° with the horizontal and rises 0.8 m vertically.

Side view of each ramp. Diagram not to scale (vertical scale exaggerated)

(a)

Calculate the angle that the first ramp makes with the horizontal. Give your answer correct to 1 decimal place.

(3)

(b)

Calculate the length of the sloping surface of the second ramp. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin x = 0.6 ÷ 4.5	✓ 1
M1 x = sin⁻¹(0.6 ÷ 4.5)	✓ 1
A1 7.7°	✓ 1
Part (b)
M1 sin 9° = 0.8 ÷ length	✓ 1
M1 length = 0.8 ÷ sin 9°	✓ 1
A1 5.11 m	✓ 1

Exam tip: Label the sides opposite, adjacent and hypotenuse before choosing sine, cosine or tangent.

+ Add to test✓ In your test131IGME3-4-11r6-06Angles, polygons and circle properties Paper FMedium5 marks

ABCDE is a pentagon. The interior angles are x°, (x + 10)°, 2x°, 110° and 120°.

Diagram not to scale

(a)

Show that the sum of the interior angles of a pentagon is 540°.

(2)

(b)

Work out the value of x.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (5 − 2) × 180 or 3 triangles	✓ 1
A1 540°	✓ 1
Part (b)
M1 x + x + 10 + 2x + 110 + 120 = 540	✓ 1
M1 4x + 240 = 540	✓ 1
A1 x = 75	✓ 1

Common error: Collect the number terms and the x terms separately before solving.

+ Add to test✓ In your test132IGME3-4-11r6-07Area, surface area and volume Paper FMedium5 marks

A data centre uses a cylindrical water tank for cooling. The tank has radius 0.8 m and height 2.5 m. It has a flat circular base and no lid.

(a)

Calculate the volume of the tank. Give your answer in m³ correct to 3 significant figures.

(2)

(b)

Calculate the total area of the outside surface of the tank, including the base. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 0.8² × 2.5	✓ 1
A1 5.03 m³	✓ 1
Part (b)
M1 curved surface 2 × π × 0.8 × 2.5 (= 12.57)	✓ 1
M1 base π × 0.8² (= 2.01)	✓ 1
A1 14.6 m²	✓ 1

Common error: An open tank has only one circular face, so do not add a second circle for a lid.

+ Add to test✓ In your test133IGME3-4-11r9-06Pythagoras' theorem and trigonometry Paper FMedium5 marks

A rail freight container is a cuboid with internal length 12.0 m, width 2.4 m and height 2.6 m.

Diagram not to scale

(a)

Calculate the length of the diagonal of the floor of the container. Give your answer correct to 3 significant figures.

(2)

(b)

A straight rigid pole is placed inside the container from one bottom corner to the opposite top corner. Calculate the length of this pole. Use the unrounded answer from (a). Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 12.0² + 2.4² (= 149.76)	✓ 1
A1 for 12.2 m	✓ 1
Part (b)
M1 for a right-angled triangle with the floor diagonal and the height 2.6 m	✓ 1
M1 for 149.76 + 2.6² (= 156.52) then square root	✓ 1
A1 for 12.5 m (ECF from (a))	✓ 1

Exam tip: For a cuboid, the space diagonal comes from two applications of Pythagoras' theorem, or from $\sqrt{l^2+w^2+h^2}$.

+ Add to test✓ In your test134IGME3-4-11r9-09Similar shapes and circle theorems Paper FMedium5 marks

A rail bridge has a triangular frame ABC with AB = AC. Angle ABC = (3x + 8)° and angle ACB = (5x − 20)°.

Diagram not to scale

(a)

Work out the value of x.

(3)

(b)

Work out the size of angle BAC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 3x + 8 = 5x − 20 (base angles of an isosceles triangle are equal)	✓ 1
M1 for 28 = 2x (or equivalent rearrangement)	✓ 1
A1 for x = 14	✓ 1
Part (b)
M1 for 180 − 2 × 50 (angle ABC = 50°)	✓ 1
A1 for 80° (ECF from (a))	✓ 1

Exam tip: Write down the geometrical fact you use (here: base angles equal) before forming the equation.

+ Add to test✓ In your test135IGME3-4-11r9-13Similar shapes and circle theorems Paper FMedium4 marks

In a sports science laboratory, a ramp frame is built as triangle ABC. Point D lies on AB and point E lies on AC so that DE is parallel to BC. AD = 6 cm, DB = 4.5 cm and DE = 7.2 cm.

Diagram not to scale

(a)

Explain why triangle ADE is similar to triangle ABC.

(1)

(b)

Calculate the length of BC.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for corresponding angles equal (DE parallel to BC, and angle at A is common), so the triangles have equal angles	✓ 1
Part (b)
M1 for AB = 6 + 4.5 = 10.5	✓ 1
M1 for BC ÷ 7.2 = 10.5 ÷ 6 (or 7.2 × 10.5 ÷ 6)	✓ 1
A1 for 12.6 cm	✓ 1

Common error: The scale factor uses AB (the whole side), not DB; compare 10.5 with 6, not 4.5 with 6.

+ Add to test✓ In your test136IGME3-4-11r9-14Pythagoras' theorem and trigonometry Paper FMedium5 marks

A mountain railway has a constant gradient of 1 in 35. This means the track rises 1 m vertically for every 35 m travelled horizontally.

(a)

Calculate the angle the track makes with the horizontal. Give your answer correct to 3 significant figures.

(2)

(b)

A train climbs a vertical height of 28 m along this track. Calculate the length of track it travels. Use the unrounded angle from (a). Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for tan x = 1 ÷ 35	✓ 1
A1 for 1.64°	✓ 1
Part (b)
M1 for sin(angle) = 28 ÷ length (or the horizontal distance 28 × 35 = 980 m, then Pythagoras)	✓ 1
M1 for length = 28 ÷ sin(1.636…°)	✓ 1
A1 for 980 m	✓ 1

Exam tip: Read gradient statements carefully: 1 in 35 compares vertical with horizontal, so use tan, not sin.

+ Add to test✓ In your test137IGME3-4-12r9-06Pythagoras' theorem and trigonometry Paper FMedium5 marks

On a map of a region, clinic B is 12 km due east of clinic A. Clinic C is 9 km due north of clinic B.

Diagram not to scale

(a)

Calculate the direct distance from A to C.

(2)

(b)

Work out the bearing of C from A. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 12² + 9² (= 225)	✓ 1
A1 for 15 km	✓ 1
Part (b)
M1 for tan(angle CAB) = 9 ÷ 12 (angle at A measured from east)	✓ 1
A1 for 36.9° (36.87…)	✓ 1
A1 for bearing 053.1° (90° − 36.9°, or 90 − 36.87…)	✓ 1

Common error: A bearing is measured clockwise from north; the angle found inside the triangle is measured from east, so subtract it from 90°.

+ Add to test✓ In your test138IGME3-4-12r9-07Similar shapes and circle theorems Paper FMedium6 marks

A shop sells two triangular pennants that are mathematically similar. Pennant 1 is triangle PQR and pennant 2 is triangle XYZ, where PQ corresponds to XY, QR to YZ and PR to XZ. PQ = 8.4 cm, QR = 10 cm, XY = 12.6 cm and XZ = 18 cm.

Diagram not to scale

(a)

Work out the scale factor from pennant 1 to pennant 2.

(2)

(b)

Calculate the length of YZ.

(2)

(c)

Calculate the length of PR.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 12.6 ÷ 8.4	✓ 1
A1 for 1.5	✓ 1
Part (b)
M1 for 10 × 1.5 (ECF from (a))	✓ 1
A1 for 15 cm	✓ 1
Part (c)
M1 for 18 ÷ 1.5 (dividing, since PR is on the smaller pennant)	✓ 1
A1 for 12 cm	✓ 1

Common error: When going from the larger shape back to the smaller one, divide by the scale factor rather than multiplying.

+ Add to test✓ In your test139IGME3-4-12r9-09Angles, polygons and circle properties Paper FMedium5 marks

The floor of a pop-up shop is tiled with regular polygons. One of the polygons has an interior angle that is 8 times the size of its exterior angle.

(a)

Work out the size of one exterior angle of this polygon.

(2)

(b)

Work out the number of sides of the polygon.

(1)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for exterior angle + interior angle = 180° (angles on a straight line), so 9 × exterior = 180	✓ 1
A1 for 20°	✓ 1
Part (b)
B1 for 360 ÷ 20 = 18 (ECF from (a))	✓ 1
Part (c)
M1 for (18 − 2) × 180	✓ 1
A1 for 2880° (ECF from (b))	✓ 1

Exam tip: Interior + exterior = 180° at each vertex, so a ratio between them gives the exterior angle directly.

+ Add to test✓ In your test140IGME3-4-12r9-14Area, surface area and volume Paper FMedium5 marks

A channel carries water to a small hydroelectric turbine. A 30 m section of the channel is a prism whose cross-section is a trapezium. The two parallel sides of the trapezium are 4.2 m (at the bottom) and 6.8 m (at the water surface), and the depth of the water is 2.5 m.

Cross-section of the water. Diagram not to scale

(a)

Calculate the area of the cross-section of the water.

(2)

(b)

Calculate the volume of water in the 30 m section.

(2)

(c)

The empty section is filled at 15 m³ per minute. Work out the time taken to fill it, in minutes.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for ½ × (4.2 + 6.8) × 2.5	✓ 1
A1 for 13.75 m²	✓ 1
Part (b)
M1 for 13.75 × 30 (ECF from (a))	✓ 1
A1 for 412.5 m³	✓ 1
Part (c)
B1 for 412.5 ÷ 15 = 27.5 minutes (ECF from (b))	✓ 1

Exam tip: The depth in the trapezium formula is the perpendicular distance between the parallel sides, not a sloping side.

+ Add to test✓ In your test141IGME3-4-12r10-02Pythagoras' theorem and trigonometry Paper FMedium5 marks

A bridge tower is vertical. A straight cable runs from the top of the tower to a point on the horizontal deck that is 21 m from the base of the tower. The top of the tower is 12 m above the deck.

Diagram not to scale

(a)

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

(3)

(b)

Cable costs £35 per metre and two identical cables are needed. Work out the total cost, to the nearest pound. Use the unrounded length from (a).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 21² + 12² (= 585)	✓ 1
M1 √585	✓ 1
A1 24.2 (m)	✓ 1
Part (b)
M1 2 × 24.18… × 35	✓ 1
A1 £1693	✓ 1

Common error: When using Pythagoras to find a longest side, add the squares; subtract only when finding a shorter side.

+ Add to test✓ In your test142IGME3-4-12r10-04Pythagoras' theorem and trigonometry Paper FMedium4 marks

A lander has a straight landing strut of length 3.2 m. The strut makes an angle of 64° with the horizontal ground.

Diagram not to scale

(a)

Work out the height of the top of the strut above the ground. Give your answer to 3 significant figures.

(2)

(b)

A second design uses a strut of length 3.2 m whose top is 2.5 m above the ground. Work out the angle this strut makes with the ground. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3.2 × sin 64°	✓ 1
A1 2.88 (m)	✓ 1
Part (b)
M1 sin x = 2.5 ÷ 3.2	✓ 1
A1 51.4°	✓ 1

Exam tip: Label the sides opposite, adjacent and hypotenuse first, then choose sin, cos or tan.

+ Add to test✓ In your test143IGME3-4-12r10-05Area, surface area and volume Paper FMedium5 marks

The cross-section of a bridge pier is a rectangle 6 m wide and 4 m high with a semicircle of diameter 6 m on top of the 6 m side. The pier is a prism of height 12 m with this cross-section.

Cross-section of the pier. Diagram not to scale

(a)

Work out the area of the cross-section. Give your answer to 3 significant figures.

(3)

(b)

Work out the volume of the pier, to the nearest m³. Use the unrounded area from (a).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6 × 4 = 24	✓ 1
M1 ½ × π × 3²	✓ 1
A1 38.1 (m²)	✓ 1
Part (b)
M1 38.137… × 12	✓ 1
A1 458 (m³)	✓ 1

Common error: The semicircle has radius 3 m, which is half of the diameter.

+ Add to test✓ In your test144IGME3-4-12r10-12Area, surface area and volume Paper FMedium5 marks

A space probe's fuel tank is a cuboid with a base measuring 2.5 m by 1.2 m. The volume of the tank is 7.2 m³.

Diagram not to scale

(a)

Work out the height of the tank.

(2)

(b)

Work out the total surface area of the tank.

(2)

(c)

One coat of paint costs £14.50 per m². Work out the cost of one coat on the whole outside of the tank. (ECF from (b))

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 7.2 ÷ (2.5 × 1.2)	✓ 1
A1 2.4 (m)	✓ 1
Part (b)
M1 2(2.5 × 1.2 + 2.5 × 2.4 + 1.2 × 2.4)	✓ 1
A1 23.76 (m²)	✓ 1
Part (c)
B1 £344.52	✓ 1

Exam tip: A cuboid has six faces, in three equal pairs; add three different face areas and double.

+ Add to test✓ In your test145IGME3-4-13r10-04Area, surface area and volume Paper FMedium6 marks

A window in a new library is made from a rectangle, 1.2 m wide and 1.8 m high, with a semicircle on top. The diameter of the semicircle is the 1.2 m width of the rectangle.

Diagram not to scale

(a)

Show that the area of the semicircle is 0.565 m² correct to 3 significant figures.

(2)

(b)

Calculate the total area of the window. Give your answer in m² correct to 3 significant figures.

(2)

(c)

Glass for the window costs £48 per m². Use the unrounded total area to work out the cost of the glass, in pounds to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × π × 0.6²	✓ 1
A1 0.5655 (m²) which rounds to 0.565	✓ 1
Part (b)
M1 1.2 × 1.8 + 0.565... (= 2.16 + semicircle area)	✓ 1
A1 2.73 (m²)	✓ 1
Part (c)
M1 total area × 48	✓ 1
A1 £130.82	✓ 1

Common error: Using the width as the radius of the semicircle: the radius is half of the 1.2 m diameter.

+ Add to test✓ In your test146IGME3-4-13r10-05Pythagoras' theorem and trigonometry Paper FMedium6 marks

A marine biologist stands on the top of a vertical cliff, 38 m above sea level, and watches a seal in the sea. The sea is horizontal and the cliff face is vertical.

Diagram not to scale

(a)

The angle of depression of the seal from the biologist is 24°. Calculate the horizontal distance of the seal from the base of the cliff. Give your answer in metres correct to 3 significant figures.

(3)

(b)

The seal swims directly towards the cliff until the angle of depression is 55°. Use unrounded values to calculate how far the seal swims. Give your answer in metres correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 24° = 38 ÷ d (angle at the seal is 24°)	✓ 1
M1 d = 38 ÷ tan 24°	✓ 1
A1 85.3 (m)	✓ 1
Part (b)
M1 38 ÷ tan 55° (= 26.61)	✓ 1
M1 38 ÷ tan 24° − 38 ÷ tan 55°	✓ 1
A1 58.7 (m)	✓ 1

Exam tip: The angle of depression is measured from the horizontal at the observer, and it equals the angle of elevation measured from the object, by alternate angles.

+ Add to test✓ In your test147IGME3-4-13r10-06Area, surface area and volume Paper FMedium7 marks

A shipping container has internal length 6.0 m, internal width 2.4 m and internal height 2.4 m. It is to be loaded with identical upright cylindrical drums. Each drum has radius 0.30 m and height 0.90 m. Drums are placed in layers on the floor and may be stacked directly on top of each other.

(a)

Calculate the volume of one drum. Give your answer in m³ correct to 3 significant figures.

(2)

(b)

The drums in each layer are arranged in a square grid, with each drum occupying a 0.60 m by 0.60 m square of floor. Work out the greatest number of drums that can be loaded into the container.

(3)

(c)

Work out the percentage of the volume of the container that is filled by the 80 drums. Use the unrounded drum volume and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 0.30² × 0.90	✓ 1
A1 0.254 (m³)	✓ 1
Part (b)
M1 6.0 ÷ 0.60 = 10 and 2.4 ÷ 0.60 = 4, so 40 drums per layer	✓ 1
M1 2.4 ÷ 0.90 = 2.67 so 2 layers	✓ 1
A1 80	✓ 1
Part (c)
M1 80 × π × 0.30² × 0.90 ÷ (6.0 × 2.4 × 2.4) × 100	✓ 1
A1 58.9 (%)	✓ 1

Common error: Dividing the container volume by the drum volume gives an impossible number of drums: the drums cannot fill every gap, so count whole drums along each edge.

+ Add to test✓ In your test148IGME3-4-13r10-11Area, surface area and volume Paper FMedium7 marks

A garden in the shape of a sector of a circle is to be built around a fountain. The sector has radius 12 m and angle 135° at the centre.

Diagram not to scale

(a)

Calculate the length of the curved edge of the garden. Give your answer in metres correct to 3 significant figures.

(2)

(b)

Calculate the area of the garden. Give your answer in m² correct to 1 decimal place.

(2)

(c)

A fence is placed along the whole perimeter of the garden (the curved edge and both straight edges). Fencing costs £14.50 per metre. Use the unrounded curved length to work out the cost of the fence, to the nearest penny.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 135 ÷ 360 × 2 × π × 12	✓ 1
A1 28.3 (m)	✓ 1
Part (b)
M1 135 ÷ 360 × π × 12²	✓ 1
A1 169.6 (m²)	✓ 1
Part (c)
M1 perimeter = curved length + 12 + 12	✓ 1
M1 perimeter × 14.50	✓ 1
A1 £757.98	✓ 1

Common error: Forgetting the two straight edges when finding the perimeter of a sector: the arc is only part of the boundary.

+ Add to test✓ In your test149IGME3-4-13r12-05Area, surface area and volume Paper FMedium8 marks

A window in a shop is made from a rectangle 1.8 m wide and 2.2 m high, with a semicircle of diameter 1.8 m fixed on top of the rectangle, so that the straight edge of the semicircle is the top edge of the rectangle.

Diagram not to scale

(a)

Calculate the area of the window. Give your answer in m² to 3 significant figures.

(3)

(b)

A rubber seal runs round the whole outside edge of the window (the two sides, the bottom and the curved top). Calculate the length of the seal. Give your answer in metres to 3 significant figures.

(3)

(c)

The seal costs £1.40 per metre. Work out the cost of the seal. Use the exact length from (b) without rounding it first, and give your answer to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for rectangle area $1.8 \times 2.2$ (= 3.96)	✓ 1
M1 for semicircle area $\frac{1}{2}\pi \times 0.9^2$ (= 1.272...)	✓ 1
A1 for 5.23 m² (5.232...)	✓ 1
Part (b)
M1 for $2 \times 2.2 + 1.8$ (= 6.2)	✓ 1
M1 for semicircle arc $\frac{1}{2} \times \pi \times 1.8$ (= 2.827...)	✓ 1
A1 for 9.03 m (9.027...)	✓ 1
Part (c)
M1 for $9.0274 \times 1.40$ (or their length from (b) × 1.40)	✓ 1
A1 for £12.64 (12.638...; 9.03 × 1.40 also gives 12.64)	✓ 1

Common error: The diameter of the semicircle is 1.8 m, so use radius 0.9 m in the area formula, and do not include the straight edge between the rectangle and semicircle in the perimeter.

+ Add to test✓ In your test150IGME3-4-13r12-06Pythagoras' theorem and trigonometry Paper FMedium5 marks

The tower of a wind turbine stands on level ground. A surveyor stands 28 m from the base of the tower. The angle of elevation of the hub (the centre of the rotor) is 61°. One blade is pointing vertically upwards from the hub, and the angle of elevation of the tip of this blade is 70°. Assume the tower and blade are vertical and in line with the surveyor, and ignore the height of the surveyor.

Diagram not to scale

(a)

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

(2)

(b)

Calculate the length of the blade. Give your answer in metres to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $28 \tan 61°$	✓ 1
A1 for 50.5 m (50.51...)	✓ 1
Part (b)
M1 for height of the tip $28 \tan 70°$ (= 76.9...)	✓ 1
M1 for subtracting the hub height, $28\tan 70° - 28 \tan 61°$	✓ 1
A1 for 26.4 m (26.41...)	✓ 1

Exam tip: Keep unrounded values in your calculator between steps. Here 76.9 − 50.5 gives the same answer to 3 significant figures, but in other questions rounding early can change the last digit.

+ Add to test✓ In your test151IGME3-4-13r12-07Area, surface area and volume Paper FMedium9 marks

A chocolate box is a prism. Its cross-section is a right-angled triangle with the two shorter sides 7.5 cm and 10 cm. The length of the prism is 30 cm.

Diagram not to scale

(a)

Work out the length of the longest side of the triangular cross-section.

(2)

(b)

Work out the volume of the box.

(2)

(c)

Work out the total surface area of the box (two triangular ends and three rectangular faces).

(3)

(d)

The card used costs £0.35 for every 100 cm². Work out the cost of the card for one box, assuming there is no waste. Give your answer to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $7.5^2 + 10^2$ (= 156.25)	✓ 1
A1 for 12.5 cm	✓ 1
Part (b)
M1 for $\frac{1}{2} \times 7.5 \times 10$ (= 37.5)	✓ 1
A1 for 1125 cm³	✓ 1
Part (c)
M1 for area of both triangles $2 \times 37.5$ (= 75)	✓ 1
M1 for rectangles $(7.5 + 10 + 12.5) \times 30$ (= 900)	✓ 1
A1 for 975 cm²	✓ 1
Part (d)
M1 for $975 \div 100 \times 0.35$	✓ 1
A1 for £3.41 (3.4125)	✓ 1

Exam tip: For a prism, the rectangular faces together form one large rectangle whose width is the perimeter of the cross-section and whose length is the length of the prism.

+ Add to test✓ In your test152IGME3-4-13r12-08Similar shapes and circle theorems Paper FMedium6 marks

A model of a rectangular solar panel is mathematically similar to the real panel. The model is 18 cm high and 24 cm wide. The real panel is 2.7 m high.

(a)

Work out the scale factor of enlargement from the model to the real panel.

(2)

(b)

Work out the width of the real panel. Give your answer in metres.

(2)

(c)

Work out the area of the front face of the real panel. Give your answer in m².

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for converting units, e.g. 2.7 m = 270 cm, and $270 \div 18$	✓ 1
A1 for 15	✓ 1
Part (b)
M1 for $24 \times$ their scale factor (= 360 cm)	✓ 1
A1 for 3.6 m (ECF from (a))	✓ 1
Part (c)
M1 for $2.7 \times 3.6$ (or their height × their width in the same unit)	✓ 1
A1 for 9.72 m²	✓ 1

Common error: Make sure both lengths are in the same unit before you divide to find a scale factor. Mixing cm and m gives a factor that is 100 times too big or too small.

+ Add to test✓ In your test153IGME3-4-12r14-05Pythagoras' theorem and trigonometry Paper FMedium7 marks

A train leaves station A and travels 24 km on a bearing of 040° to station B. It then travels 18 km on a bearing of 130° to station C.

Diagram not to scale

(a)

Show that angle ABC is 90°.

(2)

(b)

Calculate the direct distance from A to C.

(2)

(c)

Calculate the bearing of C from A. Give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Bearing of A from B is 040° + 180° = 220°	✓ 1
220° − 130° = 90°	✓ 1
Part (b)
24² + 18² = 900	✓ 1
30 km	✓ 1
Part (c)
tan (angle BAC) = 18 ÷ 24	✓ 1
Angle BAC = 36.9° (36.87°…)	✓ 1
040° + 36.87…° = 076.9°	✓ 1

Exam tip: A bearing is measured clockwise from north using three figures; to find a return bearing, add or subtract 180°.

+ Add to test✓ In your test154IGME3-4-12r14-09Area, surface area and volume Paper FMedium7 marks

A diesel storage tank at a rail depot is an upright cylinder of radius 0.9 m and height 4.2 m. 1 m³ = 1000 litres.

(a)

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

(2)

(b)

The tank is filled to a depth of 3.15 m. Calculate the volume of diesel in the tank in litres, to 3 significant figures.

(3)

(c)

A locomotive uses 85 litres of diesel per hour. Using the unrounded volume of diesel from (b), calculate for how many hours the diesel in the tank would last. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
π × 0.9² × 4.2	✓ 1
10.7 m³	✓ 1
Part (b)
π × 0.9² × 3.15 = 8.0157… m³	✓ 1
× 1000	✓ 1
8020 litres	✓ 1
Part (c)
8015.77… ÷ 85	✓ 1
94.3 hours	✓ 1

Exam tip: Keep unrounded values in your calculator for later parts, and round only the final answer.

+ Add to test✓ In your test155IGME3-4-12r17-04Angles, polygons and circle properties Paper FMedium5 marks

A roof truss is an isosceles triangle. The angle at the apex is (6x − 10)° and each of the two equal base angles is (2x + 5)°.

Diagram not to scale

(a)

Show that x = 18.

(3)

(b)

Work out the size of the apex angle and of one base angle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (6x − 10) + 2(2x + 5) = 180	✓ 1
M1 10x = 180 (simplifies the left-hand side)	✓ 1
A1 x = 18 shown	✓ 1
Part (b)
B1 apex angle 98°	✓ 1
B1 base angle 41°	✓ 1

Common error: Angles in a triangle add to 180°. Remember that there are two equal base angles, so the base-angle expression must be counted twice.

+ Add to test✓ In your test156IGME3-4-12r17-05Pythagoras' theorem and trigonometry Paper FMedium5 marks

A research boat is 140 m from the foot of a vertical sea cliff, on level water. From the boat, the angle of elevation of the top of the cliff is 18°.

Diagram not to scale

(a)

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

(2)

(b)

The boat sails directly towards the cliff until the angle of elevation of the top is 25°. Work out the distance the boat sails. Use the unrounded height from (a) and give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 140 × tan 18°	✓ 1
A1 45.5 m	✓ 1
Part (b)
M1 distance from cliff = 45.488... ÷ tan 25° (= 97.55...)	✓ 1
M1 140 − 97.55...	✓ 1
A1 42.4 m	✓ 1

Exam tip: Draw a right-angled triangle for each position of the boat, and note that the cliff height is shared by both triangles.

+ Add to test✓ In your test157IGME3-4-12r17-06Area, surface area and volume Paper FMedium5 marks

A crate has inside dimensions of 2.5 m by 1.3 m by 1.6 m. It is packed with identical boxes, each 0.6 m by 0.4 m by 0.5 m. All the boxes are placed in the same orientation, with the 0.6 m edges along the 2.5 m length, the 0.4 m edges along the 1.3 m width and the 0.5 m edges along the 1.6 m height.

(a)

Work out the greatest number of boxes that can be packed in the crate in this orientation.

(2)

(b)

Work out the percentage of the volume of the crate that is not filled by boxes. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts the number along each edge: 4 along the length, 3 along the width, 3 along the height	✓ 1
A1 36 boxes	✓ 1
Part (b)
M1 crate volume 2.5 × 1.3 × 1.6 = 5.2 m³	✓ 1
M1 volume of boxes 36 × 0.6 × 0.4 × 0.5 = 4.32 m³ (ECF from (a))	✓ 1
A1 16.9%	✓ 1

Common error: Round the number of boxes down along each edge: part of a box cannot be used. For example, 2.5 ÷ 0.6 = 4.16..., so only 4 fit.

+ Add to test✓ In your test158IGME3-4-12r17-07Area, surface area and volume Paper FMedium5 marks

A sector-shaped paved area at the entrance to a museum has radius 6 m and the angle at the centre is 150°.

Diagram not to scale

(a)

Work out the area of the paved 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 point	Mark	Notes
Part (a)
M1 (150/360) × π × 6²	✓ 1
A1 47.1 m²	✓ 1
Part (b)
M1 arc length (150/360) × 2 × π × 6 (= 15.70...)	✓ 1
M1 adds the two straight edges, 15.70... + 6 + 6	✓ 1
A1 27.7 m	✓ 1

Common error: The perimeter of a sector includes the two radii as well as the arc. Forgetting them is a very common loss of marks.

+ Add to test✓ In your test159IGME3-4-13r18-07Pythagoras' theorem and trigonometry Paper FMedium5 marks

A surveyor stands 45 m from the base of a water tower on level ground. Her eye level is 1.6 m above the ground. The angle of elevation of the top of the tower from her eye is 32°.

Diagram not to scale

(a)

Work out the height of the top of the tower above her eye level. Give your answer correct to 3 significant figures.

(3)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 32° = h ÷ 45	✓ 1
M1 h = 45 × tan 32°	✓ 1
A1 28.1 m	✓ 1
Part (b)
M1 28.1 + 1.6 (or 28.119… + 1.6) (ECF from (a))	✓ 1
A1 29.7 m	✓ 1

Common error: Forgetting to add the surveyor's eye height gives a tower that is too short.

+ Add to test✓ In your test160IGME3-4-13r18-08Area, surface area and volume Paper FMedium6 marks

A water trough for animals has a cross-section in the shape of a trapezium. The parallel sides are 0.9 m and 0.5 m long, and they are 0.4 m apart. The trough is 3 m long.

Cross-section of the trough. Diagram not to scale

(a)

Work out the area of the cross-section.

(2)

(b)

Work out the volume of the trough in m³.

(1)

(c)

1 m³ = 1000 litres. Water flows into the empty trough at 35 litres per minute. Work out the time taken to fill the trough completely.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (0.9 + 0.5) × 0.4	✓ 1
A1 0.28 m²	✓ 1
Part (b)
B1 0.28 × 3 = 0.84 m³ (ECF from (a))	✓ 1
Part (c)
B1 840 litres	✓ 1
M1 840 ÷ 35	✓ 1
A1 24 minutes	✓ 1

Exam tip: Convert units before dividing by a rate: here m³ to litres.

+ Add to test✓ In your test161IGME3-4-13r18-11Angles, polygons and circle properties Paper FMedium6 marks

A game tile is made by joining a regular pentagon ABCDE to an equilateral triangle ABF along the edge AB. The triangle lies outside the pentagon, so that angle EAF (measured through B) is the sum of an angle of the pentagon and an angle of the triangle.

Diagram not to scale

(a)

Work out the size of an interior angle of the regular pentagon.

(2)

(b)

Work out the size of angle EAF.

(2)

(c)

Triangle AEF is isosceles with AE = AF. Work out the size of angle AFE.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 360 ÷ 5 = 72 or (5 − 2) × 180 ÷ 5	✓ 1
A1 108°	✓ 1
Part (b)
M1 108 + 60 (ECF from (a))	✓ 1
A1 168°	✓ 1
Part (c)
M1 (180 − 168) ÷ 2 (ECF from (b))	✓ 1
A1 6°	✓ 1

Exam tip: All sides of the tile pieces are equal, so look for isosceles triangles hidden in the shape.

+ Add to test✓ In your test162IGME3-4-13r19-02Pythagoras' theorem and trigonometry Paper FMedium5 marks

A data centre has a rectangular server room 15 m long and 8 m wide. A network cable is run in a straight line across the floor from one corner to the opposite corner.

Diagram not to scale

(a)

Calculate the length of the cable. Show your working.

(3)

(b)

The cable costs £4.20 per metre. Work out the cost of the cable.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 15² + 8²	✓ 1
M1 √(225 + 64) = √289	✓ 1
A1 17 m	✓ 1
Part (b)
M1 17 × 4.20	✓ 1
A1 £71.40	✓ 1

Exam tip: Write down Pythagoras' theorem with the numbers substituted before using your calculator, so a slip can still earn the method mark.

+ Add to test✓ In your test163IGME3-4-13r19-03Pythagoras' theorem and trigonometry Paper FMedium5 marks

A cycle ramp is a straight slope that makes an angle of 12° with the horizontal ground. The sloping length of the ramp is 6.5 m.

Diagram not to scale

(a)

Calculate the vertical height of the top of the ramp above the ground. Give your answer correct to 3 significant figures.

(2)

(b)

A second ramp rises 0.9 m over a horizontal distance of 4.5 m. Calculate the angle this ramp makes with the horizontal. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6.5 × sin 12°	✓ 1
A1 1.35 m	✓ 1
Part (b)
M1 tan x = 0.9 ÷ 4.5	✓ 1
M1 x = tan⁻¹(0.2)	✓ 1
A1 11.3°	✓ 1

Common error: Choosing sin, cos or tan by looking at the sides you know: label opposite, adjacent and hypotenuse relative to the angle first.

+ Add to test✓ In your test164IGME3-4-13r19-07Pythagoras' theorem and trigonometry Paper FMedium6 marks

A cyclist rides from A due north for 9 km to B. She then rides due east for 12 km to C. She then rides in a straight line back to A.

Diagram not to scale

(a)

Calculate the distance from C to A.

(2)

(b)

Work out the total distance of the ride.

(1)

(c)

Work out the bearing of C from A. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(9² + 12²)	✓ 1
A1 15 km	✓ 1
Part (b)
B1 36 km	✓ 1
Part (c)
M1 tan x = 12 ÷ 9	✓ 1
M1 x = tan⁻¹(12/9)	✓ 1
A1 053.1°	✓ 1

Exam tip: Bearings are measured clockwise from north and written with three figures, so 53.1° is written 053.1°.

+ Add to test✓ In your test165IGME3-4-13r19-14Pythagoras' theorem and trigonometry Paper FMedium4 marks

A surveyor stands 40 m from the base of a tall ancient tower on level ground. Her eyes are 1.6 m above the ground. The angle of elevation from her eyes to the top of the tower is 28°.

Diagram not to scale

(a)

Calculate the height of the top of the tower above the level of her eyes. Give your answer correct to 3 significant figures.

(2)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 × tan 28°	✓ 1
A1 21.3 m	✓ 1
Part (b)
M1 21.268... + 1.6	✓ 1
A1 22.9 m	✓ 1

Common error: Forgetting to add the height of the observer's eyes above the ground to find the full height of the tower.

+ Add to test✓ In your test166IGME3-4-11r20-05Pythagoras' theorem and trigonometry Paper FMedium5 marks

An aircraft takes off and climbs in a straight line at a constant angle of 12° to the horizontal. After flying 8.5 km along its flight path it is directly above a point on the ground.

Diagram not to scale

(a)

Calculate the height of the aircraft above the ground at this point. Give your answer in km correct to 3 significant figures.

(3)

(b)

Calculate the horizontal distance of this point from where the aircraft left the ground. Give your answer in km correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\sin12^\circ = \dfrac{h}{8.5}$	✓ 1
M1 $h = 8.5\sin12^\circ$	✓ 1
A1 1.77 km	✓ 1
Part (b)
M1 $8.5\cos12^\circ$	✓ 1
A1 8.31 km	✓ 1

Common error: Make sure your calculator is in degree mode, and identify opposite, adjacent and hypotenuse before choosing sin, cos or tan.

+ Add to test✓ In your test167IGME3-4-11r20-09Angles, polygons and circle properties Paper FMedium5 marks

A regular polygon is drawn in a field as a crop pattern. Each interior angle of the polygon is 156°.

(a)

Calculate the size of one exterior angle.

(1)

(b)

Work out the number of sides of the polygon.

(2)

(c)

Calculate the sum of the interior angles of the polygon.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $180-156=24^\circ$	✓ 1
Part (b)
M1 $360\div24$	✓ 1
A1 15 (ECF from the exterior angle)	✓ 1
Part (c)
M1 $(15-2)\times180$	✓ 1
A1 $2340^\circ$	✓ 1

Exam tip: The exterior angles of any polygon sum to 360°, so for a regular polygon the number of sides is 360 divided by one exterior angle.

+ Add to test✓ In your test168IGME3-4-11r20-14Pythagoras' theorem and trigonometry Paper FMedium5 marks

A conveyor belt used to load hay bales onto a trailer is 6.5 m long and rises in a straight line to a height of 2.5 m above the ground.

Diagram not to scale

(a)

Calculate the horizontal distance covered by the conveyor belt.

(2)

(b)

Calculate the angle the belt makes with the horizontal. Give your answer in degrees correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\sqrt{6.5^2-2.5^2}$	✓ 1
A1 6 m	✓ 1
Part (b)
M1 $\sin\theta=\dfrac{2.5}{6.5}$ (or $\tan\theta=\dfrac{2.5}{6}$)	✓ 1
M1 $\theta=\sin^{-1}(0.3846\ldots)$	✓ 1
A1 $22.6^\circ$	✓ 1

Common error: Use the inverse function (sin⁻¹, cos⁻¹ or tan⁻¹) to find an angle, not the function itself.

+ Add to test✓ In your test169IGME3-4-12r21-03Pythagoras' theorem and trigonometry Paper FMedium5 marks

In a computer game, a character stands on the top of a vertical cliff 45 m high. A target on level ground at the foot of the cliff is seen at an angle of depression of 32°.

Diagram not to scale

(a)

Calculate the horizontal distance from the foot of the cliff to the target. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the direct distance from the character to the target. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 angle of depression equals angle of elevation from the target, 32° (tan 32° = 45/d)	✓ 1
M1 d = 45 ÷ tan 32°	✓ 1
A1 72.0 m	✓ 1
Part (b)
M1 45 ÷ sin 32° (or use of Pythagoras with 72.0 and 45)	✓ 1
A1 84.9 m	✓ 1

Common error: The angle of depression is measured from the horizontal at the observer, so it equals the angle at the target between the ground and the line of sight; it is not the angle at the base of the cliff.

+ Add to test✓ In your test170IGME3-4-12r21-09Pythagoras' theorem and trigonometry Paper FMedium6 marks

A pumping station P is on level ground. A treatment works W is 5.4 km from P on a bearing of 062°.

Diagram not to scale

(a)

Calculate how far north of P the treatment works is. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate how far east of P the treatment works is. Give your answer correct to 3 significant figures.

(2)

(c)

Work out the bearing of P from W.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5.4 × cos 62°	✓ 1
A1 2.54 km	✓ 1
Part (b)
M1 5.4 × sin 62°	✓ 1
A1 4.77 km	✓ 1
Part (c)
M1 62 + 180 (or 180 − 62 = 118 and 360 − 118)	✓ 1
A1 242°	✓ 1

Common error: A bearing is measured clockwise from north, so the north component uses cosine and the east component uses sine.

+ Add to test✓ In your test171IGME3-4-12r21-14Angles, polygons and circle properties Paper FMedium6 marks

The five interior angles of a pentagonal board space are (x + 10)°, 2x°, (2x + 20)°, (3x − 30)° and 100°.

Diagram not to scale

(a)

Write down the sum of the interior angles of a pentagon.

(2)

(b)

Work out the value of x.

(3)

(c)

Work out the size of the largest angle.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (5 − 2) × 180	✓ 1
A1 540°	✓ 1
Part (b)
M1 (x + 10) + 2x + (2x + 20) + (3x − 30) + 100 = 540	✓ 1
M1 8x + 100 = 540	✓ 1
A1 x = 55	✓ 1
Part (c)
B1 135° (3 × 55 − 30)	✓ 1

Common error: Collect the x terms and the numbers separately when forming the equation; a slip with the negative 30 is the most common error.

+ Add to test✓ In your test172IGME3-4-13r22-04Pythagoras' theorem and trigonometry Paper FMedium7 marks

Calculator

A telecoms mast stands on level ground and is vertical. A support cable is fixed to the mast at a point 18 m above the ground. The cable is straight and makes an angle of 62° with the ground. Give lengths to 3 significant figures and angles to 1 decimal place.

Diagram not to scale

(a)

Work out the length of the cable.

(2)

(b)

Work out the distance from the foot of the mast to the point where the cable meets the ground. Use unrounded values from your working.

(2)

(c)

A second cable is fixed to the same point on the mast and meets the ground 15 m from the foot of the mast. Work out the angle this cable makes with the ground.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin 62° = 18 ÷ length, or length = 18 ÷ sin 62°	✓ 1
A1 20.4 m (20.386...)	✓ 1
Part (b)
M1 18 ÷ tan 62°	✓ 1
A1 9.57 m (9.5707...)	✓ 1
Part (c)
M1 tan (angle) = 18 ÷ 15	✓ 1
M1 tan⁻¹(1.2)	✓ 1
A1 50.2° (50.19...)	✓ 1

Common error: Rounding the cable length early and then using it in a later calculation; keep full calculator values until the end.

+ Add to test✓ In your test173IGME3-4-13r22-07Area, surface area and volume Paper FMedium7 marks

Calculator

A food scientist stores honey in a cylindrical jar. The jar has radius 4 cm and height 12 cm. The honey has density 1.4 g/cm³ and fills the jar completely. Give answers to 3 significant figures.

(a)

Work out the volume of honey in the jar.

(2)

(b)

Work out the mass of honey in the jar. Use the unrounded volume from part (a).

(2)

(c)

The jar has a flat base and a flat lid. Work out the total outside surface area of the jar.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 4² × 12	✓ 1
A1 603 cm³ (603.18...)	✓ 1
Part (b)
M1 603.18... × 1.4	✓ 1
A1 844 g (844.46...)	✓ 1
Part (c)
M1 curved surface 2 × π × 4 × 12 (= 301.59...)	✓ 1
M1 two circles 2 × π × 4² (= 100.53...)	✓ 1
A1 402 cm² (402.12...)	✓ 1

Exam tip: Check units before using density: g/cm³ with a volume in cm³ gives a mass in grams.

+ Add to test✓ In your test174IGME3-4-13r22-11Angles, polygons and circle properties Paper FMedium6 marks

Calculator

Polygons are used in the design of food packaging.

Diagram not to scale

(a)

A regular polygon has interior angles of 156°. Work out the number of sides of the polygon.

(2)

(b)

A packaging tray has the shape of an irregular pentagon. Its interior angles, in degrees, are x, x + 20, 2x, 2x − 10 and 110. Work out the value of x.

(3)

(c)

Work out the size of the largest angle of the tray.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 exterior angle 180 − 156 = 24, then 360 ÷ 24	✓ 1
A1 15	✓ 1
Part (b)
B1 interior angles of a pentagon add up to 540°	✓ 1
M1 6x + 120 = 540	✓ 1
A1 x = 70	✓ 1
Part (c)
B1 140°	✓ 1

Exam tip: Check an algebra answer by substituting back and confirming the angles add up to the required total.

+ Add to test✓ In your test175IGME3-4-13r22-13Area, surface area and volume Paper FMedium8 marks

Calculator

A chocolate bar is a triangular prism. The cross-section is an isosceles triangle with base 3.2 cm and perpendicular height 2.8 cm. The prism is 24 cm long. The chocolate has density 1.3 g/cm³.

Triangular prism. Diagram not to scale

(a)

Work out the volume of one bar. Give your answer to 3 significant figures.

(2)

(b)

Work out the mass of one bar. Use 107.52 cm³ for the volume. Give your answer to 3 significant figures.

(2)

(c)

The whole bar is wrapped in foil with no overlap. Work out the area of foil needed. Give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × 3.2 × 2.8 × 24	✓ 1
A1 108 cm³ (107.52)	✓ 1
Part (b)
M1 107.52 × 1.3	✓ 1
A1 140 g (139.776)	✓ 1
Part (c)
M1 sloping side = √(1.6² + 2.8²) = 3.22...	✓ 1
M1 rectangular faces 24 × (3.2 + 2 × 3.22...) (= 231.6...)	✓ 1
M1 add two triangles 2 × ½ × 3.2 × 2.8 = 8.96	✓ 1
A1 241 cm² (240.6)	✓ 1

Common error: Forgetting the two triangular ends when finding the surface area of a prism.

+ Add to test✓ In your test176IGME3-4-11r24-09Pythagoras' theorem and trigonometry Paper FMedium6 marks

A probe is descending towards its landing site. At one moment it is at a height of 1800 m above level ground. The angle of depression of the landing site from the probe is 25°.

Diagram not to scale

(a)

Work out the horizontal distance from the probe to the landing site. Give your answer correct to 3 significant figures.

(3)

(b)

Work out the straight-line distance from the probe to the landing site. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 25° = 1800 ÷ d	✓ 1
M1 d = 1800 ÷ tan 25°	✓ 1
A1 3860 m	✓ 1
Part (b)
M1 sin 25° = 1800 ÷ s	✓ 1
M1 s = 1800 ÷ sin 25°	✓ 1
A1 4260 m	✓ 1

Exam tip: The angle of depression is measured down from the horizontal, so it equals the angle of elevation measured from the landing site to the probe.

+ Add to test✓ In your test177IGME3-4-11r24-11Angles, polygons and circle properties Paper FMedium6 marks

The heat shield of a probe is a pentagon. The interior angles, in degrees, are x, x + 20, 2x, 2x − 10 and 3x − 10.

Diagram not to scale

(a)

Show that 9x = 540.

(3)

(b)

Work out the value of x.

(1)

(c)

Work out the size of the largest angle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 sum of interior angles of a pentagon = 540°	✓ 1
M1 x + (x + 20) + 2x + (2x − 10) + (3x − 10) = 540	✓ 1
A1 9x + 0 = 540, so 9x = 540	✓ 1
Part (b)
A1 60	✓ 1
Part (c)
M1 3 × 60 − 10	✓ 1
A1 170°	✓ 1

Exam tip: The interior angles of an n-sided polygon add up to (n − 2) × 180°. Collect like terms carefully when you form the equation.

+ Add to test✓ In your test178IGME3-4-14r29-05Pythagoras' theorem and trigonometry Paper FMedium6 marks

A lighthouse is 45 m tall, standing on level ground at the edge of the sea. From the top of the lighthouse the angle of depression of ship A is $18^{\circ}$.

Diagram not to scale

(a)

Work out the horizontal distance from the base of the lighthouse to ship A. Give your answer correct to 3 significant figures.

(3)

(b)

Ship B is 200 m horizontally from the base of the lighthouse. Work out the angle of depression of ship B from the top of the lighthouse. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\tan18^{\circ}=\dfrac{45}{d}$ (angle of depression equals angle of elevation from ship)	✓ 1
M1 $d=\dfrac{45}{\tan18^{\circ}}$	✓ 1
A1 138 m	✓ 1
Part (b)
M1 $\tan\theta=\dfrac{45}{200}$	✓ 1
M1 $\theta=\tan^{-1}\!\left(\dfrac{45}{200}\right)$	✓ 1
A1 $12.7^{\circ}$	✓ 1

Exam tip: Angles of depression are measured down from the horizontal; the alternate angle at the ship is equal.

+ Add to test✓ In your test179IGME3-4-14r29-06Area, surface area and volume Paper FMedium5 marks

A circular revolving stage has diameter 10 m. A border of width 1 m is built around the outside of the stage, so the border and stage together form a larger circle.

Diagram not to scale

(a)

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

(2)

(b)

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

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\pi\times10$	✓ 1
A1 31.4 m	✓ 1
Part (b)
M1 outer radius 6 m and inner radius 5 m	✓ 1
M1 $\pi\times6^{2}-\pi\times5^{2}$	✓ 1
A1 34.6 m$^{2}$	✓ 1

Common error: The outer radius is 5 + 1 = 6 m, not 10 + 1; start from the radius, not the diameter.

+ Add to test✓ In your test180IGME3-4-14r29-07Angles, polygons and circle properties Paper FMedium4 marks

A triangular sail ABC has AB = AC. Angle BAC = $48^{\circ}$. The side BC is extended to a point D so that B, C and D lie on a straight line, with C between B and D.

Diagram not to scale

(a)

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

(2)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $(180-48)\div2$	✓ 1
A1 $66^{\circ}$ with reason: base angles of an isosceles triangle are equal and angles in a triangle sum to $180^{\circ}$	✓ 1
Part (b)
M1 $180-66$ (ECF from (a)) or $48+66$	✓ 1
A1 $114^{\circ}$ with reason: angles on a straight line sum to $180^{\circ}$ (or exterior angle equals sum of the two opposite interior angles)	✓ 1

Exam tip: Write each angle fact in words next to the value; the reason is usually worth a mark.

+ Add to test✓ In your test181IGME3-4-14r29-11Area, surface area and volume Paper FMedium5 marks

The raised platform on a theatre stage is a prism. Its cross-section is a trapezium with parallel sides of 9 m and 13 m, and a perpendicular distance of 6 m between them. The platform is 0.25 m thick, meaning the prism's length is 0.25 m.

Cross-section of the platform. Diagram not to scale

(a)

Work out the area of the trapezium.

(2)

(b)

Work out the volume of the platform.

(1)

(c)

The timber used costs £38 per m$^{3}$. Work out the total cost of the timber.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\tfrac{1}{2}(9+13)\times6$	✓ 1
A1 66 m$^{2}$	✓ 1
Part (b)
B1 $66\times0.25=16.5$ m$^{3}$ (ECF from (a))	✓ 1
Part (c)
M1 $16.5\times38$	✓ 1
A1 £627	✓ 1

Exam tip: Volume of a prism = area of cross-section × length; identify the cross-section first.

+ Add to test✓ In your test182IGME3-4-13r32-06Similar shapes and circle theorems Paper FMedium6 marks

Two posters are mathematically similar rectangles. The small poster measures 30 cm by 42 cm. The shorter side of the large poster is 75 cm.

(a)

Work out the scale factor of enlargement from the small poster to the large poster.

(2)

(b)

Work out the length of the longer side of the large poster.

(2)

(c)

Work out the perimeter of the large poster.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 75 ÷ 30	✓ 1
A1 2.5	✓ 1
Part (b)
M1 42 × 2.5 (or their scale factor)	✓ 1
A1 105 cm (ECF from (a))	✓ 1
Part (c)
M1 2 × (75 + 105) (or 2.5 × perimeter of small poster)	✓ 1
A1 360 cm	✓ 1

Exam tip: In similar shapes, every length is multiplied by the same scale factor, so find it from one pair of matching sides first.

+ Add to test✓ In your test183IGME3-4-13r32-07Area, surface area and volume Paper FMedium8 marks

A vaccine vial is a cylinder with radius 1.2 cm and height 4.5 cm. The vial is filled with vaccine to 90% of its volume. 1 cm³ = 1 mL.

(a)

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

(2)

(b)

Each dose is 0.5 mL. Work out the greatest number of complete doses that can be drawn from the vial. Use the unrounded volume.

(4)

(c)

After these complete doses have been drawn, work out the volume of vaccine left in the vial. Give your answer to 3 significant figures and use the unrounded volume of vaccine from part (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 1.2² × 4.5	✓ 1
A1 20.4 cm³ (20.36 seen)	✓ 1
Part (b)
M1 0.9 × 20.36 (volume of vaccine)	✓ 1
A1 18.3 mL (awrt 18.3)	✓ 1
M1 their volume ÷ 0.5	✓ 1
A1 36 (from 36.64, rounded down)	✓ 1
Part (c)
M1 their vaccine volume − 36 × 0.5 (or − 18)	✓ 1
A1 0.322 mL (0.3218 seen)	✓ 1

Common error: When a question asks how many whole items can be filled, round the answer down, not to the nearest whole number.

+ Add to test✓ In your test184IGME3-4-13r32-08Pythagoras' theorem and trigonometry Paper FMedium7 marks

A surveyor stands on level ground 85 m from the base of a vertical wind-turbine tower. Her eyes are 1.6 m above the ground. The angle of elevation of the top of the tower from her eyes is 38°.

Diagram not to scale

(a)

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

(3)

(b)

The surveyor walks directly towards the tower until the angle of elevation of the top is 52°. Work out how far she walks. Use the unrounded tower height from part (a) and give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 85 × tan 38° (height above eye level)	✓ 1
A1 66.4 m (awrt 66.4)	✓ 1
A1 68.0 m (after adding 1.6 m)	✓ 1
Part (b)
M1 their height above eye level ÷ tan 52°	✓ 1
A1 51.9 m (new distance from the tower, awrt 51.9)	✓ 1
M1 85 − their distance	✓ 1
A1 33.1 m (33.12 seen)	✓ 1

Exam tip: The angle of elevation is measured from the horizontal at the observer's eye, so add the eye height to the answer from the triangle.

+ Add to test✓ In your test185IGME3-4-12r36-03Pythagoras' theorem and trigonometry Paper FMedium5 marks

A ship leaves a port and sails 9 km due east, then 7 km due north to reach a buoy.

Diagram not to scale

(a)

Calculate the direct distance from the port to the buoy. Give your answer to 3 significant figures.

(2)

(b)

Calculate the bearing of the buoy from the port. Give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 9² + 7² (= 130)	✓ 1
A1 11.4 km	✓ 1
Part (b)
M1 tan x = 9/7 (or 9 ÷ 7)	✓ 1
M1 x = tan⁻¹(9/7)	✓ 1
A1 052.1°	✓ 1

Exam tip: A bearing is measured clockwise from north, so here the angle from the north line to the path is the one that gives the bearing.

+ Add to test✓ In your test186IGME3-4-12r36-04Area, surface area and volume Paper FMedium4 marks

A cylindrical tank has radius 1.5 m and height 2 m. It is filled with seawater at a rate of 0.5 m³ per minute.

(a)

Calculate the volume of the tank. Give your answer to 3 significant figures.

(2)

(b)

Calculate the time taken to fill the tank. Use the unrounded value of the volume from part (a). Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 1.5² × 2	✓ 1
A1 14.1 m³	✓ 1
Part (b)
M1 14.137... ÷ 0.5	✓ 1
A1 28.3 minutes	✓ 1

Common error: Use the radius, not the diameter, in the formula π r² h.

+ Add to test✓ In your test187IGME3-4-12r36-10Pythagoras' theorem and trigonometry Paper FMedium5 marks

A volcano is 600 m away from an observer, measured along level ground to the point directly below the crater rim. The angle of elevation of the crater rim from the observer is 24°.

Diagram not to scale

(a)

Calculate the height of the crater rim above the observer. Give your answer to 3 significant figures.

(2)

(b)

Calculate the direct distance from the observer to the crater rim. Use the original distance of 600 m. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 24° = height ÷ 600	✓ 1
A1 267 m	✓ 1
Part (b)
M1 cos 24° = 600 ÷ distance	✓ 1
M1 600 ÷ cos 24°	✓ 1
A1 657 m	✓ 1

Common error: Check that your calculator is in degree mode before using sin, cos or tan.

+ Add to test✓ In your test188IGME3-4-12r36-12Area, surface area and volume Paper FMedium6 marks

A shipping container is a cuboid with length 6.0 m, width 2.4 m and height 2.6 m.

(a)

Calculate the volume of the container.

(2)

(b)

Calculate the total outside surface area of the container.

(2)

(c)

All the outside faces are painted. Paint costs £4.50 for each square metre. Calculate the cost of painting the container.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6.0 × 2.4 × 2.6	✓ 1
A1 37.44 m³	✓ 1
Part (b)
M1 2 × (6.0 × 2.4 + 6.0 × 2.6 + 2.4 × 2.6)	✓ 1
A1 72.48 m²	✓ 1
Part (c)
M1 72.48 × 4.50 (ECF from (b))	✓ 1
A1 £326.16	✓ 1

Exam tip: Cuboid surface area is the sum of three different pairs of faces; list them before you add.

+ Add to test✓ In your test189IGME3-4-13r36-03Area, surface area and volume Paper FMedium6 marks

A data centre has a rectangular room measuring 18 m by 12 m, with a height of 3.5 m. The room contains 60 server racks. Each rack takes up a volume of 0.9 m³.

(a)

Work out the volume of the room.

(2)

(b)

Work out the volume of air in the room when all 60 racks are in place.

(2)

(c)

Work out the volume of the racks as a percentage of the volume of the room. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 18 × 12 × 3.5	✓ 1
A1 756 m³	✓ 1
Part (b)
M1 60 × 0.9 (= 54)	✓ 1
A1 702 m³	✓ 1
Part (c)
M1 54 ÷ 756 × 100	✓ 1
A1 7.14%	✓ 1

Exam tip: Percentage of a total means part ÷ whole × 100. Make sure both volumes are in the same units first.

+ Add to test✓ In your test190IGME3-4-13r36-04Pythagoras' theorem and trigonometry Paper FMedium4 marks

A surveyor stands 40 m from the base of a vertical wall of an ancient ruin. The ground is horizontal. The angle of elevation of the top of the wall, measured at the surveyor's eye level, is 28°.

(a)

Calculate the height of the top of the wall above the surveyor's eye level. Give your answer correct to 3 significant figures.

(2)

(b)

The surveyor moves closer to the wall until the angle of elevation of the top is 40°. Calculate the new horizontal distance from the surveyor to the wall. Use the unrounded answer from (a) and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 28° = h ÷ 40 or h = 40 tan 28°	✓ 1
A1 21.3 m	✓ 1
Part (b)
M1 d = 21.269… ÷ tan 40°	✓ 1
A1 25.3 m	✓ 1

Common error: Keep your calculator in degree mode for trigonometry, and use the unrounded value from an earlier part if the question says so.

+ Add to test✓ In your test191IGME3-4-13r36-10Area, surface area and volume Paper FMedium4 marks

A cycling water bottle is a cylinder with radius 3.5 cm and height 21 cm.

(a)

Calculate the volume of the bottle. Give your answer correct to 3 significant figures.

(2)

(b)

A cyclist pours 600 cm³ of drink into the empty bottle. Calculate the depth of drink in the bottle. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 3.5² × 21	✓ 1
A1 808 cm³	✓ 1
Part (b)
M1 600 ÷ (π × 3.5²)	✓ 1
A1 15.6 cm	✓ 1

Exam tip: Volume of a cylinder is πr²h. To find a height, divide the volume by the area of the circular base.

+ Add to test✓ In your test192IGME3-4-14r37-03Pythagoras' theorem and trigonometry Paper FMedium5 marks

A wind turbine stands on level ground. An observer stands 120 m from the foot of the tower. The observer's eye is at ground level. The angle of elevation of the top of the highest blade is 28°.

Diagram not to scale

(a)

Work out the height of the top of the blade above the ground. Give your answer correct to 1 decimal place.

(2)

(b)

The observer walks directly towards the tower until the angle of elevation of the top of the blade is 45°. Work out how far the observer walks. Use the unrounded height from (a) and give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 28° = h/120 or h = 120 × tan 28°	✓ 1
A1 63.8 m	✓ 1
Part (b)
M1 at 45° the distance from the tower equals the height (tan 45° = 1)	✓ 1
M1 120 − their 63.805...	✓ 1
A1 56.2 m	✓ 1

Exam tip: Label the sides opposite, adjacent and hypotenuse relative to the angle before choosing sin, cos or tan, and check your calculator is in degrees mode.

+ Add to test✓ In your test193IGME3-4-14r37-04Area, surface area and volume Paper FMedium5 marks

A swimming pool is 10 m wide and 25 m long. The floor slopes steadily from the shallow end to the deep end. The pool is a prism whose cross-section is a trapezium with two vertical parallel sides. The depth of the water is 1.2 m at the shallow end and 2.0 m at the deep end.

Cross-section of the pool (depths exaggerated). Diagram not to scale

(a)

Work out the area of the trapezium cross-section.

(2)

(b)

Work out the volume of water in the pool when it is full.

(1)

(c)

The pool is filled at a rate of 0.8 m³ per minute. Work out the time taken to fill the pool completely. Give your answer in hours and minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (1.2 + 2.0) × 25	✓ 1
A1 40 m²	✓ 1
Part (b)
B1 400 m³ (ECF from (a))	✓ 1
Part (c)
M1 400 ÷ 0.8 = 500 minutes	✓ 1
A1 8 hours 20 minutes	✓ 1

Common error: In a trapezium use the perpendicular distance between the parallel sides, which here are the two depths, and not the length of the sloping floor.

+ Add to test✓ In your test194IGME3-4-14r37-05Area, surface area and volume Paper FMedium6 marks

A library has a reading alcove whose floor is a semicircle of diameter 8 m.

Diagram not to scale

(a)

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

(2)

(b)

Work out the length of the curved wall of the alcove. Give your answer correct to 3 significant figures.

(2)

(c)

Carpet costs £18.50 per m². Work out the cost of carpeting the alcove floor. Use the unrounded area from (a) and give your answer to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × π × 4²	✓ 1
A1 25.1 m²	✓ 1
Part (b)
M1 ½ × π × 8 or π × 4	✓ 1
A1 12.6 m	✓ 1
Part (c)
M1 their 25.1327... × 18.50	✓ 1
A1 £464.96	✓ 1

Exam tip: A semicircle has half the area and half the circumference of the full circle, but its straight edge is not part of the curved wall.

+ Add to test✓ In your test195IGME3-4-14r37-14Pythagoras' theorem and trigonometry Paper FMedium7 marks

A library has a reading area in the shape of trapezium ABCD. AB is parallel to DC, with AB = 14 m and DC = 8 m. Angles DAB and ADC are both 90° and AD = 8 m.

Diagram not to scale

(a)

Work out the area of the reading area.

(2)

(b)

Work out the length of BC.

(3)

(c)

Skirting board is fitted along every side of the reading area and costs £3.75 per metre. Work out the total cost.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (14 + 8) × 8	✓ 1
A1 88 m²	✓ 1
Part (b)
M1 horizontal difference 14 − 8 = 6 used	✓ 1
M1 √(6² + 8²)	✓ 1
A1 10 m	✓ 1
Part (c)
M1 perimeter 14 + 8 + 8 + their 10 = 40	✓ 1
A1 £150 (ECF from (b))	✓ 1

Common error: For the sloping side of a trapezium, draw the right-angled triangle formed by the height and the difference between the parallel sides.

+ Add to test✓ In your test196IGME3-4-14r44-05Pythagoras' theorem and trigonometry Paper FMedium6 marks

An architect designs a roof truss in the shape of an isosceles triangle. The base of the triangle is 9.6 m long and each of the two equal sides is 6.0 m long.

Diagram not to scale

(a)

Work out the vertical height of the truss.

(2)

(b)

Work out the size of an angle at the base of the triangle. Give your answer to 1 decimal place.

(2)

(c)

Work out the area of the triangle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6.0² − 4.8² (= 12.96)	✓ 1
A1 3.6 m	✓ 1
Part (b)
M1 cos x = 4.8 ÷ 6.0 (or tan x = 3.6 ÷ 4.8)	✓ 1
A1 36.9°	✓ 1
Part (c)
M1 ½ × 9.6 × 3.6 (ECF from (a))	✓ 1
A1 17.28 m²	✓ 1

Exam tip: Drawing the height splits an isosceles triangle into two identical right-angled triangles, so the base is halved.

+ Add to test✓ In your test197IGME3-4-14r44-06Angles, polygons and circle properties Paper FMedium7 marks

A research boat leaves a harbour H and travels 8 km on a bearing of 060° to a reef R. It then travels 6 km on a bearing of 150° to a buoy B.

Diagram not to scale

(a)

Show that angle HRB = 90°.

(2)

(b)

Work out the distance HB.

(2)

(c)

Work out the bearing of B from H. Give your answer to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 the bearing of H from R is 240° (060° + 180°) or co-interior angles used	✓ 1
B1 240° − 150° = 90°	✓ 1
Part (b)
M1 8² + 6²	✓ 1
A1 10 km	✓ 1
Part (c)
M1 tan x = 6 ÷ 8 (angle RHB)	✓ 1
M1 60° + 36.87…°	✓ 1
A1 097°	✓ 1

Exam tip: Bearings are measured clockwise from north and written with three figures. Draw a north line at each point to work out the angles.

+ Add to test✓ In your test198IGME3-4-14r44-07Area, surface area and volume Paper FMedium7 marks

A wooden shipping crate is a cuboid measuring 1.2 m by 0.8 m by 0.6 m. The outside of each crate is covered in protective film.

(a)

Work out the total outer surface area of one crate.

(2)

(b)

The film costs £3.50 for each m². Work out the cost of the film for one crate.

(2)

(c)

The crates are packed in a container that is 12 m long, 2.4 m wide and 2.4 m high inside. All the crates are packed the same way round, with their 1.2 m edges along the length. Work out the greatest number of crates that fit in the container.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2(1.2 × 0.8 + 1.2 × 0.6 + 0.8 × 0.6) or the three face areas 0.96, 0.72 and 0.48 found	✓ 1
A1 4.32 m²	✓ 1
Part (b)
M1 4.32 × 3.5 (ECF from (a))	✓ 1
A1 £15.12	✓ 1
Part (c)
M1 12 ÷ 1.2 = 10 along the length	✓ 1
M1 2.4 ÷ 0.8 = 3 across and 2.4 ÷ 0.6 = 4 high	✓ 1
A1 120	✓ 1

Common error: Count crates along each edge separately, then multiply. Dividing the container’s volume by one crate’s volume only works when there is no wasted space.

+ Add to test✓ In your test199IGME3-4-14r44-10Pythagoras' theorem and trigonometry Paper FMedium6 marks

A cliff-top observer is 25 m above sea level and sees a seal in the sea. The angle of depression of the seal from the observer is 12°. The sea is horizontal.

Diagram not to scale

(a)

Work out the horizontal distance of the seal from the base of the cliff. Give your answer to the nearest metre.

(3)

(b)

The seal then swims a further 40 m directly away from the cliff. Work out the new angle of depression of the seal. Use the unrounded distance from (a) and give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 12° = 25 ÷ d (angle of elevation from the seal equals 12°, alternate angles)	✓ 1
M1 d = 25 ÷ tan 12°	✓ 1
A1 118 m	✓ 1
Part (b)
M1 horizontal distance 117.6… + 40 = 157.6…	✓ 1
M1 tan x = 25 ÷ 157.6… and inverse tangent	✓ 1
A1 9.0°	✓ 1

Exam tip: The angle of depression from the observer equals the angle of elevation from the seal, because they are alternate angles. Draw the right-angled triangle before choosing a ratio.

+ Add to test✓ In your test200IGME3-4-14r45-05Area, surface area and volume Paper FMedium5 marks

A fish farm has a cylindrical tank with radius 1.2 m and height 0.9 m.

(a)

Work out the volume of the tank. Give your answer in m³ to 3 significant figures.

(2)

(b)

The tank is filled to 80% of its volume. 1 m³ = 1000 litres. Work out the volume of water in the tank in litres. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 1.2² × 0.9	✓ 1
A1 4.07 m³	✓ 1
Part (b)
M1 0.8 × volume of tank	✓ 1
M1 multiplying by 1000 to convert m³ to litres	✓ 1
A1 3260 litres	✓ 1

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

+ Add to test✓ In your test201IGME3-4-14r45-06Similar shapes and circle theorems Paper FMedium5 marks

Two rectangular scarves are mathematically similar. The smaller scarf is 80 cm long and 30 cm wide. The larger scarf is 120 cm long.

(a)

Work out the width of the larger scarf.

(2)

(b)

The fabric costs £18 per m². Work out the cost of the fabric for the larger scarf. 1 m² = 10 000 cm².

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 scale factor 120 ÷ 80 = 1.5	✓ 1
A1 45 cm	✓ 1
Part (b)
M1 120 × 45 = 5400 cm² (ECF from (a))	✓ 1
M1 5400 ÷ 10 000 = 0.54 m² and × 18	✓ 1
A1 £9.72	✓ 1

Common error: To change cm² into m², divide by 100 × 100 = 10 000, not by 100.

+ Add to test✓ In your test202IGME3-4-14r45-07Angles, polygons and circle properties Paper FMedium6 marks

The boundary of a robot vacuum cleaner's work area is an irregular pentagon. The interior angles are x°, (x + 20)°, 2x°, (2x − 10)° and 110°.

Diagram not to scale

(a)

Show that the sum of the interior angles of a pentagon is 540°.

(2)

(b)

Work out the value of x.

(3)

(c)

Work out the size of the largest angle.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 splitting the pentagon into 3 triangles or using (5 − 2) × 180	✓ 1
A1 3 × 180 = 540°	✓ 1
Part (b)
M1 forming the sum of the five angles = 540	✓ 1
M1 collecting terms: 6x + 120 = 540	✓ 1
A1 x = 70	✓ 1
Part (c)
B1 140° (the angle 2x, ECF from (b))	✓ 1

Exam tip: When angles are given as expressions, form the sum equation first and then collect the like terms carefully.

+ Add to test✓ In your test203IGME3-4-14r45-08Pythagoras' theorem and trigonometry Paper FMedium6 marks

A camera is fixed at the top of a vertical mast 4.5 m high on a robot test track. A robot is on level ground 11 m from the foot of the mast.

Diagram not to scale

(a)

Calculate the angle of depression of the robot from the camera. Give your answer to 3 significant figures.

(2)

(b)

The robot moves towards the mast until the angle of depression is 40°. Calculate its distance from the foot of the mast. Give your answer to 3 significant figures.

(3)

(c)

Work out how far the robot moved. Give your answer to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan⁻¹(4.5 ÷ 11)	✓ 1
A1 22.2°	✓ 1
Part (b)
M1 tan 40° = 4.5 ÷ d	✓ 1
M1 d = 4.5 ÷ tan 40°	✓ 1
A1 5.36 m	✓ 1
Part (c)
B1 5.64 m (11 − 5.36, accept use of the unrounded value)	✓ 1

Common error: The angle of depression is measured from the horizontal at the camera, not from the vertical mast. It equals the angle of elevation from the robot.

+ Add to test✓ In your test204IGME3-4-13r46-05Area, surface area and volume Paper FMedium7 marks

The courtyard of a hospital is made from a rectangle measuring 12 m by 8 m with a semicircle added on one of the 8 m sides. The diameter of the semicircle is the full 8 m side.

Diagram not to scale

(a)

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

(4)

(b)

A railing is fitted along the whole boundary of the courtyard. Calculate the total length of the railing. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rectangle area 12 × 8 = 96	✓ 1
M1 semicircle area ½ × π × 4² (= 25.13…)	✓ 1
M1 96 + 25.13…	✓ 1
A1 121 m²	✓ 1
Part (b)
M1 straight edges 12 + 12 + 8 = 32	✓ 1
M1 curved edge ½ × π × 8 (= 12.56…)	✓ 1
A1 44.6 m	✓ 1

Common error: The 8 m side on which the semicircle sits is not part of the boundary, so it must not be included in the perimeter.

+ Add to test✓ In your test205IGME3-4-13r46-08Pythagoras' theorem and trigonometry Paper FMedium6 marks

A Formula 1 support truck leaves the garage and drives 2.4 km due east to a car park. It then drives 1.8 km due north to the paddock.

Diagram not to scale

(a)

Calculate the direct distance from the garage to the paddock.

(3)

(b)

Work out the bearing of the paddock from the garage. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2.4² + 1.8² (= 9)	✓ 1
M1 √9	✓ 1
A1 3 km	✓ 1
Part (b)
M1 tan x = 2.4 ÷ 1.8 (angle measured from north)	✓ 1
M1 x = tan⁻¹(2.4 ÷ 1.8)	✓ 1
A1 053.1°	✓ 1

Common error: Bearings are measured clockwise from north, so check which side of the right-angled triangle is opposite the angle from north.

+ Add to test✓ In your test206IGME3-4-13r46-14Angles, polygons and circle properties Paper FMedium5 marks

The floor of a hospital waiting room is a hexagon. Three of its interior angles are 110°, 125° and 140°. The other three interior angles are each x°.

Diagram not to scale

(a)

Work out the sum of the interior angles of a hexagon.

(2)

(b)

Work out the value of x.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (6 − 2) × 180	✓ 1
A1 720°	✓ 1
Part (b)
M1 110 + 125 + 140 + 3x = 720	✓ 1
M1 3x = 345	✓ 1
A1 115	✓ 1

Exam tip: The sum of the interior angles of an n-sided polygon is (n − 2) × 180°.

+ Add to test✓ In your test207IGME3-4-14r46-05Area, surface area and volume Paper FMedium6 marks

The side of a swimming pool is a trapezium. The two parallel sides are vertical and are 1 m and 2.2 m long. They are 20 m apart. The pool is 8 m wide, so it is a prism with this trapezium as its cross-section.

Cross-section of the pool (depths exaggerated). Diagram not to scale

(a)

Work out the area of the trapezium.

(2)

(b)

Work out the volume of the pool.

(2)

(c)

The pool is filled with water at a constant rate of 8 m³ per hour. Work out the time taken to fill the pool completely.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (1 + 2.2) × 20	✓ 1
A1 32 m²	✓ 1
Part (b)
M1 32 × 8	✓ 1
A1 256 m³	✓ 1
Part (c)
M1 256 ÷ 8	✓ 1
A1 32 hours	✓ 1

Common error: The two parallel sides of the trapezium are the depths of the pool, and the distance between them is the length, not the sloping floor.

+ Add to test✓ In your test208IGME3-4-14r46-08Pythagoras' theorem and trigonometry Paper FMedium5 marks

A support cable for a wind turbine mast is 13 m long. It runs from the top of the cable post to the ground, with the foot of the cable 5 m from the base of the post. The ground is level and the post is vertical.

Diagram not to scale

(a)

Calculate the height of the cable post.

(2)

(b)

Calculate the angle between the cable and the ground. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 13² − 5² (= 144)	✓ 1
A1 12 m	✓ 1
Part (b)
M1 cos x = 5 ÷ 13 (or tan x = 12 ÷ 5)	✓ 1
M1 x = cos⁻¹(5 ÷ 13)	✓ 1
A1 67.4°	✓ 1

Exam tip: Use the given lengths where possible rather than a calculated value, so that rounding errors do not build up.

+ Add to test✓ In your test209IGME3-4-14r46-10Area, surface area and volume Paper FMedium7 marks

A circular paddling pool at a swimming club has radius 1.4 m. The water in the pool is 0.35 m deep.

(a)

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

(2)

(b)

Calculate the volume of water in the pool in m³. Use the unrounded area from part (a) and give your answer correct to 3 significant figures.

(2)

(c)

Water flows into an empty pool at 15 litres per minute. Use the unrounded volume and 1 m³ = 1000 litres. Calculate the time taken to fill the pool to this depth. Give your answer to the nearest minute.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 1.4²	✓ 1
A1 6.16 m²	✓ 1
Part (b)
M1 6.157… × 0.35	✓ 1
A1 2.16 m³	✓ 1
Part (c)
M1 volume in litres = 2155.1… (2.155… × 1000)	✓ 1
M1 2155.1… ÷ 15	✓ 1
A1 144 minutes	✓ 1

Common error: A radius of 1.4 m must be squared before multiplying by π; multiplying π by 1.4 and then squaring the result is wrong.

+ Add to test✓ In your test210IGME3-4-14r46-14Area, surface area and volume Paper FMedium6 marks

The floor of a library room is a 9 m by 6 m rectangle with a rectangular corner cut out. The cut-out corner is 3 m wide (along the 9 m side) and 4 m long (along the 6 m side). The cut-out is not part of the room.

Diagram not to scale

(a)

Work out the area of the floor.

(2)

(b)

Carpet costs £18.50 per m². Work out the cost of carpeting the whole floor.

(2)

(c)

Work out the perimeter of the floor.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 9 × 6 − 3 × 4	✓ 1
A1 42 m²	✓ 1
Part (b)
M1 42 × 18.50	✓ 1
A1 £777	✓ 1
Part (c)
M1 sides 9, 2, 3, 4, 6, 6 identified	✓ 1
A1 30 m	✓ 1

Exam tip: A rectangle with a corner removed has the same perimeter as the full rectangle, which is a quick check.

+ Add to test✓ In your test211IGME-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 point	Mark	Notes
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.

+ Add to test✓ In your test212IGME-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°.

Diagram not to scale

(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 point	Mark	Notes
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°.

+ Add to test✓ In your test213IGME3-4-10d-08Pythagoras' theorem and trigonometry Paper FHard9 marks

A ship sails 12 km due north from a harbour A to a buoy B. It then sails 9 km due east from B to a lighthouse C.

Diagram not to scale

(a)

Work out the direct distance AC.

(2)

(b)

Work out the bearing of C from A. Give your answer to 1 decimal place.

(2)

(c)

Work out the bearing of A from C. Give your answer to 1 decimal place.

(2)

(d)

A rescue boat travels from B to the point H on AC so that BH is perpendicular to AC. Work out the length of BH.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12² + 9² = 225	✓ 1
A1 15 km	✓ 1
Part (b)
M1 tan⁻¹(9 ÷ 12)	✓ 1
A1 036.9°	✓ 1
Part (c)
M1 180° + 36.9° (ECF from (b))	✓ 1
A1 216.9°	✓ 1
Part (d)
M1 area of triangle ABC = ½ × 12 × 9 = 54 km²	✓ 1
M1 ½ × 15 × BH = 54	✓ 1
A1 7.2 km	✓ 1

Exam tip: A bearing is measured clockwise from north and written with three figures. Add 180° to turn a bearing around, or subtract 180° if the result is above 360°.

+ Add to test✓ In your test214IGME3-4-11r24-14Pythagoras' theorem and trigonometry Paper FHard8 marks

ABCD is a rectangular solar array with AB = 9 cm and BC = 14 cm. Point E lies on BC with BE = 5 cm. A support wire is fixed from A to E and a second wire from E to D.

Diagram not to scale

(a)

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

(2)

(b)

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

(2)

(c)

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

(2)

(d)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(9² + 5²) = √106	✓ 1
A1 10.3 cm	✓ 1
Part (b)
M1 tan BAE = 5 ÷ 9	✓ 1
A1 29.1°	✓ 1
Part (c)
M1 90 − 29.05...	✓ 1
A1 60.9°	✓ 1
Part (d)
M1 √(9² + 9²) with EC = 14 − 5 = 9 and CD = 9	✓ 1
A1 12.7 cm	✓ 1

Exam tip: Use the exact value of the angle from your calculator in part (c), not the rounded one, unless the question says otherwise.

+ Add to test✓ In your test215IGME-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.

Diagram not to scale

(a)

Work out the length of QR.

(2)

(b)

Work out the length of AC.

(2)

Show mark scheme

Marking point	Mark	Notes
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.

+ Add to test✓ In your test216IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test217IGME-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.

Diagram not to scale

(a)

Work out the area of the larger photograph.

(3)

Show mark scheme

Marking point	Mark	Notes
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.

+ Add to test✓ In your test218IGME-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°.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test219IGME-061-03-03Angles, polygons and circle properties Paper HEasy4 marks

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

Diagram not to scale

(a)

Work out the value of x.

(3)

(b)

Find the size of angle ADC.

(1)

Show mark scheme

Marking point	Mark	Notes
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°.

+ Add to test✓ In your test220IGME-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 point	Mark	Notes
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.

+ Add to test✓ In your test221IGME-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°.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test222IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test223IGME-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]

Cross-section through the axis of the cone. Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test224IGME-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°.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test225IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test226IGME-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.

Diagram not to scale

(a)

Calculate the distance AB.

(4)

Show mark scheme

Marking point	Mark	Notes
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.

+ Add to test✓ In your test227IGME-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 point	Mark	Notes
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.

+ Add to test✓ In your test228IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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°).

+ Add to test✓ In your test229IGME-061-01-10Pythagoras' theorem and trigonometry Paper HMedium6 marks

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

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test230IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test231IGME-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°.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test232IGME-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 = $(3y - 5)$°.

Diagram not to scale

(a)

Find the value of $x$.

(3)

(b)

Find the size of angle ABC.

(3)

Show mark scheme

Marking point	Mark	Notes
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.

+ Add to test✓ In your test233IGME-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 point	Mark	Notes
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.

+ Add to test✓ In your test234IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test235IGME3-4-10d-09Area, surface area and volume Paper HMedium8 marks

Two paint tins are mathematically similar. The small tin is 12 cm tall and holds 0.75 litres. The large tin is 18 cm tall.

(a)

Work out the volume of paint the large tin holds. Give your answer to 3 significant figures.

(3)

(b)

The label on the small tin has area 96 cm². Work out the area of the label on the large tin, which is a similar shape.

(2)

(c)

Two similar bottles hold 250 ml and 2 litres. The small bottle is 11 cm tall. Work out the height of the large bottle.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 length scale factor 18 ÷ 12 = 1.5	✓ 1
M1 0.75 × 1.5³	✓ 1
A1 2.53 litres	✓ 1
Part (b)
M1 area scale factor 1.5² = 2.25	✓ 1
A1 216 cm²	✓ 1
Part (c)
M1 volume ratio 2000 ÷ 250 = 8	✓ 1
M1 length scale factor ∛8 = 2	✓ 1
A1 22 cm	✓ 1

Common error: Lengths scale by k, areas by k² and volumes by k³. Take the cube root of a volume ratio to find the length ratio.

+ Add to test✓ In your test236IGME3-4-10d-11Pythagoras' theorem and trigonometry Paper HMedium6 marks

A port A has a lighthouse B on a bearing of 065° at a distance of 9.4 km. A second port C is on a bearing of 140° from A at a distance of 12.6 km.

Diagram not to scale

(a)

Write down the size of angle BAC.

(1)

(b)

Work out the distance BC. Give your answer to 3 significant figures.

(3)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 140° − 65° = 75°	✓ 1
Part (b)
M1 BC² = 9.4² + 12.6² − 2 × 9.4 × 12.6 × cos 75°	✓ 1
M1 BC² = 185.8...	✓ 1
A1 13.6 km	✓ 1
Part (c)
M1 ½ × 9.4 × 12.6 × sin 75°	✓ 1
A1 57.2 km²	✓ 1

Common error: In the cosine rule the angle must be the one between the two known sides. Check that your calculator is in degree mode.

+ Add to test✓ In your test237IGME3-4-10d-12Area, surface area and volume Paper HMedium7 marks

A car windscreen wiper arm pivots about a fixed point. The blade extends from 0.20 m to 0.55 m from the pivot. In each sweep the arm turns through 110°.

Diagram not to scale

(a)

Work out the length of the arc travelled by the outer tip of the blade in one sweep. Give your answer to 3 significant figures.

(2)

(b)

The wiper makes 40 sweeps in one minute. Work out the total distance travelled by the outer tip in one minute. Use the unrounded arc length and give your answer to 3 significant figures.

(2)

(c)

Work out the area of the windscreen cleaned by the blade in one sweep. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (110 ÷ 360) × 2 × π × 0.55	✓ 1
A1 1.06 m	✓ 1
Part (b)
M1 40 × 1.0559...	✓ 1
A1 42.2 m	✓ 1
Part (c)
M1 sector of radius 0.55 m: (110 ÷ 360) × π × 0.55²	✓ 1
M1 subtract sector of radius 0.20 m	✓ 1
A1 0.252 m²	✓ 1

Exam tip: A sector is a fraction of a whole circle: use angle ÷ 360 times the circle formula. For a ring-shaped sector, subtract the inner sector from the outer.

+ Add to test✓ In your test238IGME3-4-11r5-09Area, surface area and volume Paper HMedium8 marks

A model of a passenger aircraft is made to a scale of 1 : 40. All lengths on the model are 1/40 of the lengths on the real aircraft.

(a)

The wingspan of the model is 0.85 m. Calculate the wingspan of the real aircraft.

(2)

(b)

The area of the wings of the model is 0.0425 m². Calculate the area of the wings of the real aircraft.

(3)

(c)

The fuel tank of the real aircraft has a volume of 24 m³. Calculate the volume of the fuel tank on the model, in cm³.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.85 × 40	✓ 1
A1 34 m	✓ 1
Part (b)
M1 area scale factor = 40² (= 1600)	✓ 1
M1 0.0425 × 1600	✓ 1
A1 68 m²	✓ 1
Part (c)
M1 volume scale factor = 40³ (= 64 000)	✓ 1
M1 24 ÷ 64 000 (= 0.000375 m³)	✓ 1
A1 375 cm³	✓ 1

Exam tip: If lengths are in the ratio 1 : k, areas are in the ratio 1 : k² and volumes in the ratio 1 : k³.

+ Add to test✓ In your test239IGME3-4-11r5-10Similar shapes and circle theorems Paper HMedium7 marks

A, B, C and D are four points, in that order, on the circumference of a circle. Angle ABC = (3x + 15)° and angle CDA = (5x − 35)°.

(a)

Work out the value of x. Give a reason for the first step of your working.

(3)

(b)

Show that AC is a diameter of the circle.

(2)

(c)

AC = 13 cm and AB = 5 cm. Calculate the length of BC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 (3x + 15) + (5x − 35) = 180	✓ 1
B1 reason: opposite angles of a cyclic quadrilateral add up to 180°	✓ 1
A1 x = 25	✓ 1
Part (b)
M1 angle ABC = 3 × 25 + 15 = 90°	✓ 1
A1 angle in a semicircle is 90° (so AC subtends a right angle at the circumference and is a diameter)	✓ 1
Part (c)
M1 BC² = 13² − 5²	✓ 1
A1 12 cm	✓ 1

Exam tip: In circle-theorem questions always name the theorem you use, for example 'opposite angles in a cyclic quadrilateral add up to 180°'.

+ Add to test✓ In your test240IGME3-4-11r5-11Pythagoras' theorem and trigonometry Paper HMedium5 marks

A straight flight path touches a circular no-fly zone, centre O and radius 8 km, at the point A. The point T is on the flight path and is 17 km from O.

Diagram not to scale

(a)

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

(1)

(b)

Calculate the distance AT.

(2)

(c)

Calculate the size of angle AOT. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 90° because a tangent is perpendicular to the radius at the point of contact	✓ 1
Part (b)
M1 AT² = 17² − 8²	✓ 1
A1 15 km	✓ 1
Part (c)
M1 cos AOT = 8 ÷ 17 (or tan = 15 ÷ 8, or sin = 15 ÷ 17)	✓ 1
A1 61.9°	✓ 1

Common error: The tangent meets the radius at 90°; this makes a right-angled triangle, so Pythagoras and trigonometry both become available.

+ Add to test✓ In your test241IGME3-4-13r5-08Pythagoras' theorem and trigonometry Paper HMedium5 marks

Calculator

Three pumping stations A, B and C supply a town. AB = 6.4 km, AC = 8.1 km and angle BAC = 62°. Use unrounded values in later parts.

Diagram not to scale

(a)

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

(3)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 BC² = 6.4² + 8.1² − 2 × 6.4 × 8.1 × cos 62°	✓ 1
M1 BC² = 57.9 (or better)	✓ 1
A1 7.61 km	✓ 1
Part (b)
M1 ½ × 6.4 × 8.1 × sin 62°	✓ 1
A1 22.9 km²	✓ 1

Common error: In the cosine rule the angle must be the one between the two known sides.

+ Add to test✓ In your test242IGME3-4-13r5-10Similar shapes and circle theorems Paper HMedium5 marks

Calculator

Two medicine bottles are mathematically similar. The smaller bottle has height 10 cm and holds 250 ml. The larger bottle has height 14 cm. The label on the smaller bottle has area 80 cm².

(a)

Work out the capacity of the larger bottle.

(3)

(b)

The larger bottle has a similar label. Work out the area of this label.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 linear scale factor 1.4	✓ 1
M1 250 × 1.4³	✓ 1
A1 686 ml	✓ 1
Part (b)
M1 80 × 1.4²	✓ 1
A1 156.8 cm²	✓ 1

Common error: Area scales with the square of the length scale factor and volume with the cube; do not multiply by 1.4 only.

+ Add to test✓ In your test243IGME3-4-11r6-08Area, surface area and volume Paper HMedium6 marks

An archaeologist makes a scale replica of an ancient vase. The vase and the replica are mathematically similar. The original vase is 30 cm tall and the replica is 12 cm tall.

(a)

The outer surface area of the replica is 180 cm². Calculate the outer surface area of the original vase.

(3)

(b)

The volume of the replica is 400 cm³. Calculate the volume of the original vase.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 length scale factor 30 ÷ 12 = 2.5	✓ 1
M1 area scale factor 2.5² (= 6.25)	✓ 1
A1 1125 cm²	✓ 1
Part (b)
M1 volume scale factor 2.5³ (= 15.625)	✓ 1
M1 400 × 15.625	✓ 1
A1 6250 cm³	✓ 1

Common error: Area scales with the square of the length scale factor and volume with the cube.

+ Add to test✓ In your test244IGME3-4-11r6-09Similar shapes and circle theorems Paper HMedium6 marks

The points A, B and C lie on the circumference of a circular ring of standing stones with centre O. Angle ACB = 54°, where C is on the major arc AB. The line TA is a tangent to the circle at A, and T lies on the opposite side of AB from C.

Diagram not to scale

(a)

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

(2)

(b)

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

(2)

(c)

Work out the size of angle TAB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 108°	✓ 1
B1 the angle at the centre is twice the angle at the circumference	✓ 1
Part (b)
M1 (180 − 108) ÷ 2	✓ 1
A1 36° because triangle OAB is isosceles (OA = OB, radii)	✓ 1
Part (c)
M1 angle OAT = 90° (tangent is perpendicular to radius)	✓ 1
A1 90 − 36 = 54°	✓ 1

Exam tip: Name the circle theorem you use each time; the reason is usually worth a mark on its own.

+ Add to test✓ In your test245IGME3-4-11r6-10Pythagoras' theorem and trigonometry Paper HMedium5 marks

Three cycling checkpoints A, B and C form a triangle. AB = 7.2 km, AC = 9.5 km and angle BAC = 64°.

Diagram not to scale

(a)

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

(3)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 BC² = 7.2² + 9.5² − 2 × 7.2 × 9.5 × cos 64°	✓ 1
M1 BC² = 82.12...	✓ 1
A1 9.06 km	✓ 1
Part (b)
M1 ½ × 7.2 × 9.5 × sin 64°	✓ 1
A1 30.7 km²	✓ 1

Exam tip: Use the sine area formula only with the angle between the two known sides.

+ Add to test✓ In your test246IGME3-4-11r9-05Area, surface area and volume Paper HMedium6 marks

A javelin landing area is modelled as a sector of a circle. The sector has radius 90 m and angle 40° at the centre.

Diagram not to scale

(a)

Calculate the length of the curved edge of the sector. Give your answer correct to 3 significant figures.

(2)

(b)

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

(2)

(c)

Work out the total length of the boundary of the sector, including both straight edges. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 40/360 × 2 × π × 90	✓ 1
A1 for 62.8 m (20π)	✓ 1
Part (b)
M1 for 40/360 × π × 90²	✓ 1
A1 for 2830 m² (900π = 2827.4…)	✓ 1
Part (c)
M1 for 2 × 90 + arc length (unrounded or 62.8)	✓ 1
A1 for 243 m (242.8…)	✓ 1

Common error: The boundary of a sector includes the two radii as well as the arc; many answers stop at the arc length.

+ Add to test✓ In your test247IGME3-4-11r9-07Area, surface area and volume Paper HMedium8 marks

A scale model of a rocket nose cone is mathematically similar to the full-size nose cone. The model is 18 cm high and the full-size cone is 2.7 m high. The model has a volume of 240 cm³ and a surface area of 192 cm².

(a)

Show that the length scale factor from the model to the full-size cone is 15.

(2)

(b)

Calculate the volume of the full-size nose cone in m³.

(3)

(c)

Calculate the surface area of the full-size nose cone in m². Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 270 ÷ 18 (or 2.7 m = 270 cm)	✓ 1
A1 for 15 shown (conversion of units evident)	✓ 1
Part (b)
M1 for 15³ (= 3375)	✓ 1
M1 for 240 × 3375 (= 810 000 cm³)	✓ 1
A1 for 0.81 m³	✓ 1
Part (c)
M1 for 15² (= 225)	✓ 1
M1 for 192 × 225 (= 43 200 cm²)	✓ 1
A1 for 4.32 m²	✓ 1

Common error: Areas scale with the square of the length scale factor and volumes with the cube; using the plain scale factor for either gives a far-too-small answer.

+ Add to test✓ In your test248IGME3-4-11r9-08Angles, polygons and circle properties Paper HMedium4 marks

A running track bend is drawn as part of a circle with centre O. Points A, B and C lie on the circle, with C on the major arc AB. Angle AOB = 118°. Point D lies on the minor arc AB.

Diagram not to scale

(a)

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

(2)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 118 ÷ 2	✓ 1
A1 for 59° with reason: the angle at the centre is twice the angle at the circumference (standing on the same arc)	✓ 1
Part (b)
M1 for 180 − 59	✓ 1
A1 for 121° with reason: opposite angles of a cyclic quadrilateral ACBD add up to 180° (ECF from (a))	✓ 1

Common error: Every circle-theorem answer needs the named theorem as well as the angle; a correct angle with no reason earns only the method mark.

+ Add to test✓ In your test249IGME3-4-12r9-05Angles, polygons and circle properties Paper HMedium5 marks

The blades of a wind turbine are each 45 m long. As a blade turns, its tip moves along an arc of a circle centred on the hub. In a short time interval the tip moves 30 m along the arc.

Diagram not to scale

(a)

Calculate the angle through which the blade turns. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the area swept by the blade in this time interval (the area of the sector). Use the unrounded angle from (a). Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for circumference = 2 × π × 45 (= 282.7…)	✓ 1
M1 for 30 ÷ 282.7… × 360	✓ 1
A1 for 38.2°	✓ 1
Part (b)
M1 for (their angle) ÷ 360 × π × 45²	✓ 1
A1 for 675 m²	✓ 1

Exam tip: As a check, the sector area is also ½ × radius × arc length = ½ × 45 × 30.

+ Add to test✓ In your test250IGME3-4-12r9-08Similar shapes and circle theorems Paper HMedium6 marks

A circular solar-farm fence has centre O. A straight footpath touches the fence at point T (it is a tangent). TA is a chord of the circle, and the angle between the footpath and the chord TA is 48°. Point B lies on the major arc TA.

Diagram not to scale

(a)

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

(2)

(b)

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

(2)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for angle between tangent and radius OT = 90° (a tangent is perpendicular to the radius)	✓ 1
A1 for 90 − 48 = 42°	✓ 1
Part (b)
M1 for OT = OA (radii), so triangle OTA is isosceles with base angles 42°	✓ 1
A1 for 180 − 2 × 42 = 96°	✓ 1
Part (c)
M1 for 96 ÷ 2	✓ 1
A1 for 48° (the angle at the centre is twice the angle at the circumference; this also matches the alternate segment theorem)	✓ 1

Exam tip: The angle between a tangent and a chord equals the angle in the alternate segment, so the answer to (c) can be checked against the 48° given.

+ Add to test✓ In your test251IGME3-4-12r9-11Angles, polygons and circle properties Paper HMedium7 marks

Four testing stations A, B, C and D lie in order on a circular ring road, forming cyclic quadrilateral ABCD. Angle DAB = (2x + 10)°, angle BCD = (4x + 14)°, angle ABC = (3y + 2)° and angle CDA = (5y + 2)°.

Diagram not to scale

(a)

Show that x = 26. Give a reason for your first step.

(3)

(b)

Work out the value of y.

(3)

(c)

Work out the size of the largest angle of the quadrilateral.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 2x + 10 + 4x + 14 = 180 (opposite angles of a cyclic quadrilateral add up to 180°)	✓ 1
M1 for 6x + 24 = 180	✓ 1
A1 for x = 26 shown	✓ 1
Part (b)
M1 for 3y + 2 + 5y + 2 = 180	✓ 1
M1 for 8y = 176	✓ 1
A1 for y = 22	✓ 1
Part (c)
B1 for 118° (angle BCD = 4 × 26 + 14; the other angles are 62°, 68° and 112°)	✓ 1

Common error: Opposite angles of a cyclic quadrilateral add to 180° (not 360°), and adjacent angles do not follow this rule.

+ Add to test✓ In your test252IGME3-4-12r10-07Area, surface area and volume Paper HMedium4 marks

Two coffee-roasting drums are mathematically similar. The small drum has height 50 cm, surface area 0.85 m² and capacity 12 litres. The large drum has height 70 cm.

(a)

Work out the surface area of the large drum. Give your answer to 3 significant figures.

(2)

(b)

Work out the capacity of the large drum. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (70 ÷ 50)² (= 1.96)	✓ 1
A1 1.67 (m²)	✓ 1
Part (b)
M1 12 × (70 ÷ 50)³	✓ 1
A1 32.9 (litres)	✓ 1

Common error: Areas scale with the square of the length ratio and volumes with the cube.

+ Add to test✓ In your test253IGME3-4-12r10-08Similar shapes and circle theorems Paper HMedium6 marks

A, B, C and D are points on the circular rim of an antenna dish with centre O. Angle AOB = 112°. Point C is on the major arc AB and point D is on the minor arc AB. A straight line touches the circle at B; P is a point on this tangent on the same side of line OB as A.

Diagram not to scale

(a)

Work out angle ACB. Give a reason for your answer.

(2)

(b)

Work out angle ADB. Give a reason for your answer.

(2)

(c)

Work out angle PBA.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 56°	✓ 1
B1 the angle at the centre is twice the angle at the circumference	✓ 1
Part (b)
B1 124°	✓ 1
B1 opposite angles of a cyclic quadrilateral add up to 180°	✓ 1
Part (c)
M1 angle OBA = (180 − 112) ÷ 2 = 34° (or use of the alternate segment theorem)	✓ 1
A1 56°	✓ 1

Exam tip: Whenever a circle theorem is used, write its name or rule next to the angle you found.

+ Add to test✓ In your test254IGME3-4-12r10-09Pythagoras' theorem and trigonometry Paper HMedium6 marks

Two ground stations A and B are 800 km apart on level ground. (Treat the ground as flat.) A space probe P is in the same vertical plane as A and B, above the ground between them. The angle of elevation of P from A is 52° and from B is 71°.

Side view; ground treated as flat. Diagram not to scale

(a)

Work out angle APB.

(1)

(b)

Work out the distance AP. Give your answer to 3 significant figures.

(3)

(c)

Work out the height of P above the ground. Use the unrounded value of AP and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 57°	✓ 1
Part (b)
M1 sine rule with AB = 800 and angle APB = 57°	✓ 1
M1 AP = 800 × sin 71° ÷ sin 57°	✓ 1
A1 902 (km)	✓ 1
Part (c)
M1 AP × sin 52°	✓ 1
A1 711 (km)	✓ 1

Exam tip: In the sine rule, each side is paired with the angle opposite it.

+ Add to test✓ In your test255IGME3-4-13r10-07Pythagoras' theorem and trigonometry Paper HMedium6 marks

A storage room in a warehouse is a cuboid, 8 m long, 6 m wide and 4 m high. A cable runs in a straight line from one corner of the floor to the opposite corner of the ceiling.

Diagram not to scale

(a)

Calculate the length of the cable. Give your answer in metres correct to 3 significant figures.

(3)

(b)

Calculate the angle between the cable and the floor. Give your answer in degrees correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 8² + 6² + 4² (= 116) or floor diagonal = 10	✓ 1
M1 √116 or √(10² + 4²)	✓ 1
A1 10.8 (m)	✓ 1
Part (b)
M1 floor diagonal = √(8² + 6²) = 10	✓ 1
M1 tan x = 4 ÷ 10 (or sin x = 4 ÷ √116)	✓ 1
A1 21.8 (°)	✓ 1

Exam tip: In three-dimensional problems, find the diagonal of the floor first, then use it as one side of a second right-angled triangle.

+ Add to test✓ In your test256IGME3-4-13r10-08Similar shapes and circle theorems Paper HMedium7 marks

Two fish tanks are mathematically similar. The larger tank has volume 250 litres and the smaller tank has volume 128 litres.

(a)

Show that the ratio of corresponding lengths, larger : smaller, is 1.25 : 1.

(3)

(b)

The length of the larger tank is 1.5 m. Work out the length of the smaller tank.

(2)

(c)

The total area of glass in the smaller tank is 1.44 m². Work out the total area of glass in the larger tank.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume ratio 250 ÷ 128 (= 1.953125)	✓ 1
M1 length ratio = cube root of the volume ratio	✓ 1
A1 ∛1.953125 = 1.25	✓ 1
Part (b)
M1 1.5 ÷ 1.25	✓ 1
A1 1.2 (m)	✓ 1
Part (c)
M1 1.44 × 1.25² (area scale factor 1.5625)	✓ 1
A1 2.25 (m²)	✓ 1

Common error: Using the length scale factor for areas or volumes: area scales with the square of the length scale factor and volume with the cube.

+ Add to test✓ In your test257IGME3-4-13r10-09Similar shapes and circle theorems Paper HMedium5 marks

In the diagram description below, O is the centre of a circular lagoon. Points A and B lie on the shore (the circumference). The tangents to the circle at A and at B meet at a point T outside the lagoon. Angle AOB = 118°. Point C lies on the major arc AB of the circle.

Diagram not to scale

(a)

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

(1)

(b)

Work out the size of angle ATB.

(2)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 90° because a tangent is perpendicular to the radius at the point of contact	✓ 1
Part (b)
M1 360 − 90 − 90 − 118 (angles in quadrilateral OATB)	✓ 1
A1 62 (°)	✓ 1
Part (c)
M1 118 ÷ 2	✓ 1
A1 59 (°) because the angle at the centre is twice the angle at the circumference	✓ 1

Exam tip: Give the full circle theorem as the reason, such as the angle at the centre being twice the angle at the circumference, subtended by the same arc.

+ Add to test✓ In your test258IGME3-4-13r12-09Similar shapes and circle theorems Paper HMedium6 marks

A pharmacy sells hand sanitiser in two bottles that are mathematically similar. The small bottle is 12 cm tall and holds 200 ml. The large bottle is 18 cm tall.

(a)

Work out the volume of the large bottle.

(3)

(b)

The label on the large bottle has an area of 150 cm². The label on the small bottle is similar to it. Work out the area of the label on the small bottle. Give your answer in cm² to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for linear scale factor $\frac{18}{12} = 1.5$	✓ 1
M1 for volume scale factor $1.5^3$ (= 3.375)	✓ 1
A1 for 675 ml	✓ 1
Part (b)
M1 for area scale factor $1.5^2$ (= 2.25) used correctly, i.e. dividing by it	✓ 1
M1 for $150 \div 2.25$	✓ 1
A1 for 66.7 cm² (66.66...)	✓ 1

Exam tip: Lengths scale by $k$, areas by $k^2$ and volumes by $k^3$. When going from the large shape to the small shape, divide rather than multiply.

+ Add to test✓ In your test259IGME3-4-13r12-10Angles, polygons and circle properties Paper HMedium6 marks

In this question O is the centre of a circle. Points A, B, C and D lie on the circle in the order A, B, C, D going round the circle, so that ABCD is a cyclic quadrilateral. The chord AC does not pass through O, and angle AOC = 136°, where this is the angle at the centre subtended by the minor arc AC. Point B lies on the major arc AC and point D lies on the minor arc AC.

Diagram not to scale

(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)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 68°	✓ 1
B1 for reason: the angle at the centre is twice the angle at the circumference (when both stand on the same arc)	✓ 1
Part (b)
B1 for 112° (ECF from (a))	✓ 1
B1 for reason: opposite angles of a cyclic quadrilateral add up to 180°	✓ 1
Part (c)
M1 for triangle OAC is isosceles with OA = OC (radii) and the angles in a triangle sum to 180°, e.g. $(180 - 136) \div 2$	✓ 1
A1 for 22°	✓ 1

Exam tip: Whenever a circle theorem is used, write its reason in words, for example “angle at the centre is twice the angle at the circumference”. Marks are tied to the reason as well as the angle.

+ Add to test✓ In your test260IGME3-4-12r14-06Similar shapes and circle theorems Paper HMedium6 marks

Points A, B, C and D lie on a circle with centre O, in the order A, B, C, D around the circle. B lies on the major arc AC and D lies on the minor arc AC. The angle AOC is 132°. The points mark the edge of a circular training area.

Diagram not to scale

(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)

(c)

Find the size of angle OAC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
66°	✓ 1
The angle at the centre is twice the angle at the circumference	✓ 1
Part (b)
114° (ECF from (a))	✓ 1
Opposite angles of a cyclic quadrilateral add up to 180°	✓ 1
Part (c)
Triangle OAC is isosceles because OA = OC (radii), so angle OAC = (180° − 132°) ÷ 2	✓ 1
24°	✓ 1

Common error: Each circle theorem needs its reason stated in full; giving only the angle value earns no reason mark.

+ Add to test✓ In your test261IGME3-4-12r14-08Similar shapes and circle theorems Paper HMedium6 marks

Two protein-powder tubs are mathematically similar. The smaller tub is 18 cm tall and the larger tub is 27 cm tall. The smaller tub has a volume of 480 cm³.

(a)

Calculate the volume of the larger tub.

(3)

(b)

The total outer surface area of the larger tub is 1530 cm². Calculate the total outer surface area of the smaller tub.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Length scale factor 27 ÷ 18 = 1.5	✓ 1
Volume scale factor 1.5³ = 3.375	✓ 1
1620 cm³	✓ 1
Part (b)
Area scale factor 1.5² = 2.25	✓ 1
1530 ÷ 2.25	✓ 1
680 cm²	✓ 1

Common error: Areas scale by the square of the length scale factor and volumes by the cube; multiplying by the length factor alone underestimates both.

+ Add to test✓ In your test262IGME3-4-12r14-11Area, surface area and volume Paper HMedium6 marks

A hammer-throw landing area is a sector of a circle with its centre at the throwing circle. The sector has radius 80 m and angle 35°.

Diagram not to scale

(a)

Calculate the arc length of the sector. Give your answer to 3 significant figures.

(2)

(b)

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

(2)

(c)

The sector is re-seeded at €3.60 per m². Using the unrounded area, calculate the cost to the nearest euro.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(35 ÷ 360) × 2 × π × 80	✓ 1
48.9 m	✓ 1
Part (b)
(35 ÷ 360) × π × 80²	✓ 1
1950 m²	✓ 1
Part (c)
1954.76… × 3.60	✓ 1
€7037	✓ 1

Exam tip: The fraction of the circle used is angle ÷ 360 for both arc length and sector area.

+ Add to test✓ In your test263IGME3-4-12r14-12Pythagoras' theorem and trigonometry Paper HMedium6 marks

A rectangular equipment crate on a flatbed rail wagon is 6.0 m long, 3.0 m wide and 2.0 m high.

Diagram not to scale

(a)

Calculate the length of the longest straight rod that fits inside the crate.

(3)

(b)

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

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
6.0² + 3.0² + 2.0² = 49	✓ 1
√49	✓ 1
7.0 m	✓ 1
Part (b)
Base diagonal = √(6.0² + 3.0²) = √45	✓ 1
tan x = 2.0 ÷ √45	✓ 1
16.6°	✓ 1

Common error: In 3D, find the diagonal of the base first; the angle with the base uses the base diagonal, not an edge.

+ Add to test✓ In your test264IGME3-4-12r17-09Area, surface area and volume Paper HMedium5 marks

Two sculptures for a courtyard are mathematically similar. The small sculpture has surface area 360 cm², height 12 cm and volume 540 cm³. The large sculpture has surface area 1000 cm².

(a)

Work out the height of the large sculpture.

(3)

(b)

Work out the volume of the large sculpture.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area scale factor 1000/360 = 25/9	✓ 1
M1 length scale factor √(25/9) = 5/3	✓ 1
A1 20 cm	✓ 1
Part (b)
M1 540 × (5/3)³ (or 540 × (1000/360)^1.5)	✓ 1
A1 2500 cm³	✓ 1

Exam tip: Lengths scale by k, areas by k² and volumes by k³. Always convert between them through the length scale factor.

+ Add to test✓ In your test265IGME3-4-12r17-10Similar shapes and circle theorems Paper HMedium7 marks

A, B and C are points on a circle. AC is a diameter of the circle. Angle BAC = 34°. The tangent to the circle at B meets the line AC extended beyond C at the point T.

Diagram not to scale

(a)

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

(2)

(b)

Work out the size of angle ACB.

(1)

(c)

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

(2)

(d)

Work out the size of angle BTC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 90°	✓ 1
B1 the angle in a semicircle is a right angle	✓ 1
Part (b)
B1 56°	✓ 1
Part (c)
B1 34°	✓ 1
B1 the angle between a tangent and a chord is equal to the angle in the alternate segment	✓ 1
Part (d)
M1 180 − (90 + 34) − 34 (angle sum of triangle ABT, ECF from (a) and (c))	✓ 1
A1 22°	✓ 1

Common error: Name the circle theorem in full: a bare statement such as 'alternate segment' without the angle and chord relationship is often not enough for the reason mark.

+ Add to test✓ In your test266IGME3-4-12r17-11Pythagoras' theorem and trigonometry Paper HMedium6 marks

Two research buoys, A and B, are 3.8 km apart on level water. A whale is spotted at W. Angle WAB = 52° and angle WBA = 71°.

Diagram not to scale

(a)

Work out the size of angle AWB.

(1)

(b)

Work out the distance AW. Give your answer correct to 3 significant figures.

(3)

(c)

Work out the area of triangle ABW. Use the unrounded value of AW and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 57°	✓ 1
Part (b)
M1 AW/sin 71° = 3.8/sin 57°	✓ 1
M1 AW = 3.8 × sin 71° ÷ sin 57°	✓ 1
A1 4.28 km	✓ 1
Part (c)
M1 ½ × 3.8 × 4.284... × sin 52°	✓ 1
A1 6.41 km²	✓ 1

Exam tip: To use the sine rule you need a side and its opposite angle. Here, AB is opposite angle W, which is why finding angle W first is useful.

+ Add to test✓ In your test267IGME3-4-13r18-05Similar shapes and circle theorems Paper HMedium6 marks

Two medicine bottles are mathematically similar. The smaller bottle is 8 cm tall and holds 150 ml. The larger bottle is 12 cm tall.

(a)

Work out the volume of the larger bottle, correct to 3 significant figures.

(3)

(b)

The label on the larger bottle has area 189 cm². Work out the area of the label on the smaller bottle.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 length scale factor 12 ÷ 8 = 1.5	✓ 1
M1 volume scale factor 1.5³ (= 3.375)	✓ 1
A1 506 ml (from 506.25)	✓ 1
Part (b)
M1 area scale factor 1.5² = 2.25	✓ 1
M1 189 ÷ 2.25	✓ 1
A1 84 cm²	✓ 1

Common error: Use the length scale factor squared for areas and cubed for volumes, and divide when going from larger to smaller.

+ Add to test✓ In your test268IGME3-4-13r18-06Similar shapes and circle theorems Paper HMedium5 marks

A, B and C are points on the circumference of a circle with centre O. C lies on the major arc AB. Angle ACB = 37°. The tangents to the circle at A and at B meet at T.

Diagram not to scale

(a)

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

(2)

(b)

Work out the size of angle ATB. Give reasons for your working.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 74°	✓ 1
B1 the angle at the centre is twice the angle at the circumference	✓ 1
Part (b)
B1 angles OAT and OBT are 90° because a tangent is perpendicular to the radius	✓ 1
M1 angles in quadrilateral OATB sum to 360°: 360 − 90 − 90 − 74 (ECF from (a))	✓ 1
A1 106°	✓ 1

Exam tip: Name each circle theorem in words next to the angle it gives you; the reason is a separate mark.

+ Add to test✓ In your test269IGME3-4-13r18-09Area, surface area and volume Paper HMedium5 marks

A circular radar display in a game has radius 6.0 cm. A sector of the display, with sector angle 50°, is highlighted.

Diagram not to scale

(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 point	Mark	Notes
Part (a)
M1 (50 ÷ 360) × π × 6.0²	✓ 1
A1 15.7 cm²	✓ 1
Part (b)
M1 arc length = (50 ÷ 360) × 2 × π × 6.0 (= 5.236…)	✓ 1
M1 add two radii: 5.236… + 12	✓ 1
A1 17.2 cm	✓ 1

Common error: A sector's perimeter includes the two straight edges as well as the curved arc.

+ Add to test✓ In your test270IGME3-4-13r19-08Area, surface area and volume Paper HMedium4 marks

An archaeologist makes a small replica of an amphora (a clay storage jar). The replica is mathematically similar to the original. The original is 18 cm tall and the replica is 12 cm tall.

(a)

The surface area of the replica is 240 cm². Calculate the surface area of the original.

(2)

(b)

The volume of the replica is 400 cm³. Calculate the volume of the original.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 240 × (18 ÷ 12)²	✓ 1
A1 540 cm²	✓ 1
Part (b)
M1 400 × (18 ÷ 12)³	✓ 1
A1 1350 cm³	✓ 1

Common error: Using the length scale factor for areas or volumes; areas use the square and volumes use the cube of the scale factor.

+ Add to test✓ In your test271IGME3-4-13r19-09Similar shapes and circle theorems Paper HMedium4 marks

A, B, C and D are points on a circle with centre O, in the order A, C, B, D around the circle, so ACBD is a cyclic quadrilateral. Angle AOB = 112°, and AOB stands on the minor arc AB. Point D lies on the minor arc AB and point C lies on the major arc AB.

Diagram not to scale

(a)

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

(2)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 56°	✓ 1
C1 the angle at the centre is twice the angle at the circumference	✓ 1
Part (b)
B1 124°	✓ 1
C1 opposite angles of a cyclic quadrilateral add up to 180°	✓ 1

Exam tip: Always name the circle theorem in words when asked for a reason; a correct angle without a reason does not earn the communication mark.

+ Add to test✓ In your test272IGME3-4-13r19-13Area, surface area and volume Paper HMedium6 marks

A cyclist's drinks flask is a solid shape made of a cylinder with a hemisphere attached to the top. The cylinder has radius 3.4 cm and height 18 cm. The hemisphere has radius 3.4 cm. The flask is filled completely.

Cross-section. Diagram not to scale

(a)

Calculate the volume of the cylinder. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the volume of the hemisphere. Give your answer correct to 3 significant figures.

(2)

(c)

Work out the capacity of the flask in litres. Use the unrounded volumes from your calculator and give your answer correct to 3 significant figures. (1 litre = 1000 cm³)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 3.4² × 18	✓ 1
A1 654 cm³	✓ 1
Part (b)
M1 ⅔ × π × 3.4³	✓ 1
A1 82.3 cm³	✓ 1
Part (c)
M1 (653.70... + 82.318...) ÷ 1000	✓ 1
A1 0.736 litres	✓ 1

Exam tip: Keep full calculator values between parts and round only the final answer, as the question instructs.

+ Add to test✓ In your test273IGME3-4-11r20-07Area, surface area and volume Paper HMedium6 marks

Two cylindrical grain silos, P and Q, are mathematically similar. Silo P has height 4 m, volume 30 m³ and total outer surface area 52 m². Silo Q has height 6 m.

(a)

Calculate the volume of silo Q.

(3)

(b)

Calculate the total outer surface area of silo Q.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 length scale factor $\tfrac64=1.5$	✓ 1
M1 volume scale factor $1.5^3$ so $30\times1.5^3$	✓ 1
A1 101.25 m³	✓ 1
Part (b)
M1 area scale factor $1.5^2=2.25$	✓ 1
M1 $52\times2.25$	✓ 1
A1 117 m²	✓ 1

Common error: Lengths scale by k, areas by k² and volumes by k³; do not use the length scale factor for areas or volumes.

+ Add to test✓ In your test274IGME3-4-11r20-08Similar shapes and circle theorems Paper HMedium7 marks

A festival logo is based on a circle with centre O. Points A, B and C lie on the circle, with C on the major arc AB, and angle ACB = 64°. The lines TA and TB are tangents to the circle at A and B.

Diagram not to scale

(a)

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

(2)

(b)

Find the size of angle ATB. Give reasons for your working.

(3)

(c)

Calculate the size of angle OAB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $128^\circ$	✓ 1
B1 the angle at the centre is twice the angle at the circumference	✓ 1
Part (b)
M1 angles OAT and OBT are $90^\circ$, since a tangent is perpendicular to the radius at the point of contact	✓ 1
M1 angles in quadrilateral OATB sum to $360^\circ$: $360-90-90-128$	✓ 1
A1 $52^\circ$ (ECF from the first part)	✓ 1
Part (c)
M1 triangle OAB is isosceles as OA = OB (radii), so $(180-128)\div2$	✓ 1
A1 $26^\circ$ (ECF from the first part)	✓ 1

Exam tip: Name each circle theorem in full when you use it, for example “tangent is perpendicular to the radius”.

+ Add to test✓ In your test275IGME3-4-11r20-10Pythagoras' theorem and trigonometry Paper HMedium6 marks

An aircraft leaves airport A and flies 80 km on a bearing of 070° to a point B. It then flies 60 km on a bearing of 150° to a point C.

Diagram not to scale

(a)

Show that angle ABC = 100°.

(2)

(b)

Calculate the distance AC. Give your answer in km correct to 1 decimal place.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the bearing of A from B is $070+180=250^\circ$ (co-interior angles with north lines)	✓ 1
A1 $250-150=100^\circ$	✓ 1
Part (b)
M1 cosine rule $AC^2=80^2+60^2-2\times80\times60\times\cos100^\circ$	✓ 1
M1 $AC^2=10000+1667.0\ldots=11667.0\ldots$	✓ 1
M1 square root	✓ 1
A1 108.0 km	✓ 1

Common error: Bearings are measured clockwise from north; find the back-bearing by adding or subtracting 180° before working out the angle at B.

+ Add to test✓ In your test276IGME3-4-12r21-05Similar shapes and circle theorems Paper HMedium7 marks

Two medicine bottles are mathematically similar. The small bottle is 8 cm tall and holds 120 ml. The large bottle is 12 cm tall.

(a)

Work out the volume of the large bottle.

(2)

(b)

The label on the small bottle has an area of 64 cm². Work out the area of the label on the large bottle.

(2)

(c)

A third bottle, similar to the small bottle, holds 960 ml. Work out its height.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 scale factor 12/8 = 1.5 cubed (1.5³)	✓ 1
A1 405 ml	✓ 1
Part (b)
M1 64 × 1.5²	✓ 1
A1 144 cm²	✓ 1
Part (c)
M1 volume scale factor 960 ÷ 120 = 8	✓ 1
M1 length scale factor cube root of 8 = 2	✓ 1
A1 16 cm	✓ 1

Common error: Lengths scale by k, areas by k² and volumes by k³; going from a volume ratio back to a length ratio needs a cube root.

+ Add to test✓ In your test277IGME3-4-12r21-10Pythagoras' theorem and trigonometry Paper HMedium8 marks

In a computer game, three towers are at points A, B and C on flat ground. AB = 120 m, AC = 85 m and angle BAC = 48°.

Diagram not to scale

(a)

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

(3)

(b)

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

(2)

(c)

Calculate the size of angle ABC. Use the unrounded value of BC. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 BC² = 120² + 85² − 2 × 120 × 85 × cos 48°	✓ 1
M1 BC² = 7974.7...	✓ 1
A1 89.3 m	✓ 1
Part (b)
M1 1/2 × 120 × 85 × sin 48°	✓ 1
A1 3790 m²	✓ 1
Part (c)
M1 sin B / 85 = sin 48° / BC	✓ 1
M1 B = sin⁻¹(85 sin 48° ÷ BC)	✓ 1
A1 45.0°	✓ 1

Exam tip: If two sides and the angle between them are known, start with the cosine rule; after that the sine rule is simpler for another angle.

+ Add to test✓ In your test278IGME3-4-12r21-12Similar shapes and circle theorems Paper HMedium6 marks

Two solid game figurines are mathematically similar. The smaller figurine has a surface area of 45 cm² and a volume of 54 cm³. The larger figurine has a surface area of 80 cm².

(a)

Work out the ratio of the heights of the two figurines in the form a : b, where a and b are integers.

(2)

(b)

Work out the volume of the larger figurine.

(2)

(c)

The larger figurine is made of plastic of density 1.2 g/cm³. Work out its mass.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area scale factor 80/45 = 16/9 and square root	✓ 1
A1 3 : 4	✓ 1
Part (b)
M1 54 × (4/3)³	✓ 1
A1 128 cm³	✓ 1
Part (c)
M1 1.2 × 128	✓ 1
A1 153.6 g	✓ 1

Common error: The ratio of areas is the square of the ratio of lengths, so take a square root to get the length ratio before finding the volume ratio.

+ Add to test✓ In your test279IGME3-4-13r22-02Area, surface area and volume Paper HMedium8 marks

Calculator

Two cylindrical tins of soup stock are mathematically similar. The smaller tin has height 8 cm, holds 250 ml and has a curved label of area 192 cm². The larger tin has height 10 cm.

(a)

Work out the volume of the larger tin. Give your answer to the nearest millilitre.

(3)

(b)

Work out the area of the label on the larger tin.

(2)

(c)

A third tin, similar to the smaller tin, holds 2000 ml. Work out its height.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 linear scale factor 10 ÷ 8 = 1.25	✓ 1
M1 250 × 1.25³	✓ 1
A1 488 ml (488.28...)	✓ 1
Part (b)
M1 192 × 1.25²	✓ 1
A1 300 cm²	✓ 1
Part (c)
M1 volume scale factor 2000 ÷ 250 = 8	✓ 1
M1 linear scale factor = cube root of 8 = 2	✓ 1
A1 16 cm	✓ 1

Exam tip: For similar shapes, lengths scale by k, areas by k² and volumes by k³; write the scale factor down before using it.

+ Add to test✓ In your test280IGME3-4-13r22-08Angles, polygons and circle properties Paper HMedium6 marks

Calculator

A, B, C and D are points on a circle with centre O, in that order round the circle. Angle DAB = 74° and angle ABC = 108°.

Diagram not to scale

(a)

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

(2)

(b)

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

(2)

(c)

Work out the size of angle BOD, the angle at the centre that is less than 180°.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 106°	✓ 1
B1 opposite angles of a cyclic quadrilateral add up to 180°	✓ 1
Part (b)
A1 72°	✓ 1
B1 opposite angles of a cyclic quadrilateral add up to 180°	✓ 1
Part (c)
M1 the angle at the centre is twice the angle at the circumference, 2 × 74	✓ 1
A1 148°	✓ 1

Common error: Doubling the wrong angle: the angle at the centre must stand on the same arc as the angle at the circumference.

+ Add to test✓ In your test281IGME3-4-13r22-09Similar shapes and circle theorems Paper HMedium6 marks

Calculator

In triangle ABC, D lies on AB and E lies on AC, and DE is parallel to BC. AD = 6 cm, DB = 4 cm and DE = 9 cm. The area of triangle ADE is 27 cm².

Diagram not to scale

(a)

Work out the length of BC.

(3)

(b)

Work out the area of the trapezium DBCE.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 triangles ADE and ABC are similar (angles equal because DE ∥ BC)	✓ 1
M1 scale factor AB ÷ AD = 10 ÷ 6	✓ 1
A1 15 cm	✓ 1
Part (b)
M1 area scale factor (10 ÷ 6)² (= 25/9)	✓ 1
M1 area of ABC = 27 × 25/9 = 75	✓ 1
A1 48 cm²	✓ 1

Common error: Using the scale factor 10 ÷ 6 for area without squaring it, or using AB = 4 instead of 10 for the large triangle.

+ Add to test✓ In your test282IGME3-4-11r24-05Area, surface area and volume Paper HMedium8 marks

A scale model of a space probe is made to a scale of 1 : 25. The model is similar to the real probe.

(a)

The model is 18 cm long. Work out the length of the real probe. Give your answer in metres.

(2)

(b)

A solar panel on the model has area 40 cm². Work out the area of the corresponding panel on the real probe. Give your answer in m².

(3)

(c)

A fuel tank on the model has volume 240 cm³. Work out the volume of the real tank. Give your answer in m³.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 18 × 25 (= 450)	✓ 1
A1 4.5 m	✓ 1
Part (b)
M1 area scale factor 25² (= 625)	✓ 1
M1 40 × 625 = 25 000 cm²	✓ 1
A1 2.5 m²	✓ 1
Part (c)
M1 volume scale factor 25³ (= 15 625)	✓ 1
M1 240 × 15 625 = 3 750 000 cm³	✓ 1
A1 3.75 m³	✓ 1

Exam tip: If the length scale factor is k, areas scale by k² and volumes by k³. Convert units only at the end.

+ Add to test✓ In your test283IGME3-4-11r24-06Area, surface area and volume Paper HMedium7 marks

The curved arch of a bridge is an arc of a circle of radius 40 m. The arc subtends an angle of 50° at the centre of the circle.

Diagram not to scale

(a)

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

(2)

(b)

Work out the area of the sector bounded by the arc and the two radii. Give your answer correct to 3 significant figures.

(2)

(c)

A different arch is an arc of a circle of radius 40 m and length 20 m. Work out the angle it subtends at the centre. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (50 ÷ 360) × 2 × π × 40	✓ 1
A1 34.9 m	✓ 1
Part (b)
M1 (50 ÷ 360) × π × 40²	✓ 1
A1 698 m²	✓ 1
Part (c)
M1 20 ÷ (2 × π × 40) or 20 ÷ 251.3...	✓ 1
M1 × 360	✓ 1
A1 28.6°	✓ 1

Common error: For a sector use the fraction angle ÷ 360. Do not use the radius where the full circumference or area formula needs it.

+ Add to test✓ In your test284IGME3-4-11r24-08Similar shapes and circle theorems Paper HMedium8 marks

The circular lens of a probe camera has centre O and radius 9 cm. Points P and Q lie on the circumference. The tangents to the circle at P and Q meet at point T. Angle PTQ = 48°.

Diagram not to scale

(a)

Work out the size of angle POQ. Give a reason for each step of your working.

(3)

(b)

Point R lies on the major arc PQ. Work out angle PRQ. Give a reason for your answer.

(2)

(c)

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

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 angle OPT = angle OQT = 90° (tangent is perpendicular to radius)	✓ 1
M1 360 − 90 − 90 − 48 (angles in a quadrilateral sum to 360°)	✓ 1
A1 132°	✓ 1
Part (b)
M1 132 ÷ 2	✓ 1
A1 66° (angle at the centre is twice the angle at the circumference)	✓ 1
Part (c)
B1 angle OTP = 24° (OT bisects angle PTQ)	✓ 1
M1 sin 24° = 9 ÷ OT	✓ 1
A1 22.1 cm	✓ 1

Common error: Always state the circle theorem in words next to each angle you find; an angle value alone does not give a reason.

+ Add to test✓ In your test285IGME3-4-11r24-10Pythagoras' theorem and trigonometry Paper HMedium5 marks

Two straight cables run from the top T of a bridge tower to anchor points A and B on level ground. TA = 52 m, TB = 65 m and angle ATB = 38°.

Diagram not to scale

(a)

Work out the distance AB. Give your answer correct to 3 significant figures.

(3)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 AB² = 52² + 65² − 2 × 52 × 65 × cos 38°	✓ 1
M1 AB = √(1602.0...)	✓ 1
A1 40.0 m	✓ 1
Part (b)
M1 ½ × 52 × 65 × sin 38°	✓ 1
A1 1040 m²	✓ 1

Common error: In the cosine rule the angle must be the one between the two known sides, and the subtraction happens after multiplying out 2bc cos A.

+ Add to test✓ In your test286IGME3-4-14r29-09Angles, polygons and circle properties Paper HMedium4 marks

The points A, B and C lie on a circle with centre O, where B lies on the major arc AC. Angle AOC = $112^{\circ}$. The point D lies on the minor arc AC, so that ABCD is a cyclic quadrilateral.

Diagram not to scale

(a)

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

(2)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $112\div2$	✓ 1
A1 $56^{\circ}$ with reason: the angle at the centre is twice the angle at the circumference	✓ 1
Part (b)
M1 $180-56$ (ECF from (a))	✓ 1
A1 $124^{\circ}$ with reason: opposite angles of a cyclic quadrilateral sum to $180^{\circ}$	✓ 1

Exam tip: Quote the full circle theorem, including which angles it relates; a name alone such as 'cyclic' is not enough.

+ Add to test✓ In your test287IGME3-4-14r29-10Pythagoras' theorem and trigonometry Paper HMedium6 marks

Two ships leave port P at the same time. Ship A sails 14 km on a bearing of $040^{\circ}$. Ship B sails 22 km on a bearing of $115^{\circ}$.

Diagram not to scale

(a)

Write down the size of angle APB.

(1)

(b)

Work out the distance AB between the two ships. Give your answer correct to 3 significant figures.

(3)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $75^{\circ}$	✓ 1
Part (b)
M1 $AB^{2}=14^{2}+22^{2}-2\times14\times22\times\cos75^{\circ}$	✓ 1
M1 $AB^{2}=520.6\ldots$ (accept 520 to 521)	✓ 1
A1 22.8 km	✓ 1
Part (c)
M1 $\tfrac{1}{2}\times14\times22\times\sin75^{\circ}$	✓ 1
A1 149 km$^{2}$	✓ 1

Common error: Set your calculator to degrees before using $\cos75^{\circ}$ and $\sin75^{\circ}$.

+ Add to test✓ In your test288IGME3-4-13r32-09Area, surface area and volume Paper HMedium8 marks

Two bottles of hand sanitiser are mathematically similar. The small bottle has height 12 cm and capacity 250 mL. The large bottle has height 18 cm.

(a)

Work out the capacity of the large bottle. Give your answer to the nearest mL.

(3)

(b)

The label on the small bottle has area 90 cm². The label on the large bottle is similar. Work out the area of the label on the large bottle.

(2)

(c)

A refill drum is also similar to the small bottle and has capacity 6750 mL. Work out the height of the drum.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 length scale factor 18 ÷ 12 = 1.5	✓ 1
M1 250 × 1.5³	✓ 1
A1 844 mL (843.75 seen)	✓ 1
Part (b)
M1 90 × 1.5²	✓ 1
A1 202.5 cm²	✓ 1
Part (c)
M1 6750 ÷ 250 = 27 (volume scale factor)	✓ 1
M1 cube root of 27 = 3	✓ 1
A1 36 cm	✓ 1

Common error: Lengths, areas and volumes scale by k, k² and k³; using the length scale factor for volume gives a far too small answer.

+ Add to test✓ In your test289IGME3-4-13r32-10Similar shapes and circle theorems Paper HMedium7 marks

A circular reservoir has centre O. Two straight paths, TA and TB, touch the reservoir at A and B. Angle ATB = 48°.

Diagram not to scale

(a)

Work out the size of angle AOB. Give a reason for each step of your working.

(3)

(b)

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

(2)

(c)

The point C lies on the major arc AB. Work out the size of angle ACB. Give a reason for your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 angle OAT = angle OBT = 90° (tangent and radius meet at 90°)	✓ 1
M1 360 − 90 − 90 − 48 (angles in a quadrilateral add up to 360°)	✓ 1
A1 132°	✓ 1
Part (b)
M1 (180 − 132) ÷ 2 (triangle OAB is isosceles because OA = OB, both radii)	✓ 1
A1 24°	✓ 1
Part (c)
M1 132 ÷ 2	✓ 1
A1 66° (angle at the centre is twice the angle at the circumference)	✓ 1

Exam tip: In circle-theorem questions, always write the reason in words next to each angle you find, because the reason earns a mark alongside the value.

+ Add to test✓ In your test290IGME3-4-12r36-08Similar shapes and circle theorems Paper HMedium7 marks

A, B and C are points on the circumference of a circular stage with centre O. C lies on the major arc AB. Angle ACB = 54°. AT is the tangent to the circle at A, and T is on the opposite side of AB from C.

Diagram not to scale

(a)

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

(2)

(b)

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

(2)

(c)

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

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 108°	✓ 1
B1 the angle at the centre is twice the angle at the circumference	✓ 1
Part (b)
M1 (180 − 108) ÷ 2 (ECF from (a))	✓ 1
A1 36° because triangle OAB is isosceles (OA = OB are radii)	✓ 1
Part (c)
M1 90 − 36 (ECF from (b))	✓ 1
A1 54°	✓ 1
B1 the tangent is perpendicular to the radius at A (or the alternate segment theorem)	✓ 1

Exam tip: Give each reason in words using the standard theorem name or statement; an angle with no reason earns only the answer mark.

+ Add to test✓ In your test291IGME3-4-12r36-11Area, surface area and volume Paper HMedium6 marks

A revolving theatre stage is a sector of a circle of radius 6 m with sector angle 140°.

Diagram not to scale

(a)

Calculate the arc length of the sector. Give your answer to 3 significant figures.

(2)

(b)

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

(2)

(c)

Calculate the perimeter of the sector. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 140/360 × 2 × π × 6	✓ 1
A1 14.7 m	✓ 1
Part (b)
M1 140/360 × π × 6²	✓ 1
A1 44.0 m²	✓ 1
Part (c)
M1 arc length + 6 + 6 (14.66... + 12)	✓ 1
A1 26.7 m	✓ 1

Common error: The perimeter of a sector includes both straight radii as well as the arc.

+ Add to test✓ In your test292IGME3-4-13r36-05Area, surface area and volume Paper HMedium4 marks

Two cooling tanks in a data centre are mathematically similar. The smaller tank has height 1.2 m, volume 640 litres and a total surface area of 4.8 m². The larger tank has height 1.8 m.

(a)

Work out the volume of the larger tank.

(2)

(b)

Work out the total surface area of the larger tank.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 length scale factor 1.8 ÷ 1.2 = 1.5 and 640 × 1.5³	✓ 1
A1 2160 litres	✓ 1
Part (b)
M1 4.8 × 1.5²	✓ 1
A1 10.8 m²	✓ 1

Common error: Area scales with the square of the length scale factor and volume with its cube. Do not multiply by the length scale factor alone.

+ Add to test✓ In your test293IGME3-4-13r36-06Similar shapes and circle theorems Paper HMedium6 marks

A circular stone ring has centre O and radius 5 m. A surveyor stands at point P, 13 m from O. The straight line from P just touches the ring at point T (it is a tangent to the circle at T).

Diagram not to scale

(a)

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

(2)

(b)

Calculate the length PT.

(2)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 90°	✓ 1
B1 The tangent is perpendicular to the radius at the point of contact	✓ 1
Part (b)
M1 13² − 5² (= 144)	✓ 1
A1 12 m	✓ 1
Part (c)
M1 cos(TOP) = 5 ÷ 13	✓ 1
A1 67.4°	✓ 1

Exam tip: A radius meeting a tangent always gives a right angle, which lets you use Pythagoras' theorem or trigonometry.

+ Add to test✓ In your test294IGME3-4-13r36-09Area, surface area and volume Paper HMedium4 marks

A curved fragment of a pottery plate is a sector of a circle of radius 15 cm. The angle at the centre of the sector is 72°.

Diagram not to scale

(a)

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

(2)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 72 ÷ 360 × 2 × π × 15	✓ 1
A1 18.8 cm	✓ 1
Part (b)
M1 72 ÷ 360 × π × 15²	✓ 1
A1 141 cm²	✓ 1

Common error: Do not confuse the formulas: arc length uses the circumference formula, whereas sector area uses πr².

+ Add to test✓ In your test295IGME3-4-13r36-13Similar shapes and circle theorems Paper HMedium4 marks

Two replica vases are mathematically similar. The smaller vase has a surface area of 90 cm² and a volume of 400 cm³. The larger vase has a surface area of 202.5 cm².

(a)

Work out the length scale factor from the smaller vase to the larger vase.

(2)

(b)

Work out the volume of the larger vase.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(202.5 ÷ 90) = √2.25	✓ 1
A1 1.5	✓ 1
Part (b)
M1 400 × 1.5³	✓ 1
A1 1350 cm³	✓ 1

Exam tip: Take the square root of an area ratio to get the length scale factor, then cube it for the volume ratio.

+ Add to test✓ In your test296IGME3-4-14r37-06Area, surface area and volume Paper HMedium6 marks

A model of a wind turbine is made to a scale of 1 : 50. The model blade has length 0.36 m. The model blade has surface area 0.12 m². The model nacelle has volume 80 cm³.

(a)

Work out the length of the real blade.

(1)

(b)

Work out the surface area of the real blade.

(2)

(c)

Work out the volume of the real nacelle. Give your answer in m³.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 18 m	✓ 1
Part (b)
M1 0.12 × 50²	✓ 1
A1 300 m²	✓ 1
Part (c)
M1 80 × 50³	✓ 1
A1 10 000 000 cm³	✓ 1
A1 10 m³	✓ 1

Common error: Areas scale by the square of the length scale factor and volumes by the cube; convert units only after scaling.

+ Add to test✓ In your test297IGME3-4-14r37-08Similar shapes and circle theorems Paper HMedium6 marks

A, B and C are points on a circle with centre O. C lies on the major arc AB. Angle ACB = 37°. The line AT is the tangent to the circle at A, with T on the opposite side of AB from C.

Diagram not to scale

(a)

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

(2)

(b)

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

(2)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 74°	✓ 1
B1 the angle at the centre is twice the angle at the circumference	✓ 1
Part (b)
M1 (180 − 74) ÷ 2	✓ 1
A1 53° because triangle OAB is isosceles (OA = OB, radii)	✓ 1
Part (c)
M1 90 − 53 using angle OAT = 90° (tangent perpendicular to radius), or the alternate segment theorem	✓ 1
A1 37° (ECF from (b) if the method is correct)	✓ 1

Common error: Always name the theorem in a circle-theorem question, and check the reason matches the angles used: centre, circumference, tangent or radius.

+ Add to test✓ In your test298IGME3-4-14r37-13Area, surface area and volume Paper HMedium7 marks

A swimming pool cover is in the shape of a sector of a circle with radius 6 m and angle 100°.

Diagram not to scale

(a)

Work out the length of the curved edge of the cover. Give your answer correct to 3 significant figures.

(2)

(b)

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

(2)

(c)

A second sector has radius 6 m and area 48 m². Work out the angle of this sector. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 100/360 × 2 × π × 6	✓ 1
A1 10.5 m	✓ 1
Part (b)
M1 100/360 × π × 6²	✓ 1
A1 31.4 m²	✓ 1
Part (c)
M1 48 = θ/360 × π × 6²	✓ 1
M1 θ = 48 × 360 ÷ (π × 36)	✓ 1
A1 152.8°	✓ 1

Exam tip: A sector is a fraction angle/360 of the whole circle; use the same fraction for both its arc length and its area.

+ Add to test✓ In your test299IGME3-4-14r44-08Angles, polygons and circle properties Paper HMedium6 marks

Points A, B, C and D lie in that order on the circumference of a circle with centre O, so ABCD is a cyclic quadrilateral. The angle AOC at the centre is 148°, and the minor arc AC contains the point D.

Diagram not to scale

(a)

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

(2)

(b)

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

(2)

(c)

Work out the size of angle OAC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 74°	✓ 1
B1 the angle at the centre is twice the angle at the circumference	✓ 1
Part (b)
B1 106° (ECF from (a))	✓ 1
B1 opposite angles in a cyclic quadrilateral add up to 180°	✓ 1
Part (c)
M1 triangle OAC is isosceles (OA = OC), so (180 − 148) ÷ 2	✓ 1
A1 16°	✓ 1

Exam tip: When a question says ‘give a reason’, write the full circle theorem in words, for example ‘angle at the centre is twice the angle at the circumference’.

+ Add to test✓ In your test300IGME3-4-14r44-09Area, surface area and volume Paper HMedium8 marks

An architect makes a scale model of a tower. The scale of the model is 1 : 80.

(a)

The model is 45 cm tall. Work out the height of the real tower in metres.

(2)

(b)

The area of glass on one face of the real tower is 5.76 m². Work out the area of glass on the same face of the model, in cm².

(3)

(c)

The volume of the model is 1.5 litres. Work out the volume of the real tower in m³. (1 litre = 1000 cm³)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 45 × 80 (= 3600 cm)	✓ 1
A1 36 m	✓ 1
Part (b)
M1 5.76 m² = 57 600 cm²	✓ 1
M1 57 600 ÷ 80²	✓ 1
A1 9 cm²	✓ 1
Part (c)
M1 volume scale factor 80³ = 512 000	✓ 1
M1 1500 × 512 000 = 768 000 000 cm³	✓ 1
A1 768 m³	✓ 1

Common error: Lengths scale by k, areas by k² and volumes by k³. Convert units after scaling, taking care that 1 m³ = 1 000 000 cm³.

+ Add to test✓ In your test301IGME3-4-14r44-11Similar shapes and circle theorems Paper HMedium5 marks

Two similar storage tanks are carried on a cargo ship. The smaller tank has a surface area of 3.6 m² and a volume of 0.8 m³. The larger tank has a surface area of 8.1 m².

(a)

Work out the length scale factor from the smaller tank to the larger tank.

(2)

(b)

Work out the volume of the larger tank.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area scale factor 8.1 ÷ 3.6 = 2.25 and square root	✓ 1
A1 1.5	✓ 1
Part (b)
M1 volume scale factor 1.5³ (ECF from (a))	✓ 1
M1 0.8 × 3.375	✓ 1
A1 2.7 m³	✓ 1

Common error: Take the square root of an area ratio to get the length scale factor, then cube it for volume. Do not cube the area ratio.

+ Add to test✓ In your test302IGME3-4-14r45-09Area, surface area and volume Paper HMedium5 marks

A fish farm uses a net that covers a sector of a circle. The sector has radius 30 m and angle 140°.

Diagram not to scale

(a)

Work out the arc length of the sector. Give your answer to 3 significant figures.

(2)

(b)

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

(2)

(c)

A fence runs along the whole boundary of the sector (the arc and the two radii). Work out the length of the fence to the nearest metre.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{140}{360}\times 2\times\pi\times 30$	✓ 1
A1 73.3 m	✓ 1
Part (b)
M1 $\frac{140}{360}\times\pi\times 30^2$	✓ 1
A1 1100 m²	✓ 1
Part (c)
B1 133 m (73.3 + 60)	✓ 1

Common error: The boundary of a sector includes the two straight radii as well as the arc, so remember to add 2 × the radius.

+ Add to test✓ In your test303IGME3-4-14r45-10Angles, polygons and circle properties Paper HMedium6 marks

A robotic arm moves along a circular track with centre O. The points P, Q, R and S lie on the track in that order and angle PQR = 108°.

Diagram not to scale

(a)

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

(2)

(b)

Work out the size of the angle POR that is less than 180°. Give a reason for your answer.

(2)

(c)

Work out the size of angle OPR.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 72°	✓ 1
B1 opposite angles of a cyclic quadrilateral add up to 180°	✓ 1
Part (b)
B1 144°	✓ 1
B1 the angle at the centre is twice the angle at the circumference (angle PSR stands on the same arc)	✓ 1
Part (c)
M1 triangle OPR is isosceles since OP = OR (radii), so $\frac{180-144}{2}$	✓ 1
A1 18°	✓ 1

Exam tip: In circle theorem questions, name the theorem in words, for example 'angle at the centre is twice the angle at the circumference'. A calculation alone does not earn the reason mark.

+ Add to test✓ In your test304IGME3-4-14r45-13Pythagoras' theorem and trigonometry Paper HMedium6 marks

A fish transport crate is a cuboid measuring 60 cm by 40 cm by 30 cm. A thin sensor rod stretches from one bottom corner to the opposite top corner.

Diagram not to scale

(a)

Work out the length of the diagonal of the base of the crate. Give your answer to 3 significant figures.

(2)

(b)

Work out the length of the sensor rod. Give your answer to 3 significant figures.

(2)

(c)

Work out the angle between the sensor rod and the base of the crate. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\sqrt{60^2+40^2}$	✓ 1
A1 72.1 cm	✓ 1
Part (b)
M1 $\sqrt{60^2+40^2+30^2}$ or $\sqrt{72.1^2+30^2}$	✓ 1
A1 78.1 cm	✓ 1
Part (c)
M1 tan⁻¹(30 ÷ 72.1) (or sin⁻¹(30 ÷ 78.1))	✓ 1
A1 22.6°	✓ 1

Exam tip: For a 3D problem, find a right-angled triangle that contains the line and the angle you need. The base diagonal is often the first step.

+ Add to test✓ In your test305IGME3-4-13r46-07Angles, polygons and circle properties Paper HMedium6 marks

The circular test track at a Formula 1 factory has marker posts A, B, C and D on its edge. Give a reason for each of your answers.

(a)

ABCD is a cyclic quadrilateral with angle ABC = 112°. Work out the size of angle ADC.

(2)

(b)

The tangent at point T and the chord TU make an angle of 38°. V is a point on the circle in the alternate segment. Work out the size of angle TVU.

(2)

(c)

P, Q and R are points on a different circle with centre O. Angle PQR = 41° and Q lies on the major arc PR. Work out the size of angle POR.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 68°	✓ 1
B1 Opposite angles of a cyclic quadrilateral add up to 180°	✓ 1
Part (b)
B1 38°	✓ 1
B1 The angle between a tangent and a chord is equal to the angle in the alternate segment	✓ 1
Part (c)
B1 82°	✓ 1
B1 The angle at the centre is twice the angle at the circumference	✓ 1

Exam tip: A reason must name the circle theorem in full; a calculation alone does not earn the reason mark.

+ Add to test✓ In your test306IGME3-4-13r46-10Area, surface area and volume Paper HMedium7 marks

A beekeeper uses a conical funnel to fill jars with honey. The cone has base radius 6 cm and perpendicular height 15 cm. The funnel is completely full of honey.

Cross-section. Diagram not to scale

(a)

Show that the volume of honey in the funnel is 180π cm³.

(2)

(b)

All the honey is poured into an empty cylindrical jar of radius 4 cm. Work out the depth of honey in the jar.

(3)

(c)

The density of the honey is 1.4 g/cm³. Calculate the mass of honey in the funnel. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ⅓ × π × 6² × 15	✓ 1
A1 180π cm³ shown	✓ 1
Part (b)
M1 π × 4² × h = 180π	✓ 1
M1 h = 180 ÷ 16	✓ 1
A1 11.25 cm	✓ 1
Part (c)
M1 180π × 1.4 (= 791.6…)	✓ 1
A1 792 g	✓ 1

Exam tip: Keep π in your working until the last step so the 180π cancels cleanly when the volumes are equated.

+ Add to test✓ In your test307IGME3-4-13r46-12Angles, polygons and circle properties Paper HMedium5 marks

Four marker posts A, B, C and D, in that order, lie on the edge of a circular Formula 1 test circuit. In the cyclic quadrilateral ABCD, angle A = (3x + 10)°, angle B = (2x + 20)° and angle C = (5x − 30)°.

Diagram not to scale

(a)

Show that x = 25. Give a reason for your working.

(3)

(b)

Work out the size of angle D.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 Opposite angles of a cyclic quadrilateral add up to 180°	✓ 1
M1 (3x + 10) + (5x − 30) = 180, so 8x − 20 = 180	✓ 1
A1 8x = 200, so x = 25 shown	✓ 1
Part (b)
M1 angle B = 2 × 25 + 20 = 70°	✓ 1
A1 180 − 70 = 110°	✓ 1

Common error: Opposite angles of a cyclic quadrilateral sum to 180°; adjacent angles do not.

+ Add to test✓ In your test308IGME3-4-14r46-06Similar shapes and circle theorems Paper HMedium7 marks

A model of a wind turbine is mathematically similar to the real turbine. A blade of the model is 0.35 m long and the corresponding blade of the real turbine is 42 m long.

(a)

Work out the length scale factor from the model to the real turbine.

(2)

(b)

A blade of the model has a surface area of 0.05 m². Work out the surface area of the corresponding real blade.

(2)

(c)

The hub of the model has a volume of 250 cm³. Work out the volume of the real hub in m³.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 42 ÷ 0.35	✓ 1
A1 120	✓ 1
Part (b)
M1 0.05 × 120²	✓ 1
A1 720 m²	✓ 1
Part (c)
M1 250 × 120³ = 432 000 000 cm³	✓ 1
M1 ÷ 1 000 000	✓ 1
A1 432 m³	✓ 1

Common error: Convert cm³ to m³ by dividing by 1 000 000, not by 100 or 1000.

+ Add to test✓ In your test309IGME3-4-14r46-07Similar shapes and circle theorems Paper HMedium6 marks

A circular fence surrounds a wind farm. Give a reason for each of your answers.

(a)

A, B and C are points on the circle and AB is a diameter. Angle CAB = 28°. Work out the size of angle ABC.

(3)

(b)

Tangents from an external point T touch another circle at P and Q. Angle PTQ = 50°. Work out the size of angle TPQ.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 angle ACB = 90° because the angle in a semicircle is a right angle	✓ 1
M1 180 − 90 − 28	✓ 1
A1 62°	✓ 1
Part (b)
B1 TP = TQ because tangents from an external point are equal in length	✓ 1
M1 (180 − 50) ÷ 2	✓ 1
A1 65°	✓ 1

Exam tip: Write the circle theorem as a short sentence beside each angle you find so that the reason is clear.

+ Add to test✓ In your test310IGME-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 point	Mark	Notes
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.

+ Add to test✓ In your test311IGME-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 point	Mark	Notes
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.

+ Add to test✓ In your test312IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test313IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test314IGME-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 point	Mark	Notes
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.

+ Add to test✓ In your test315IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test316IGME-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.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test317IGME-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 = $\frac{1}{3}\pi r^2 h$]

Cross-section through the axis. Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test318IGME-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)°.

Diagram not to scale

(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 point	Mark	Notes
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.

+ Add to test✓ In your test319IGME3-4-10d-13Similar shapes and circle theorems Paper HHard11 marks

From a point P outside a circle with centre O, tangents are drawn to the circle, touching it at A and at B. Angle APB = 50°. The radius of the circle is 7 cm. The point C lies on the major arc AB.

Diagram not to scale

(a)

Work out the size of angle AOB. Give reasons.

(3)

(b)

Work out the size of angle PAB. Give a reason.

(3)

(c)

Work out the size of angle ACB. Give a reason, and say how your answer links to your answer to (b).

(2)

(d)

Work out the length of the tangent PA. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 angles OAP and OBP are 90° (a tangent is perpendicular to the radius)	✓ 1
M1 360° − 90° − 90° − 50° (angles in a quadrilateral)	✓ 1
A1 130°	✓ 1
Part (b)
B1 PA = PB (tangents from an external point are equal), so triangle PAB is isosceles	✓ 1
M1 (180 − 50) ÷ 2	✓ 1
A1 65°	✓ 1
Part (c)
B1 65° (ECF from (a)) with reason: the angle at the centre is twice the angle at the circumference	✓ 1
B1 equal to angle PAB by the alternate segment theorem	✓ 1
Part (d)
M1 angle APO = 25° because OP bisects angle APB	✓ 1
M1 tan 25° = 7 ÷ PA	✓ 1
A1 15.0 cm	✓ 1

Common error: A radius meets a tangent at 90°. Mark this right angle on your sketch first; it unlocks most of the other angles.

+ Add to test✓ In your test320IGME3-4-10d-14Pythagoras' theorem and trigonometry Paper HHard10 marks

A model of a glass roof is a square-based pyramid VABCD. The base ABCD is a square of side 10 cm. The apex V is directly above the centre of the base, at a height of 12 cm.

Diagram not to scale

(a)

Work out the volume of the pyramid.

(2)

(b)

Work out the length of the diagonal AC. Give your answer to 3 significant figures.

(2)

(c)

Work out the length of the sloping edge VA. Give your answer to 3 significant figures.

(2)

(d)

Work out the angle between the edge VA and the base. Give your answer to 1 decimal place.

(2)

(e)

M is the midpoint of AB and N is the centre of the base. Work out the angle between the triangular face VAB and the base, that is angle VMN. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ⅓ × 10² × 12	✓ 1
A1 400 cm³	✓ 1
Part (b)
M1 √(10² + 10²)	✓ 1
A1 14.1 cm	✓ 1
Part (c)
M1 VA² = 12² + (½ × AC)² = 144 + 50	✓ 1
A1 13.9 cm	✓ 1
Part (d)
M1 tan x = 12 ÷ 7.07... (half of AC)	✓ 1
A1 59.5°	✓ 1
Part (e)
M1 tan VMN = 12 ÷ 5 (NM = 5 cm)	✓ 1
A1 67.4°	✓ 1

Exam tip: In 3D problems, draw the right-angled triangle you are using as a flat sketch and label the lengths you know before applying Pythagoras or trigonometry.

+ Add to test✓ In your test321IGME3-4-11r5-12Area, surface area and volume Paper HHard6 marks

A grain hopper is a frustum: a solid cone with its pointed end cut off by a plane parallel to the base. The original cone had a base radius of 15 cm and a height of 24 cm. The small cone that was removed has a height of 9.6 cm. Volume of a cone = (1/3)πr²h.

Cross-section through the axis. Diagram not to scale

(a)

Show that the small cone that was removed has a radius of 6 cm.

(2)

(b)

Calculate the volume of the frustum. Give your answer to the nearest cm³.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the cones are similar, so scale factor = 9.6 ÷ 24 = 0.4	✓ 1
A1 0.4 × 15 = 6 cm	✓ 1
Part (b)
M1 volume of original cone = (1/3) × π × 15² × 24 (= 5654.8…)	✓ 1
M1 volume of small cone = (1/3) × π × 6² × 9.6 (= 361.9…)	✓ 1
M1 subtract the two volumes	✓ 1
A1 5293 cm³	✓ 1

Exam tip: For a frustum, find the volume of the full cone, subtract the volume of the cone removed, and use the scale factor from similar triangles to find the missing dimensions.

+ Add to test✓ In your test322IGME3-4-11r5-13Pythagoras' theorem and trigonometry Paper HHard6 marks

A rectangular crate for transporting an organ pipe is 3.0 m long, 1.5 m wide and 1.0 m high (inside measurements).

Diagram not to scale

(a)

Calculate the length of the diagonal of the base of the crate. Give your answer to 3 decimal places.

(2)

(b)

Calculate the length of the longest straight pipe that can lie inside the crate, from one bottom corner to the opposite top corner.

(2)

(c)

Calculate the angle between this longest pipe and the base of the crate. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3.0² + 1.5² (= 11.25)	✓ 1
A1 3.354 m	✓ 1
Part (b)
M1 3.0² + 1.5² + 1.0² (= 12.25) or (base diagonal)² + 1.0²	✓ 1
A1 3.5 m	✓ 1
Part (c)
M1 tan (angle) = 1.0 ÷ 3.354… (or sin = 1.0 ÷ 3.5)	✓ 1
A1 16.6°	✓ 1

Exam tip: In 3D problems, draw out the right-angled triangle you need, often using the diagonal of the base as one of its sides.

+ Add to test✓ In your test323IGME3-4-13r5-12Area, surface area and volume Paper HHard9 marks

Calculator

A solid toy is made from a cone joined to a hemisphere. The common circular face has radius 4 cm. The cone has perpendicular height 9 cm. The material has density 2.7 g/cm³. Use unrounded values in later parts.

Cross-section through the axis. Diagram not to scale

(a)

Show that the volume of the toy is 285 cm³ correct to 3 significant figures.

(3)

(b)

Calculate the mass of the toy. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the total surface area of the toy. Give your answer correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cone volume (1/3) × π × 4² × 9 (= 48π)	✓ 1
M1 hemisphere volume (2/3) × π × 4³	✓ 1
A1 total 284.8... which rounds to 285 cm³	✓ 1
Part (b)
M1 284.8 × 2.7	✓ 1
A1 769 g	✓ 1
Part (c)
M1 slant height = √(4² + 9²) = 9.85	✓ 1
M1 curved area of cone π × 4 × 9.85 = 123.8	✓ 1
M1 curved area of hemisphere 2 × π × 4² = 100.5	✓ 1
A1 224 cm²	✓ 1

Common error: The flat circular face where the shapes join is not part of the outer surface, so it is not included.

+ Add to test✓ In your test324IGME3-4-13r5-13Pythagoras' theorem and trigonometry Paper HHard6 marks

Calculator

A rectangular water tank has internal length 1.8 m, width 1.2 m and height 0.9 m. A thin rod rests with one end at a bottom corner A and the other end at the opposite top corner B, on the far side of the tank.

Diagram not to scale

(a)

Calculate the length of the diagonal of the base of the tank. Give your answer correct to 3 significant figures.

(2)

(b)

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

(2)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(1.8² + 1.2²)	✓ 1
A1 2.16 m	✓ 1
Part (b)
M1 √(1.8² + 1.2² + 0.9²) (or √(2.1633² + 0.9²))	✓ 1
A1 2.34 m	✓ 1
Part (c)
M1 tan(angle) = 0.9 ÷ 2.1633	✓ 1
A1 22.6°	✓ 1

Exam tip: In 3D problems, find a right-angled triangle that contains both the line and its projection on the plane.

+ Add to test✓ In your test325IGME3-4-13r5-14Pythagoras' theorem and trigonometry Paper HHard6 marks

Calculator

A circular dial in a game has centre O and radius 10 cm. A chord AB subtends an angle of 80° at O. The region between the chord and the minor arc AB is shaded.

Diagram not to scale

(a)

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

(2)

(b)

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

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × 10 × sin 40° (or cosine rule)	✓ 1
A1 12.9 cm	✓ 1
Part (b)
M1 sector area (80/360) × π × 10² = 69.8	✓ 1
M1 triangle area ½ × 10 × 10 × sin 80° = 49.2	✓ 1
M1 subtraction of triangle from sector	✓ 1
A1 20.6 cm²	✓ 1

Common error: Subtract the triangle OAB from the sector, not from the whole circle.

+ Add to test✓ In your test326IGME3-4-11r6-11Area, surface area and volume Paper HHard7 marks

A cooling funnel for a data centre is made from a solid cone of height 24 cm and base radius 9 cm. A smaller cone of height 8 cm is removed from the top (the apex) by a cut parallel to the base. The solid that remains is a frustum.

Cross-section through the axis. Diagram not to scale

(a)

Show that the radius of the small cone that is removed is 3 cm.

(2)

(b)

Calculate the volume of the original cone, in terms of π.

(2)

(c)

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

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 scale factor 8 ÷ 24 = 1/3 (similar cones)	✓ 1
A1 9 × 1/3 = 3 cm	✓ 1
Part (b)
M1 ⅓ × π × 9² × 24	✓ 1
A1 648π cm³	✓ 1
Part (c)
M1 volume of small cone ⅓ × π × 3² × 8 (= 24π)	✓ 1
M1 648π − 24π (ECF from (b))	✓ 1
A1 1960 cm³	✓ 1

Common error: Subtract the volume of the small cone, not its height or radius, from the original cone.

+ Add to test✓ In your test327IGME3-4-11r6-12Angles, polygons and circle properties Paper HHard6 marks

ABCD is a cyclic quadrilateral. Angle DAB = (3x + 10)°, angle BCD = (5x − 30)° and angle ABC = (4x − 35)°.

Diagram not to scale

(a)

Write down the circle theorem that gives the relationship between angle DAB and angle BCD.

(1)

(b)

Find the value of x.

(3)

(c)

Work out the size of angle ADC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 opposite angles of a cyclic quadrilateral add up to 180°	✓ 1
Part (b)
M1 3x + 10 + 5x − 30 = 180	✓ 1
M1 8x = 200	✓ 1
A1 x = 25	✓ 1
Part (c)
M1 angle ABC = 4 × 25 − 35 = 65° (ECF from (b))	✓ 1
A1 180 − 65 = 115°	✓ 1

Exam tip: Check your value of x by substituting back; the opposite pair should total exactly 180°.

+ Add to test✓ In your test328IGME3-4-11r6-13Pythagoras' theorem and trigonometry Paper HHard7 marks

A cable is run inside a data centre room that is a cuboid 12 m long, 9 m wide and 4 m high. The cable goes in a straight line from a floor corner A to the diagonally opposite corner G at ceiling level.

Diagram not to scale

(a)

Calculate the length of the diagonal of the floor of the room.

(2)

(b)

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

(2)

(c)

Calculate the angle that the cable makes with the floor. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12² + 9² (= 225)	✓ 1
A1 15 m	✓ 1
Part (b)
M1 15² + 4² (= 241) (ECF from (a))	✓ 1
A1 15.5 m	✓ 1
Part (c)
M1 tan θ = 4 ÷ 15	✓ 1
M1 θ = tan⁻¹(4 ÷ 15)	✓ 1
A1 14.9°	✓ 1

Exam tip: For an angle with a plane, use the right-angled triangle made by the line, its shadow on the plane and the vertical.

+ Add to test✓ In your test329IGME3-4-11r9-10Pythagoras' theorem and trigonometry Paper HHard9 marks

Two straight railway lines leave a junction J with an angle of 52° between them. A goods train is 8.4 km from J on one line (at point G) and a passenger train is 6.5 km from J on the other line (at point P).

Diagram not to scale

(a)

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

(4)

(b)

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

(2)

(c)

Calculate the size of angle JGP. Give your answer correct to 1 decimal place. Use the unrounded value of GP.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for GP² = 8.4² + 6.5² − 2 × 8.4 × 6.5 × cos 52°	✓ 1
M1 for evaluating to 45.6 (45.58…)	✓ 1
M1 for square root of their value	✓ 1
A1 for 6.75 km	✓ 1
Part (b)
M1 for ½ × 8.4 × 6.5 × sin 52°	✓ 1
A1 for 21.5 km²	✓ 1
Part (c)
M1 for sin JGP ÷ 6.5 = sin 52° ÷ GP	✓ 1
M1 for sin JGP = 6.5 × sin 52° ÷ GP	✓ 1
A1 for 49.3° (ECF from (a))	✓ 1

Common error: Do not use the sine rule with a pair that does not include a known angle opposite a known side; check that each ratio uses a side with its opposite angle.

+ Add to test✓ In your test330IGME3-4-11r9-11Similar shapes and circle theorems Paper HHard7 marks

In an astronomy exercise, a circular crater rim has centre O and radius 9 km. From a point P outside the crater, two straight sight lines PA and PB just touch the rim at A and B, so they are tangents to the circle. Angle APB = 54°.

Diagram not to scale

(a)

Work out the size of angle PAB. Give a reason for each step.

(2)

(b)

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

(2)

(c)

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

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for PA = PB (tangents from an external point are equal in length), so triangle PAB is isosceles	✓ 1
A1 for (180 − 54) ÷ 2 = 63°	✓ 1
Part (b)
M1 for angle OAP = 90° (a tangent is perpendicular to the radius at the point of contact)	✓ 1
A1 for 90 − 63 = 27° (ECF from (a))	✓ 1
Part (c)
M1 for angle APO = 27° (OP bisects angle APB) or angle AOP = 63°	✓ 1
M1 for tan 27° = 9 ÷ PA (or PA = 9 × tan 63°)	✓ 1
A1 for 17.7 km	✓ 1

Exam tip: Tangent facts do the heavy lifting: equal tangent lengths give an isosceles triangle, and a right angle appears at each point of contact.

+ Add to test✓ In your test331IGME3-4-11r9-12Area, surface area and volume Paper HHard6 marks

A water tank at a rail depot is a cylinder with a hemisphere fixed on top, sharing the same circular face. The radius of the cylinder and of the hemisphere is 3.5 m. The total volume of the tank is 390 m³.
Volume of a sphere = $\frac{4}{3}\pi r^3$. Volume of a cylinder = $\pi r^2 h$.

Cross-section. Diagram not to scale

(a)

Calculate the volume of the hemisphere. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the height of the cylindrical part of the tank. Give your answer correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for ½ × 4/3 × π × 3.5³	✓ 1
A1 for 89.8 m³ (89.79…)	✓ 1
Part (b)
M1 for 390 − 89.79… (volume of cylinder, 300.2…)	✓ 1
M1 for π × 3.5² × h = their cylinder volume	✓ 1
M1 for dividing by 38.48… (π × 3.5²)	✓ 1
A1 for 7.80 m (ECF from (a))	✓ 1

Common error: A hemisphere is half a sphere: forgetting to halve $\frac{4}{3}\pi r^3$ is a frequent slip.

+ Add to test✓ In your test332IGME3-4-12r9-12Area, surface area and volume Paper HHard7 marks

A conical hopper stores wood pellets for a biomass heating plant. The hopper is a cone with its point at the bottom and a circular top. It is 6 m tall (vertical height) and holds 28 m³ when full.
Volume of a cone = $\frac{1}{3}\pi r^2 h$. Curved surface area of a cone = $\pi r l$, where l is the slant height.

Cross-section through the axis. Diagram not to scale

(a)

Calculate the radius of the top of the hopper. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the slant height of the hopper. Use the unrounded radius from (a). Give your answer correct to 3 significant figures.

(2)

(c)

The curved surface of the hopper is to be coated with a protective paint. Calculate the area to be painted. Use unrounded values and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{1}{3}\pi r^2 \times 6 = 28$ (or 2πr² = 28)	✓ 1
M1 for r² = 14 ÷ π (= 4.456…)	✓ 1
A1 for 2.11 m	✓ 1
Part (b)
M1 for l² = r² + 6²	✓ 1
A1 for 6.36 m (ECF from (a))	✓ 1
Part (c)
M1 for π × r × l	✓ 1
A1 for 42.2 m² (ECF from (a) and (b))	✓ 1

Exam tip: Slant height and vertical height are different: the volume formula uses the vertical height, but the curved-area formula needs the slant height.

+ Add to test✓ In your test333IGME3-4-12r9-13Similar shapes and circle theorems Paper HHard9 marks

Two vaccine vials are mathematically similar in shape. The smaller vial has a surface area of 90 cm² and the larger has a surface area of 202.5 cm². The larger vial holds 135 mL when full. The smaller vial is 6.4 cm tall.

(a)

Show that the length scale factor from the smaller vial to the larger vial is 1.5.

(2)

(b)

Work out how much the smaller vial holds when full.

(3)

(c)

Work out the height of the larger vial.

(2)

(d)

The liquid costs €0.12 per mL. Work out how much more it costs to fill the larger vial than the smaller vial.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for area scale factor 202.5 ÷ 90 = 2.25	✓ 1
A1 for √2.25 = 1.5 shown	✓ 1
Part (b)
M1 for volume scale factor 1.5³ = 3.375	✓ 1
M1 for 135 ÷ 3.375	✓ 1
A1 for 40 mL	✓ 1
Part (c)
M1 for 6.4 × 1.5	✓ 1
A1 for 9.6 cm	✓ 1
Part (d)
M1 for 135 − 40 = 95 mL (ECF from (b))	✓ 1
A1 for €11.40	✓ 1

Common error: Going from areas to lengths needs a square root, and going from lengths to volumes needs a cube; mixing the two up is the usual cause of wrong answers.

+ Add to test✓ In your test334IGME3-4-12r10-10Pythagoras' theorem and trigonometry Paper HHard8 marks

A bridge arch is an arc of a circle of radius 25 m. The two ends of the arch, A and B, are on the level deck and 40 m apart. The centre O of the circle is below the deck, and the arch is the shorter arc AB.

Diagram not to scale

(a)

Work out angle AOB. Give your answer to 1 decimal place.

(3)

(b)

Work out the length of the arch. Use the unrounded angle and give your answer to 3 significant figures.

(2)

(c)

Work out the area between the arch and the deck. Use the unrounded angle and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 right-angled triangle with half of AB = 20 and hypotenuse 25	✓ 1
M1 sin(½AOB) = 20 ÷ 25	✓ 1
A1 106.3°	✓ 1
Part (b)
M1 (106.26… ÷ 360) × 2 × π × 25	✓ 1
A1 46.4 (m)	✓ 1
Part (c)
M1 sector area (106.26… ÷ 360) × π × 25²	✓ 1
M1 subtract triangle area ½ × 25² × sin 106.26… (= 300)	✓ 1
A1 280 (m²)	✓ 1

Common error: The area between an arc and its chord is the sector area minus the triangle area, not plus.

+ Add to test✓ In your test335IGME3-4-12r10-11Area, surface area and volume Paper HHard7 marks

A coffee storage hopper is a frustum, formed by removing a small cone from the pointed end of a large cone. The large cone has base radius 0.9 m and height 1.2 m. The small cone removed has height 0.4 m and is mathematically similar to the large cone. The volume of a cone is $\frac{1}{3}\pi r^2 h$.

Cross-section through the axis. Diagram not to scale

(a)

Work out the base radius of the small cone.

(2)

(b)

Work out the volume of the hopper, in m³, to 3 significant figures.

(3)

(c)

The hopper is full of coffee beans of density 650 kg/m³. Work out the mass of the beans, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.9 × (0.4 ÷ 1.2)	✓ 1
A1 0.3 (m)	✓ 1
Part (b)
M1 large cone volume ⅓π × 0.9² × 1.2 (= 1.0179…)	✓ 1
M1 subtract small cone volume ⅓π × 0.3² × 0.4 (= 0.0377…)	✓ 1
A1 0.980 (m³)	✓ 1
Part (c)
M1 650 × 0.980…	✓ 1
A1 637 (kg)	✓ 1

Common error: Do not subtract the heights or radii to get the frustum volume; subtract the two cone volumes.

+ Add to test✓ In your test336IGME3-4-12r10-13Similar shapes and circle theorems Paper HHard6 marks

A circular pillar of a bridge has centre O and radius 7 m. A cable is anchored at point P, which is 25 m from O. Two straight cables from P each just touch the pillar, at points T and U. Use unrounded values from earlier parts in later parts.

Diagram not to scale

(a)

Work out the length PT.

(2)

(b)

Work out angle TOP. Give your answer to 1 decimal place.

(2)

(c)

Work out angle TPU. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 angle OTP = 90° (tangent meets radius at right angles) and PT² = 25² − 7²	✓ 1
A1 24 (m)	✓ 1
Part (b)
M1 cos TOP = 7 ÷ 25	✓ 1
A1 73.7°	✓ 1
Part (c)
M1 angle TPO = 90° − 73.74…° and angle TPU = 2 × angle TPO (symmetry)	✓ 1
A1 32.5°	✓ 1

Common error: The two tangents from an external point are equal in length and the line from that point to the centre bisects the angle between them.

+ Add to test✓ In your test337IGME3-4-13r10-13Pythagoras' theorem and trigonometry Paper HHard7 marks

A triangular plot of land in a new housing development has sides of length 24 m, 31 m and 38 m.

Diagram not to scale

(a)

Calculate the size of the largest angle of the plot. Give your answer in degrees correct to 1 decimal place.

(3)

(b)

Use your unrounded angle to calculate the area of the plot. Give your answer in m² correct to 4 significant figures.

(2)

(c)

A path is laid from the vertex opposite the 38 m side to meet that side at right angles. Calculate the length of this path, using the unrounded area. Give your answer in metres correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 largest angle is opposite the 38 m side	✓ 1
M1 cos x = (24² + 31² − 38²) ÷ (2 × 24 × 31) (= 0.0625)	✓ 1
A1 86.4 (°)	✓ 1
Part (b)
M1 ½ × 24 × 31 × sin(86.4...°)	✓ 1
A1 371.3 (m²)	✓ 1
Part (c)
M1 ½ × 38 × h = area	✓ 1
A1 19.5 (m)	✓ 1

Common error: Using the cosine rule with the wrong angle: the angle found is always opposite the side whose square is subtracted at the start of the numerator.

+ Add to test✓ In your test338IGME3-4-13r10-14Area, surface area and volume Paper HHard8 marks

A marker buoy is a solid made from a hemisphere joined to a cone. The flat circular face of the hemisphere, radius 0.35 m, is joined exactly to the circular base of the cone. The slant height of the cone is 0.85 m. The volume of a cone is $\frac{1}{3}\pi r^2 h$, the volume of a sphere is $\frac{4}{3}\pi r^3$, the curved surface area of a cone is $\pi r l$ and the surface area of a sphere is $4\pi r^2$.

Cross-section through the axis. Diagram not to scale

(a)

Show that the perpendicular height of the cone is 0.775 m correct to 3 significant figures.

(2)

(b)

Calculate the total volume of the buoy. Use the unrounded height and give your answer in m³ correct to 3 significant figures.

(3)

(c)

The outside of the buoy is to be painted. Calculate the total area to be painted, which is the curved surface of the hemisphere and the curved surface of the cone. Give your answer in m² correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 h² = 0.85² − 0.35² (= 0.6)	✓ 1
A1 √0.6 = 0.7746... which rounds to 0.775	✓ 1
Part (b)
M1 hemisphere volume ½ × 4/3 × π × 0.35³ (= 0.0898...)	✓ 1
M1 cone volume 1/3 × π × 0.35² × √0.6 (= 0.0993...) and add	✓ 1
A1 0.189 (m³)	✓ 1
Part (c)
M1 hemisphere curved area ½ × 4 × π × 0.35² (= 0.770...)	✓ 1
M1 cone curved area π × 0.35 × 0.85 (= 0.934...) and add	✓ 1
A1 1.70 (m²)	✓ 1

Common error: Using the vertical height instead of the slant height in the curved surface area of a cone: the formula needs the slant length.

+ Add to test✓ In your test339IGME3-4-13r12-12Area, surface area and volume Paper HHard9 marks

A biogas store is made from a cylinder of radius 4 m and vertical height h metres, with a hemisphere of radius 4 m joined to its top. There is no base on top of the cylinder under the hemisphere. The total volume of the store is 420 m³.

Cross-section. Diagram not to scale

(a)

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

(2)

(b)

Calculate the value of h. Give your answer in metres to 3 significant figures.

(3)

(c)

The outer surface of the store (the curved surface of the cylinder plus the curved surface of the hemisphere, with no base) is to be coated. Calculate the area to be coated. Give your answer in m² to the nearest whole number. Use the unrounded value of h from (b).

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{2}{3}\pi \times 4^3$	✓ 1
A1 for 134 m³ (134.04...)	✓ 1
Part (b)
M1 for cylinder volume $= 420 - 134.04...$ (= 285.96...)	✓ 1
M1 for $\pi \times 4^2 \times h = $ their cylinder volume	✓ 1
A1 for 5.69 m (5.689...)	✓ 1
Part (c)
M1 for curved surface of cylinder $2 \times \pi \times 4 \times h$ (= 142.98...)	✓ 1
M1 for curved surface of hemisphere $2\pi \times 4^2$ (= 100.53...)	✓ 1
A1 for adding the two areas	✓ 1
A1 for 244 m² (243.5...)	✓ 1

Common error: The curved surface of a hemisphere is half of $4\pi r^2$, which is $2\pi r^2$. Do not add a circular base unless the question says the shape is closed.

+ Add to test✓ In your test340IGME3-4-13r12-13Area, surface area and volume Paper HHard8 marks

A paper cup is in the shape of a frustum. It is made by taking a solid cone of height 24 cm and base radius 9 cm and cutting off the top part with a plane parallel to the base, 8 cm above the base. The part removed is a smaller cone. [Volume of a cone = $\frac{1}{3}\pi r^2 h$]

Cross-section through the axis; the cup is the lower part below the dashed cut. Diagram not to scale

(a)

Show that the radius of the top of the cup is 6 cm.

(3)

(b)

Calculate the volume of the cup (the frustum). Give your answer in cm³ to 3 significant figures.

(3)

(c)

The volume of the removed cone is $\frac{8}{27}$ of the volume of the original cone. Show that this is consistent with your answer to (b) by finding the fraction of the original cone's volume that the cup holds. Give the fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for the removed cone has height $24 - 8 = 16$ cm and is similar to the original cone	✓ 1
M1 for $\frac{r}{9} = \frac{16}{24}$ or $9 \times \frac{16}{24}$	✓ 1
A1 for $r = 6$ cm (shown, with the working visible)	✓ 1
Part (b)
M1 for volume of original cone $\frac{1}{3}\pi \times 9^2 \times 24$ (= 648π = 2035.75...)	✓ 1
M1 for volume of removed cone $\frac{1}{3}\pi \times 6^2 \times 16$ (= 192π = 603.19...)	✓ 1
A1 for 1430 cm³ (456π = 1432.6...)	✓ 1
Part (c)
M1 for cup volume as a fraction of the original, e.g. $\frac{456\pi}{648\pi}$ or $1 - \frac{8}{27}$	✓ 1
A1 for $\frac{19}{27}$	✓ 1

Exam tip: If you keep $\pi$ in your working until the last step, the numbers stay exact and the fractions in later parts cancel neatly.

+ Add to test✓ In your test341IGME3-4-12r14-07Similar shapes and circle theorems Paper HHard8 marks

A rover is at point T outside a circular crater with centre O and radius 6 km. The straight lines TA and TB are tangents to the crater rim at A and B. Angle ATB is 56°.

Diagram not to scale

(a)

Give a reason why angle OAT is 90°.

(1)

(b)

Work out the size of angle AOB.

(2)

(c)

Work out the size of angle OAB.

(2)

(d)

Calculate the distance OT. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
The tangent is perpendicular to the radius at the point of contact	✓ 1
Part (b)
360° − 90° − 90° − 56°	✓ 1
124°	✓ 1
Part (c)
Triangle OAB is isosceles with OA = OB: (180° − 124°) ÷ 2 (ECF from (b))	✓ 1
28°	✓ 1
Part (d)
Angle OTA = 28° (OT bisects angle ATB)	✓ 1
sin 28° = 6 ÷ OT	✓ 1
12.8 km	✓ 1

Exam tip: Tangents drawn from the same external point are equal in length, so OT bisects the angle between them.

+ Add to test✓ In your test342IGME3-4-12r14-10Pythagoras' theorem and trigonometry Paper HHard8 marks

Two straight railway lines leave a junction J. Station A is 6.2 km from J along one line and station B is 9.5 km from J along the other. The angle AJB is 68°.

Diagram not to scale

(a)

Calculate the distance AB. Give your answer to 3 significant figures.

(3)

(b)

Calculate the size of angle JBA. Use the unrounded value of AB. Give your answer to 1 decimal place.

(3)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
AB² = 6.2² + 9.5² − 2 × 6.2 × 9.5 × cos 68°	✓ 1
AB² = 84.56…	✓ 1
9.20 km	✓ 1
Part (b)
sin B ÷ 6.2 = sin 68° ÷ AB	✓ 1
sin B = 0.6251… (or any valid method such as the cosine rule)	✓ 1
38.7°	✓ 1
Part (c)
½ × 6.2 × 9.5 × sin 68°	✓ 1
27.3 km²	✓ 1

Common error: In the cosine rule, work out the whole of 2ab cos C before subtracting it from a² + b².

+ Add to test✓ In your test343IGME3-4-12r14-14Area, surface area and volume Paper HHard6 marks

A sand hopper is made by removing a small cone from the top of a large cone, leaving a frustum. The large cone has a base radius of 1.2 m and a height of 3.0 m. The cone that is removed has a height of 1.2 m and is mathematically similar to the large cone.

Cross-section through the axis. Diagram not to scale

(a)

Work out the radius of the cone that is removed.

(2)

(b)

Calculate the volume of the hopper. Give your answer to 3 significant figures. The volume of a cone is ⅓πr²h.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Scale factor 1.2 ÷ 3.0 = 0.4	✓ 1
1.2 × 0.4 = 0.48 m	✓ 1
Part (b)
Large cone: ⅓ × π × 1.2² × 3.0 = 4.524…	✓ 1
Small cone: ⅓ × π × 0.48² × 1.2 = 0.2895…	✓ 1
Subtract the two volumes	✓ 1
4.23 m³	✓ 1

Common error: Subtract the volume of the removed cone from the whole cone; do not use the average of the two radii as if the shape were a cylinder.

+ Add to test✓ In your test344IGME3-4-12r17-12Angles, polygons and circle properties Paper HHard7 marks

ABCD is a cyclic quadrilateral. Angle ABC = (3x + 10)° and angle ADC = (5x − 30)°. Angle BAD = (2x + 10)°. The sides BC and CD are equal in length.

Diagram not to scale

(a)

Work out the value of x.

(3)

(b)

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

(2)

(c)

Work out the size of angle ABD.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (3x + 10) + (5x − 30) = 180 (opposite angles of a cyclic quadrilateral)	✓ 1
M1 8x = 200	✓ 1
A1 x = 25	✓ 1
Part (b)
B1 120° (180 − 60 where angle BAD = 60°)	✓ 1
B1 opposite angles in a cyclic quadrilateral add up to 180°	✓ 1
Part (c)
M1 angle CBD = (180 − 120) ÷ 2 = 30° (isosceles triangle BCD)	✓ 1
A1 85 − 30 = 55°	✓ 1

Common error: In a cyclic quadrilateral it is opposite angles that add to 180°, not adjacent ones. Mark the equal angles of the isosceles triangle on your sketch.

+ Add to test✓ In your test345IGME3-4-12r17-13Area, surface area and volume Paper HHard7 marks

Two freight containers, A and B, are mathematically similar. The surface areas are 5.6 m² for A and 12.6 m² for B. The volume of A is 0.8 m³ and the height of A is 1.2 m.

(a)

Work out the ratio of corresponding lengths, A : B, in the form 1 : n.

(2)

(b)

Work out the volume of B.

(2)

(c)

Work out the height of B.

(1)

(d)

Container B is completely filled with fertiliser of density 640 kg per m³. Work out the mass of fertiliser.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(12.6 ÷ 5.6) = √2.25	✓ 1
A1 1 : 1.5	✓ 1
Part (b)
M1 0.8 × 1.5³ (ECF from (a))	✓ 1
A1 2.7 m³	✓ 1
Part (c)
B1 1.8 m (ECF from (a))	✓ 1
Part (d)
M1 2.7 × 640	✓ 1
A1 1728 kg (ECF from (b))	✓ 1

Exam tip: Area ratios need a square root to give the length ratio; volume ratios need a cube root. Decide which you have before you calculate.

+ Add to test✓ In your test346IGME3-4-12r17-14Area, surface area and volume Paper HHard7 marks

A marine navigation buoy is made from a solid hemisphere of radius 0.6 m joined to a solid cone. The flat circular face of the hemisphere is exactly joined to the base of the cone, which also has radius 0.6 m. The cone has perpendicular height h metres. The total volume of the buoy is 1.2 m³.
[Volume of sphere = (4/3)πr³; volume of cone = (1/3)πr²h; curved surface area of cone = πrl, where l is the slant height; surface area of sphere = 4πr²]

Cross-section. Diagram not to scale

(a)

Work out h. Give your answer correct to 3 significant figures.

(3)

(b)

Work out the total outer curved surface area of the buoy (the curved surface of the hemisphere and the curved surface of the cone). Use the unrounded value of h and give your answer correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume of hemisphere (2/3)π(0.6)³ (= 0.144π)	✓ 1
M1 (2/3)π(0.6)³ + (1/3)π(0.6)²h = 1.2	✓ 1
A1 h = 1.98 m	✓ 1
Part (b)
M1 slant height l = √(0.6² + 1.983...²) = 2.071...	✓ 1
M1 curved area of hemisphere 2π(0.6)² (= 2.261...)	✓ 1
M1 curved area of cone π × 0.6 × 2.071...	✓ 1
A1 6.17 m²	✓ 1

Common error: Do not include the flat circular face of the hemisphere: it is joined to the cone, so it is not on the outside of the buoy.

+ Add to test✓ In your test347IGME3-4-13r18-12Angles, polygons and circle properties Paper HHard9 marks

A, B, C and D lie on the circumference of a circle, in that order, so that ABCD is a cyclic quadrilateral. Angle ABC = (3x + 12)° and angle ADC = (5x − 8)°. AD = DC. Angle CAB = 35°.

Diagram not to scale

(a)

Work out the value of x.

(3)

(b)

Write down the size of angle ABC.

(1)

(c)

Work out the size of angle ACB.

(2)

(d)

Work out the size of angle BCD.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 opposite angles of a cyclic quadrilateral sum to 180°: 3x + 12 + 5x − 8 = 180	✓ 1
M1 8x + 4 = 180	✓ 1
A1 x = 22	✓ 1
Part (b)
A1 78°	✓ 1
Part (c)
M1 180 − 78 − 35 (ECF)	✓ 1
A1 67°	✓ 1
Part (d)
M1 angle ADC = 102°, so angle DCA = (180 − 102) ÷ 2	✓ 1
M1 angle BCD = 39 + 67 (ECF)	✓ 1
A1 106°	✓ 1

Common error: Opposite angles of a cyclic quadrilateral sum to 180°; adjacent angles do not.

+ Add to test✓ In your test348IGME3-4-13r18-13Area, surface area and volume Paper HHard7 marks

Two bottles of cough syrup are mathematically similar. The total outer surface areas are 240 cm² (smaller) and 540 cm² (larger). The larger bottle holds 1350 ml and is 21 cm tall.

(a)

Work out the length scale factor from the smaller bottle to the larger bottle.

(2)

(b)

Work out the capacity of the smaller bottle.

(3)

(c)

Work out the height of the smaller bottle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(540 ÷ 240) (or √2.25)	✓ 1
A1 1.5	✓ 1
Part (b)
M1 volume scale factor 1.5³ = 3.375 (ECF from (a))	✓ 1
M1 1350 ÷ 3.375	✓ 1
A1 400 ml	✓ 1
Part (c)
M1 21 ÷ 1.5 (ECF from (a))	✓ 1
A1 14 cm	✓ 1

Exam tip: From areas, take a square root to get the length scale factor; cube it for volumes.

+ Add to test✓ In your test349IGME3-4-13r18-14Pythagoras' theorem and trigonometry Paper HHard8 marks

A rectangular water tank is a cuboid ABCDEFGH. The base ABCD measures 8.0 m by 6.0 m with AB = 8.0 m, and the tank is 5.0 m high. AC is a diagonal of the base and G is the top corner directly above C. A sensor cable runs in a straight line from A to G.

Diagram not to scale

(a)

Work out the length of AC.

(2)

(b)

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

(3)

(c)

Work out the angle between AG and the base of the tank. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 8.0² + 6.0² = 100	✓ 1
A1 10 m	✓ 1
Part (b)
M1 AG² = AC² + CG²	✓ 1
M1 √(10² + 5.0²) = √125 (ECF)	✓ 1
A1 11.2 m	✓ 1
Part (c)
M1 tan x = 5.0 ÷ 10 (ECF)	✓ 1
M1 x = tan⁻¹(0.5)	✓ 1
A1 26.6°	✓ 1

Exam tip: For 3D problems, pick out a right-angled triangle and redraw it flat before using Pythagoras or trigonometry.

+ Add to test✓ In your test350IGME3-4-13r19-10Similar shapes and circle theorems Paper HHard6 marks

O is the centre of a circular enclosure of radius 7.5 m. An observer stands at point P, 12.5 m from O. A straight line from P just touches the circle at T, so that PT is a tangent.

Diagram not to scale

(a)

Write down the size of angle OTP, and give a reason.

(1)

(b)

Calculate the length of PT.

(3)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
C1 90° because a tangent is perpendicular to the radius at the point of contact	✓ 1
Part (b)
M1 PT² = 12.5² − 7.5²	✓ 1
M1 √(156.25 − 56.25) = √100	✓ 1
A1 10 m	✓ 1
Part (c)
M1 cos TOP = 7.5 ÷ 12.5	✓ 1
A1 53.1°	✓ 1

Common error: Treating triangle OTP as if the right angle were at O; the right angle is at T, so OP (12.5 m) is the hypotenuse.

+ Add to test✓ In your test351IGME3-4-13r19-11Pythagoras' theorem and trigonometry Paper HHard5 marks

In a data centre, three cooling units are at points A, B and C on a flat floor. AB = 14 m, AC = 9 m and angle BAC = 62°.

Diagram not to scale

(a)

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

(3)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 BC² = 14² + 9² − 2 × 14 × 9 × cos 62°	✓ 1
M1 √158.69...	✓ 1
A1 12.6 m	✓ 1
Part (b)
M1 ½ × 14 × 9 × sin 62°	✓ 1
A1 55.6 m²	✓ 1

Exam tip: Make sure your calculator is in degree mode before using sin or cos with angles given in degrees.

+ Add to test✓ In your test352IGME3-4-13r19-12Similar shapes and circle theorems Paper HHard5 marks

Two cylindrical cooling tanks, P and Q, are mathematically similar. The surface area of P is 36 m² and the surface area of Q is 81 m². The volume of P is 48 m³.

(a)

Work out the ratio of corresponding lengths of P to Q. Give your answer in the form 2 : n.

(2)

(b)

Calculate the volume of Q.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(81 ÷ 36) or 9 : 6	✓ 1
A1 2 : 3	✓ 1
Part (b)
M1 volume scale factor (3 ÷ 2)³	✓ 1
M1 48 × 27 ÷ 8	✓ 1
A1 162 m³	✓ 1

Common error: Cubing the area ratio instead of first taking its square root to get the length ratio.

+ Add to test✓ In your test353IGME3-4-11r20-11Similar shapes and circle theorems Paper HHard9 marks

A concert hall has a curved sound reflector. Its cross-section is a sector OAB of a circle with centre O, radius 12 m and angle AOB = 150°. The chord AB cuts off a segment of the circle.

Diagram not to scale

(a)

Calculate the length of the arc AB. Give your answer in m correct to 1 decimal place.

(2)

(b)

Calculate the area of the segment between the arc AB and the chord AB. Give your answer in m² correct to 1 decimal place.

(4)

(c)

Calculate the perimeter of the segment. Give your answer in m correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\dfrac{150}{360}\times2\pi\times12$	✓ 1
A1 31.4 m	✓ 1
Part (b)
M1 sector area $\dfrac{150}{360}\times\pi\times12^2=188.4\ldots$	✓ 1
M1 triangle area $\tfrac12\times12\times12\times\sin150^\circ=36$	✓ 1
M1 sector minus triangle	✓ 1
A1 152.5 m²	✓ 1
Part (c)
M1 chord $AB^2=12^2+12^2-2\times12\times12\cos150^\circ$ or $2\times12\sin75^\circ$	✓ 1
A1 $AB=23.2\ldots$ m	✓ 1
A1 $31.4\ldots+23.2\ldots=54.6$ m	✓ 1

Exam tip: Segment area = sector area minus triangle area; use $\tfrac12ab\sin C$ for the triangle when the angle at O is not 90°.

+ Add to test✓ In your test354IGME3-4-11r20-12Area, surface area and volume Paper HHard9 marks

A grain hopper is made from a solid cone of height 30 cm and base radius 12 cm. The top part, a smaller cone of height 18 cm that is similar to the original cone, is removed by a cut parallel to the base, leaving a frustum.

Cross-section. Diagram not to scale

(a)

Show that the radius of the circular cut face is 7.2 cm.

(2)

(b)

Calculate the volume of the frustum. Give your answer in cm³ correct to 3 significant figures.

(4)

(c)

The frustum is filled with grain of density 0.75 g/cm³. Calculate the mass of the grain in kg. Use the unrounded volume and give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 scale factor $\tfrac{18}{30}=0.6$	✓ 1
A1 $12\times0.6=7.2$ cm	✓ 1
Part (b)
M1 volume of large cone $\tfrac13\pi\times12^2\times30=1440\pi$	✓ 1
M1 volume of small cone $\tfrac13\pi\times7.2^2\times18=311.04\pi$	✓ 1
M1 subtract	✓ 1
A1 3550 cm³	✓ 1
Part (c)
M1 mass in g $=0.75\times3546.7\ldots$	✓ 1
M1 divide by 1000 to convert to kg	✓ 1
A1 2.66 kg	✓ 1

Common error: Remember that volume, not length, scales by k³; here it is safest to subtract the two cone volumes directly.

+ Add to test✓ In your test355IGME3-4-11r20-13Pythagoras' theorem and trigonometry Paper HHard7 marks

A music studio is a cuboid room 8 m long, 6 m wide and 3 m high. A cable is run in a straight line from one bottom corner of the room to the opposite top corner.

Diagram not to scale

(a)

Calculate the length of the diagonal of the floor.

(2)

(b)

Calculate the length of the cable. Give your answer in m correct to 3 significant figures.

(2)

(c)

Calculate the angle the cable makes with the floor. Give your answer in degrees correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\sqrt{8^2+6^2}$	✓ 1
A1 10 m	✓ 1
Part (b)
M1 $\sqrt{10^2+3^2}$ (or $\sqrt{8^2+6^2+3^2}$)	✓ 1
A1 10.4 m	✓ 1
Part (c)
M1 right-angled triangle with the floor diagonal and the height as sides	✓ 1
M1 $\tan\theta=\dfrac{3}{10}$	✓ 1
A1 $16.7^\circ$	✓ 1

Exam tip: For a 3D problem, draw the right-angled triangle you are using as a separate flat sketch and label the sides.

+ Add to test✓ In your test356IGME3-4-12r21-07Angles, polygons and circle properties Paper HHard7 marks

P, Q and R are points on the circumference of a circle with centre O, where R lies on the major arc PQ. Angle POQ = (6x − 20)° and angle PRQ = (2x + 5)°. The line PT is the tangent to the circle at P, and T and R lie on opposite sides of line PQ.

Diagram not to scale

(a)

Work out the value of x. Give a reason for your first step.

(3)

(b)

Work out the size of angle OPQ.

(2)

(c)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6x − 20 = 2(2x + 5) because the angle at the centre is twice the angle at the circumference	✓ 1
M1 6x − 20 = 4x + 10, so 2x = 30	✓ 1
A1 x = 15	✓ 1
Part (b)
M1 angle POQ = 70° and triangle OPQ is isosceles (OP = OQ), (180 − 70) ÷ 2	✓ 1
A1 55°	✓ 1
Part (c)
M1 90 − 55 (tangent is perpendicular to the radius OP) or angle in the alternate segment = angle PRQ	✓ 1
A1 35°	✓ 1

Common error: The angle at the centre is twice the angle at the circumference only when both angles stand on the same arc; check the arc before doubling.

+ Add to test✓ In your test357IGME3-4-12r21-11Area, surface area and volume Paper HHard6 marks

A water tower tank is made from a cylinder of radius 2.5 m and height h metres, with a solid hemisphere of radius 2.5 m on top. The total volume of the tank is 100 m³.

Cross-section. Diagram not to scale

(a)

Calculate the volume of the hemisphere. Give your answer correct to 3 significant figures.

(2)

(b)

Work out the value of h. Use the unrounded volume of the hemisphere and give your answer correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2/3 × π × 2.5³	✓ 1
A1 32.7 m³	✓ 1
Part (b)
M1 volume of cylinder = π × 2.5² × h	✓ 1
M1 π × 2.5² × h + 32.72... = 100	✓ 1
M1 h = (100 − 32.72...) ÷ (π × 2.5²)	✓ 1
A1 3.43 m	✓ 1

Common error: The volume of a hemisphere is half of 4/3 πr³; forgetting to halve the sphere formula doubles that part of the volume.

+ Add to test✓ In your test358IGME3-4-13r22-06Pythagoras' theorem and trigonometry Paper HHard8 marks

Calculator

A mountain tent is a square-based pyramid. The square base ABCD has sides of length 8 m on level ground. The vertex V is vertically above the centre M of the base, and VM = 6 m. Give lengths to 3 significant figures and angles to 1 decimal place.

Diagram not to scale

(a)

Work out the length of the sloping edge VA.

(3)

(b)

Work out the angle between VA and the base.

(3)

(c)

Work out the angle between the sloping face VAB and the base.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 AM = half of the diagonal = √(8² + 8²) ÷ 2 = √32	✓ 1
M1 VA² = 32 + 6²	✓ 1
A1 8.25 m (√68 = 8.246...)	✓ 1
Part (b)
M1 angle VAM is the required angle	✓ 1
M1 tan (VAM) = 6 ÷ √32	✓ 1
A1 46.7° (46.69...)	✓ 1
Part (c)
M1 tan (angle) = 6 ÷ 4, using the midpoint of AB	✓ 1
A1 56.3° (56.31...)	✓ 1

Exam tip: In 3D problems, draw the right-angled triangle you need as a flat picture and mark the lengths you know.

+ Add to test✓ In your test359IGME3-4-13r22-12Pythagoras' theorem and trigonometry Paper HHard8 marks

Calculator

Three mountain peaks A, B and C form a triangle on a map. AB = 7.2 km, AC = 5.4 km and angle BAC = 62°. Give lengths and areas to 3 significant figures and angles to 1 decimal place.

Diagram not to scale

(a)

Work out the distance BC.

(3)

(b)

Work out the area of triangle ABC.

(2)

(c)

Work out the size of angle ABC. Use the unrounded value of BC.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 BC² = 7.2² + 5.4² − 2 × 7.2 × 5.4 × cos 62°	✓ 1
M1 BC² = 44.49...	✓ 1
A1 6.67 km (6.670...)	✓ 1
Part (b)
M1 ½ × 7.2 × 5.4 × sin 62°	✓ 1
A1 17.2 km² (17.16...)	✓ 1
Part (c)
M1 sin B ÷ 5.4 = sin 62° ÷ BC	✓ 1
M1 sin B = 5.4 × sin 62° ÷ 6.670... (= 0.7148...)	✓ 1
A1 45.6° (45.62...)	✓ 1

Exam tip: Use the cosine rule when you know two sides and the angle between them; use the sine rule when you have a side and its opposite angle.

+ Add to test✓ In your test360IGME3-4-13r22-14Similar shapes and circle theorems Paper HHard7 marks

Calculator

A telecoms engineer models the coverage of a rotating antenna as a circle with centre O and radius 5 cm on a plan. PT is a tangent to the circle at P. Q is a point on the circle such that angle TPQ = 38°. R is a point on the major arc PQ (the longer arc).

Diagram not to scale

(a)

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

(2)

(b)

Work out the size of angle POQ.

(2)

(c)

Work out the length of the chord PQ. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 38°	✓ 1
B1 angle between tangent and chord equals the angle in the alternate segment	✓ 1
Part (b)
M1 angle at the centre is twice the angle at the circumference	✓ 1
A1 76° (ECF from (a))	✓ 1
Part (c)
M1 triangle OPQ is isosceles; split it or use cosine/sine rule with OP = OQ = 5	✓ 1
M1 PQ = 2 × 5 × sin 38° or √(5² + 5² − 2 × 5 × 5 × cos 76°)	✓ 1
A1 6.16 cm (6.157...)	✓ 1

Exam tip: Name the circle theorem you use each time; the alternate segment theorem links a tangent-chord angle to an angle on the circumference.

+ Add to test✓ In your test361IGME3-4-11r24-12Area, surface area and volume Paper HHard11 marks

The re-entry capsule of a probe is made from a hemisphere of radius 0.9 m with a cone of the same base radius fixed on top. The cone has vertical height 1.2 m. The flat circular face of the cone coincides with the flat face of the hemisphere.
Volume of a cone = ⅓πr²h. Curved surface area of a cone = πrl, where l is the slant height. Volume of a sphere = (4/3)πr³. Surface area of a sphere = 4πr².

Cross-section. Diagram not to scale

(a)

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

(4)

(b)

Work out the slant height of the cone.

(2)

(c)

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

(3)

(d)

One litre of heat-resistant paint covers 6.5 m². Work out the number of litres needed for one coat on the curved surface. Use the unrounded area from part (c). Give your answer correct to 2 decimal places.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 hemisphere volume ½ × (4/3) × π × 0.9³ (= 0.486π)	✓ 1
M1 cone volume ⅓ × π × 0.9² × 1.2 (= 0.324π)	✓ 1
M1 addition of the two volumes (= 0.81π)	✓ 1
A1 2.54 m³	✓ 1
Part (b)
M1 √(0.9² + 1.2²)	✓ 1
A1 1.5 m	✓ 1
Part (c)
M1 hemisphere curved area ½ × 4 × π × 0.9²	✓ 1
M1 cone curved area π × 0.9 × 1.5 and addition	✓ 1
A1 9.33 m²	✓ 1
Part (d)
M1 9.3305... ÷ 6.5	✓ 1
A1 1.44 litres	✓ 1

Common error: The flat face where the shapes join is not part of the outer surface, so do not include it in the surface area.

+ Add to test✓ In your test362IGME3-4-11r24-13Area, surface area and volume Paper HHard8 marks

A rectangular test panel for a probe has length (w + 5) cm and width w cm. Its diagonal is 25 cm.

Diagram not to scale

(a)

Show that w² + 5w − 300 = 0.

(3)

(b)

Solve w² + 5w − 300 = 0 to find the width of the panel.

(3)

(c)

Work out the area of the panel.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 w² + (w + 5)² = 25²	✓ 1
M1 w² + w² + 10w + 25 = 625	✓ 1
A1 2w² + 10w − 600 = 0, divide by 2 to give w² + 5w − 300 = 0	✓ 1
Part (b)
M1 (w + 20)(w − 15) = 0 or correct use of the formula	✓ 1
A1 w = 15 or w = −20	✓ 1
A1 width = 15 cm (reject −20 as a length cannot be negative)	✓ 1
Part (c)
M1 15 × 20	✓ 1
A1 300 cm² (ECF from (b))	✓ 1

Exam tip: After solving a quadratic from a geometry problem, reject any solution that gives a negative length.

+ Add to test✓ In your test363IGME3-4-14r29-12Area, surface area and volume Paper HHard6 marks

A solid cone has radius 5 cm and perpendicular height 12 cm. A smaller cone of height 4 cm is cut from the top by a plane parallel to the base. The part that remains is a frustum.

Cross-section. Diagram not to scale

(a)

Work out the radius of the smaller cone that was removed.

(2)

(b)

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

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\tfrac{4}{12}\times5$ (similar cones)	✓ 1
A1 $\tfrac{5}{3}$ cm (or 1.67 cm)	✓ 1
Part (b)
M1 volume of large cone $\tfrac{1}{3}\pi\times5^{2}\times12=100\pi$	✓ 1
M1 volume of small cone $\tfrac{1}{3}\pi\left(\tfrac{5}{3}\right)^{2}\times4=\tfrac{100\pi}{27}$ (or ratio $\tfrac{1}{27}$)	✓ 1
M1 subtract: $100\pi-\tfrac{100\pi}{27}$	✓ 1
A1 303 cm$^{3}$	✓ 1

Exam tip: The volume of a cone is $\frac{1}{3}\pi r^{2}h$; a frustum's volume is the large cone minus the small cone.

+ Add to test✓ In your test364IGME3-4-14r29-13Pythagoras' theorem and trigonometry Paper HHard7 marks

A theatre rehearsal room is a cuboid 12 m long, 9 m wide and 5 m high. A cable runs in a straight line from one corner of the floor to the opposite corner of the ceiling.

Diagram not to scale

(a)

Work out the length of the diagonal of the floor.

(2)

(b)

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

(2)

(c)

Work out the angle the cable makes with the floor. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\sqrt{12^{2}+9^{2}}$	✓ 1
A1 15 m	✓ 1
Part (b)
M1 $\sqrt{15^{2}+5^{2}}$ (ECF from (a))	✓ 1
A1 15.8 m	✓ 1
Part (c)
M1 $\tan\theta=\dfrac{5}{15}$	✓ 1
M1 $\theta=\tan^{-1}\!\left(\dfrac{5}{15}\right)$	✓ 1
A1 $18.4^{\circ}$	✓ 1

Common error: The angle with the floor uses the floor diagonal (15 m) as the adjacent side, not an edge of the room.

+ Add to test✓ In your test365IGME3-4-14r29-14Angles, polygons and circle properties Paper HHard8 marks

A ship leaves harbour H and sails 18 km on a bearing of $060^{\circ}$ to a buoy B. It then sails 25 km on a bearing of $150^{\circ}$ to a point C.

Diagram not to scale

(a)

Show that angle HBC = $90^{\circ}$.

(3)

(b)

Work out the distance HC. Give your answer correct to 3 significant figures.

(2)

(c)

Work out the bearing of C from H. Give your answer to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of H from B is $060+180=240^{\circ}$ (or co-interior angles $180-60=120^{\circ}$)	✓ 1
M1 angle between the directions $240-150$ (or $120-30$)	✓ 1
A1 $90^{\circ}$	✓ 1
Part (b)
M1 $HC^{2}=18^{2}+25^{2}=949$	✓ 1
A1 30.8 km	✓ 1
Part (c)
M1 $\tan(\angle BHC)=\dfrac{25}{18}$	✓ 1
M1 $\angle BHC=54.2^{\circ}$ and add to 060	✓ 1
A1 $114^{\circ}$	✓ 1

Exam tip: Draw a north line at each point and use parallel lines to find the angles between the legs.

+ Add to test✓ In your test366IGME3-4-13r32-13Angles, polygons and circle properties Paper HHard8 marks

Four wind turbines A, B, C and D stand on a circular track with centre O, in that order round the track. Angle BAD = (3x + 10)° and angle BCD = (5x − 30)°. A straight road touches the track at B.

Diagram not to scale

(a)

Work out the value of x. Give a reason for your first step.

(3)

(b)

Work out the size of angle BOD, the angle at O that faces the arc BCD (the arc that does not contain A). Give a reason.

(2)

(c)

The angle between the road and the chord BA, measured on the side of the line AB that does not contain C and D, is 40°. Work out the size of angle ABD. Give reasons for your working.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 (3x + 10) + (5x − 30) = 180 (opposite angles of a cyclic quadrilateral add up to 180°)	✓ 1
M1 8x − 20 = 180	✓ 1
A1 x = 25	✓ 1
Part (b)
M1 angle BAD = 3 × 25 + 10 = 85°	✓ 1
A1 170° (angle at the centre is twice the angle at the circumference, 2 × 85)	✓ 1
Part (c)
B1 angle ADB = 40° (angle between tangent and chord equals the angle in the alternate segment)	✓ 1
M1 180 − 85 − 40	✓ 1
A1 55°	✓ 1

Exam tip: Write each circle-theorem reason by name, such as 'alternate segment theorem', and give the angle value it produces.

+ Add to test✓ In your test367IGME3-4-13r32-14Pythagoras' theorem and trigonometry Paper HHard10 marks

A cold-storage box for vaccines is a cuboid with a rectangular base 60 cm by 45 cm and height 36 cm. A thin temperature probe is placed along the line from one bottom corner A to the opposite top corner G, passing through the inside of the box.

Diagram not to scale

(a)

Work out the length of the probe AG. Give your answer to 3 significant figures.

(3)

(b)

Work out the angle between AG and the base of the box. Give your answer to 1 decimal place.

(3)

(c)

A larger box is mathematically similar to this box and has volume 328 050 cm³. Work out the length of the longest straight probe that fits inside the larger box. Give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 60² + 45² + 36² (= 6921) or diagonal of the base = 75	✓ 1
M1 square root of their sum	✓ 1
A1 83.2 cm (83.19 seen)	✓ 1
Part (b)
M1 tan x = 36 ÷ 75 (using the base diagonal 75)	✓ 1
M1 inverse tangent of their ratio	✓ 1
A1 25.6° (25.64 seen)	✓ 1
Part (c)
M1 volume of the small box = 60 × 45 × 36 = 97 200 cm³	✓ 1
M1 328 050 ÷ 97 200 = 3.375	✓ 1
M1 cube root = 1.5 (length scale factor)	✓ 1
A1 125 cm (124.79 seen, using unrounded AG; ECF from (a))	✓ 1

Common error: For similar solids, find the length scale factor by taking the cube root of the volume ratio; do not use the volume ratio itself as the multiplier for lengths.

+ Add to test✓ In your test368IGME3-4-12r36-09Angles, polygons and circle properties Paper HHard9 marks

A ship leaves port P and sails 18 km on a bearing of 070° to a point A. It then sails 25 km on a bearing of 150° to a point B.

Diagram not to scale

(a)

Show that angle PAB = 100°.

(2)

(b)

Calculate the distance PB. Give your answer to 3 significant figures.

(3)

(c)

Calculate the bearing of B from P. Use the unrounded value of PB. Give your answer to the nearest degree.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 the bearing of P from A is 250° (070° + 180°, using alternate or co-interior angles)	✓ 1
A1 250° − 150° = 100°	✓ 1
Part (b)
M1 PB² = 18² + 25² − 2 × 18 × 25 × cos 100°	✓ 1
M1 PB² = 1105.28... correctly evaluated	✓ 1
A1 33.2 km	✓ 1
Part (c)
M1 sin APB ÷ 25 = sin 100° ÷ 33.2458...	✓ 1
M1 angle APB = 47.8° (47.78...)	✓ 1
M1 070° + 47.8°	✓ 1
A1 118°	✓ 1

Exam tip: Draw the north line at every turning point and mark each bearing on it before you decide which angle the cosine or sine rule needs.

+ Add to test✓ In your test369IGME3-4-12r36-13Pythagoras' theorem and trigonometry Paper HHard10 marks

A circular crater lake has radius 130 m. A straight bridge crosses the lake as a chord of length 200 m.

Diagram not to scale

(a)

Calculate the shortest distance from the centre of the lake to the bridge. Give your answer to 3 significant figures.

(3)

(b)

Calculate the angle that the bridge subtends at the centre of the lake. Give your answer to the nearest degree.

(3)

(c)

Calculate the area of the segment of the lake between the bridge and the minor arc. Use unrounded values from earlier parts. Give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 half the chord = 100 m	✓ 1
M1 130² − 100² (= 6900)	✓ 1
A1 83.1 m	✓ 1
Part (b)
M1 sin x = 100 ÷ 130 (or cos x = 83.07... ÷ 130)	✓ 1
M1 2 × 50.28...°	✓ 1
A1 101°	✓ 1
Part (c)
M1 sector area = (100.57.../360) × π × 130² (= 14 832...)	✓ 1
M1 triangle area = ½ × 200 × 83.07... (= 8306...)	✓ 1
M1 sector area − triangle area	✓ 1
A1 6530 m²	✓ 1

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

+ Add to test✓ In your test370IGME3-4-12r36-14Similar shapes and circle theorems Paper HHard5 marks

Two mathematically similar shipping crates have outside surface areas of 180 cm² and 405 cm². The volume of the larger crate is 2700 cm³.

(a)

Find the ratio of the lengths of the smaller crate to the larger crate in the form 1 : n.

(2)

(b)

Calculate the volume of the smaller crate.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(405 ÷ 180) = √2.25	✓ 1
A1 1 : 1.5	✓ 1
Part (b)
M1 volume ratio = 1 : 1.5³ (= 1 : 3.375)	✓ 1
M1 2700 ÷ 3.375 (ECF from (a))	✓ 1
A1 800 cm³	✓ 1

Common error: Take the square root of the area ratio to get the length ratio before cubing for volume.

+ Add to test✓ In your test371IGME3-4-13r36-07Angles, polygons and circle properties Paper HHard5 marks

A, B, C and D are four points on a circle with centre O, in that order round the circle, so that ABCD is a cyclic quadrilateral. Angle ABC = (2x + 10)° and angle ADC = (3x − 25)°.

Diagram not to scale

(a)

Show that x = 39. Give a reason for your working.

(3)

(b)

The angle at the centre subtended by the arc AC that does not contain B is angle AOC. Work out the size of this angle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (2x + 10) + (3x − 25) = 180	✓ 1
R1 Opposite angles of a cyclic quadrilateral add up to 180°	✓ 1
A1 5x − 15 = 180 so x = 39	✓ 1
Part (b)
M1 angle at the centre = 2 × angle at the circumference, with angle ABC = 88°	✓ 1
A1 176°	✓ 1

Common error: In a cyclic quadrilateral it is the opposite angles that add to 180°, not adjacent ones.

+ Add to test✓ In your test372IGME3-4-13r36-11Pythagoras' theorem and trigonometry Paper HHard5 marks

A triangular cycle track ABC is on level ground. AB = 14 m, AC = 9 m and angle BAC = 55°.

Diagram not to scale

(a)

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

(3)

(b)

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

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 BC² = 14² + 9² − 2 × 14 × 9 × cos 55°	✓ 1
M1 BC² = 132.459…	✓ 1
A1 11.5 m	✓ 1
Part (b)
M1 ½ × 14 × 9 × sin 55°	✓ 1
A1 51.6 m²	✓ 1

Common error: In the cosine rule, work out the whole product 2bc cos A before subtracting it from the sum of the squares.

+ Add to test✓ In your test373IGME3-4-13r36-14Pythagoras' theorem and trigonometry Paper HHard7 marks

A square-based pyramid has a horizontal base of side 230 m. Its apex is vertically above the centre of the base at a height of 146 m.

Diagram not to scale

(a)

Show that the distance from the centre of the base to a corner of the base is 162.6 m, correct to 4 significant figures.

(2)

(b)

Calculate the length of a sloping edge from the apex to a corner of the base. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate the angle between a sloping edge and the base. Use unrounded values and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 diagonal² = 230² + 230² (= 105 800), or half-diagonal² = 26 450	✓ 1
A1 √26 450 = 162.6… m	✓ 1
Part (b)
M1 146² + 26 450 (= 47 766)	✓ 1
M1 square root of 47 766	✓ 1
A1 219 m	✓ 1
Part (c)
M1 tan θ = 146 ÷ 162.63…	✓ 1
A1 41.9°	✓ 1

Common error: In three-dimensional problems, draw the right-angled triangle you are using. A common slip is using the full diagonal or the base side instead of half the diagonal.

+ Add to test✓ In your test374IGME3-4-14r37-07Area, surface area and volume Paper HHard6 marks

Two swimming pools, A and B, are mathematically similar. Pool A has a volume of 60 m³ and a water-surface area of 50 m². Pool B has a volume of 202.5 m³.

(a)

Work out the scale factor of the lengths from pool A to pool B.

(2)

(b)

Work out the water-surface area of pool B.

(2)

(c)

The surface of pool B is covered with a safety sheet costing £42 per m². Work out the cost of the sheet.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ∛(202.5 ÷ 60) or ∛3.375	✓ 1
A1 1.5	✓ 1
Part (b)
M1 50 × (their 1.5)²	✓ 1
A1 112.5 m²	✓ 1
Part (c)
M1 their 112.5 × 42	✓ 1
A1 £4725	✓ 1

Exam tip: For similar shapes, convert between a volume ratio and a length scale factor with a cube root, and between a length scale factor and an area ratio with a square.

+ Add to test✓ In your test375IGME3-4-14r37-09Pythagoras' theorem and trigonometry Paper HHard8 marks

Three wind turbines A, B and C stand on level ground. AB = 450 m, AC = 380 m and angle BAC = 62°.

Diagram not to scale

(a)

Work out the distance BC. Give your answer correct to 3 significant figures.

(3)

(b)

Work out the size of angle ABC. Use the unrounded value of BC and give your answer correct to 1 decimal place.

(3)

(c)

Work out the area of triangle ABC in hectares (1 hectare = 10 000 m²). Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 BC² = 450² + 380² − 2 × 450 × 380 × cos 62°	✓ 1
M1 √(their 186 340...)	✓ 1
A1 432 m	✓ 1
Part (b)
M1 sin B/380 = sin 62°/BC	✓ 1
M1 B = sin⁻¹(380 sin 62° ÷ 431.67...)	✓ 1
A1 51.0°	✓ 1
Part (c)
M1 ½ × 450 × 380 × sin 62° (= 75 492...)	✓ 1
A1 7.55 hectares	✓ 1

Exam tip: Use the cosine rule when two sides and the included angle are known, and the sine rule once a side and its opposite angle are known.

+ Add to test✓ In your test376IGME3-4-14r37-10Pythagoras' theorem and trigonometry Paper HHard6 marks

A library storeroom is a cuboid. The floor measures 6 m by 4.5 m and the room is 3 m high. A cable runs in a straight line from one corner of the floor to the opposite corner of the ceiling.

Diagram not to scale

(a)

Work out the length of the diagonal of the floor.

(2)

(b)

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

(2)

(c)

Work out the angle between the cable and the floor. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6² + 4.5² (= 56.25)	✓ 1
A1 7.5 m	✓ 1
Part (b)
M1 √(7.5² + 3²) or √(65.25)	✓ 1
A1 8.08 m	✓ 1
Part (c)
M1 tan⁻¹(3 ÷ 7.5) or sin⁻¹(3 ÷ their 8.0777...)	✓ 1
A1 21.8°	✓ 1

Common error: In three dimensions, find a diagonal across the floor first, then use it as one side of a second right-angled triangle that includes the height.

+ Add to test✓ In your test377IGME3-4-14r44-12Pythagoras' theorem and trigonometry Paper HHard8 marks

A triangular plot of land PQR has PQ = 85 m and PR = 62 m. The angle QPR is 48°.

(a)

Work out the length of QR. Give your answer to 1 decimal place.

(3)

(b)

Work out the size of angle PQR. Use the unrounded value of QR and give your answer to 1 decimal place.

(3)

(c)

Work out the area of the plot. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 QR² = 85² + 62² − 2 × 85 × 62 × cos 48°	✓ 1
M1 QR² = 4016.4… and square root	✓ 1
A1 63.4 m	✓ 1
Part (b)
M1 sin Q ÷ 62 = sin 48° ÷ QR (QR = 63.37…)	✓ 1
M1 sin⁻¹(62 × sin 48° ÷ 63.37…)	✓ 1
A1 46.6°	✓ 1
Part (c)
M1 ½ × 85 × 62 × sin 48°	✓ 1
A1 1958.2 m²	✓ 1

Exam tip: Use the cosine rule when you know two sides and the angle between them. Once you have the third side, the sine rule gives the other angles.

+ Add to test✓ In your test378IGME3-4-14r44-13Area, surface area and volume Paper HHard8 marks

A marker buoy is made from a hemisphere of radius 0.6 m joined to a cone of base radius 0.6 m and vertical height 1.6 m. The flat face of the hemisphere is joined to the base of the cone.
Volume of a sphere = ⁴/₃πr³; volume of a cone = ¹/₃πr²h; curved surface area of a cone = πrl where l is the slant height.

Cross-section. Diagram not to scale

(a)

Work out the total volume of the buoy. Give your answer in m³ to 3 significant figures.

(3)

(b)

Work out the slant height of the cone. Give your answer in metres to 3 significant figures.

(2)

(c)

Work out the total area of the curved surfaces of the buoy. Use the unrounded slant height and give your answer in m² to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 hemisphere ½ × ⁴/₃ × π × 0.6³ (= 0.144π)	✓ 1
M1 cone ¹/₃ × π × 0.6² × 1.6 (= 0.192π)	✓ 1
A1 1.06 m³	✓ 1
Part (b)
M1 √(0.6² + 1.6²)	✓ 1
A1 1.71 m	✓ 1
Part (c)
M1 hemisphere curved surface 2π × 0.6²	✓ 1
M1 cone curved surface π × 0.6 × 1.7088…	✓ 1
A1 5.48 m²	✓ 1

Common error: The curved surface of a cone uses the slant height, not the vertical height. Find the slant height with Pythagoras’ theorem first.

+ Add to test✓ In your test379IGME3-4-14r44-14Area, surface area and volume Paper HHard8 marks

A conical sample cup is made from a sector of a circle of radius 12 cm. The angle of the sector is 150°. The straight edges of the sector are joined together, without overlap, to form the cone, so the radius of the sector becomes the slant height of the cone.

Diagram not to scale

(a)

Show that the arc length of the sector is 10π cm.

(2)

(b)

Work out the base radius of the cone.

(2)

(c)

Work out the vertical height of the cone. Give your answer in cm to 3 significant figures.

(2)

(d)

Work out the volume of the cone. Use the unrounded height and give your answer in cm³ to 3 significant figures. [Volume of a cone = ¹/₃πr²h]

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ¹⁵⁰/₃₆₀ × 2 × π × 12	✓ 1
A1 10π shown	✓ 1
Part (b)
M1 2πr = 10π	✓ 1
A1 5 cm	✓ 1
Part (c)
M1 √(12² − 5²) (= √119)	✓ 1
A1 10.9 cm	✓ 1
Part (d)
M1 ¹/₃ × π × 5² × √119 (ECF from (b) and (c))	✓ 1
A1 286 cm³	✓ 1

Exam tip: When a sector is rolled into a cone, the sector’s arc becomes the circumference of the base and the sector’s radius becomes the slant height.

+ Add to test✓ In your test380IGME3-4-14r45-11Pythagoras' theorem and trigonometry Paper HHard9 marks

A circular fish pond has radius 6.5 m and centre O. A straight wooden walkway AB of length 12 m has both ends on the edge of the pond.

Diagram not to scale

(a)

Work out the shortest distance from O to the walkway AB.

(3)

(b)

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

(3)

(c)

Work out the area of the region between the walkway and the edge of the pond on the smaller side of AB (the minor segment). Give your answer to 3 significant figures. Use the unrounded value of angle AOB if needed.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the perpendicular from O to AB bisects the chord, so half the walkway is 6 m	✓ 1
M1 $\sqrt{6.5^2-6^2}$	✓ 1
A1 2.5 m	✓ 1
Part (b)
M1 $\sin(\tfrac{1}{2}\angle AOB)=\frac{6}{6.5}$ or $\cos=\frac{2.5}{6.5}$	✓ 1
M1 doubling the half-angle (67.38...)	✓ 1
A1 134.8°	✓ 1
Part (c)
M1 area of sector = $\frac{134.76}{360}\times\pi\times 6.5^2$ (= 49.69...)	✓ 1
M1 area of triangle AOB = $\frac{1}{2}\times 12\times 2.5$ = 15 and subtracting	✓ 1
A1 34.7 m²	✓ 1

Exam tip: A segment is a sector minus a triangle. Draw the radii to the end points of the chord to see both shapes.

+ Add to test✓ In your test381IGME3-4-14r45-14Pythagoras' theorem and trigonometry Paper HHard7 marks

A robotic lawnmower works on a triangular lawn ABC. AB = 14 m, BC = 9 m and angle ABC = 40°.

Diagram not to scale

(a)

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

(2)

(b)

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

(3)

(c)

Work out the shortest distance from B to the side AC. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{1}{2}\times 14\times 9\times\sin 40^\circ$	✓ 1
A1 40.5 m²	✓ 1
Part (b)
M1 $AC^2=14^2+9^2-2\times 14\times 9\times\cos 40^\circ$	✓ 1
M1 $AC^2=83.96...$ and square root	✓ 1
A1 9.16 m	✓ 1
Part (c)
M1 area = $\frac{1}{2}\times AC\times h$, so $h=\frac{2\times 40.5}{9.16}$ (accept unrounded values)	✓ 1
A1 8.84 m	✓ 1

Exam tip: When two sides and the included angle are known, use the area formula $\frac{1}{2}ab\sin C$. The same area can then give a height by rearranging $\frac{1}{2}\times\text{base}\times h$.

+ Add to test✓ In your test382IGME3-4-13r46-09Pythagoras' theorem and trigonometry Paper HHard7 marks

Three hospitals A, B and C are the corners of a triangle on a map. AB = 12 km, AC = 9 km and angle BAC = 64°.

Diagram not to scale

(a)

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

(3)

(b)

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

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area = ½ ab sin C	✓ 1
M1 ½ × 12 × 9 × sin 64°	✓ 1
A1 48.5 km²	✓ 1
Part (b)
M1 cosine rule BC² = 12² + 9² − 2 × 12 × 9 × cos 64°	✓ 1
M1 BC² = 130.3…	✓ 1
M1 √130.3…	✓ 1
A1 11.4 km	✓ 1

Exam tip: Set your calculator to degrees and keep the full calculator value for BC² before taking the square root.

+ Add to test✓ In your test383IGME3-4-13r46-11Similar shapes and circle theorems Paper HHard6 marks

Two hospital medicine bottles are mathematically similar. The larger bottle has a label area of 90 cm² and holds 540 ml. The smaller bottle has a label area of 40 cm².

(a)

Work out the ratio of the heights of the bottles, smaller : larger. Give your answer in its simplest form.

(3)

(b)

Work out the volume of liquid that the smaller bottle holds.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area ratio 40 : 90	✓ 1
M1 length ratio √(40/90) = 2/3	✓ 1
A1 2 : 3	✓ 1
Part (b)
M1 volume scale factor (2/3)³ = 8/27	✓ 1
M1 540 × 8/27 (ECF from (a))	✓ 1
A1 160 ml	✓ 1

Common error: Do not square the length ratio to go from volume to volume; volumes use the cube of the length scale factor.

+ Add to test✓ In your test384IGME3-4-13r46-13Pythagoras' theorem and trigonometry Paper HHard6 marks

A beehive body is a cuboid 50 cm long, 40 cm wide and 30 cm high. A thin straight rod runs inside the hive from one bottom corner to the opposite top corner.

Diagram not to scale

(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 hive. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 50² + 40² + 30² (= 5000)	✓ 1
M1 √5000	✓ 1
A1 70.7 cm	✓ 1
Part (b)
M1 base diagonal √(50² + 40²) (= 64.03…)	✓ 1
M1 tan x = 30 ÷ 64.03…	✓ 1
A1 25.1°	✓ 1

Exam tip: Draw the right-angled triangle formed by the base diagonal, the vertical edge and the rod before using trigonometry.

+ Add to test✓ In your test385IGME3-4-14r46-11Similar shapes and circle theorems Paper HHard6 marks

Two swimming trophies are made from the same metal and are mathematically similar. The larger trophy has a mass of 4 kg and a height of 30 cm. The smaller trophy has a mass of 0.5 kg.

(a)

Work out the height of the smaller trophy.

(3)

(b)

The larger trophy has a surface area of 900 cm². Work out the surface area of the smaller trophy.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume (mass) ratio 0.5 : 4 = 1 : 8	✓ 1
M1 length ratio ∛(1/8) = 1/2	✓ 1
A1 15 cm	✓ 1
Part (b)
M1 area scale factor (1/2)² = 1/4	✓ 1
M1 900 × 1/4	✓ 1
A1 225 cm²	✓ 1

Exam tip: For objects made of the same material, mass is proportional to volume, so the mass ratio equals the volume ratio.

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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.)	—

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
Part (a)
M1 for \(2.4^2 - 0.9^2\) or \(\sqrt{2.4^2-0.9^2}\)	✓ 1
M1 for square root of their value, e.g. \(\sqrt{4.95}\)	✓ 1
A1 for 2.22 m (2.2248...)	✓ 1
Part (b)
M1 for a correct trigonometric ratio, e.g. \(\sin x = \frac{0.9}{2.4}\) (or \(\cos x = \frac{2.2249}{2.4}\) or \(\tan x = \frac{0.9}{2.2249}\))	✓ 1
A1 for 22.0° (22.02...)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\pi \times 3.5^2 \times 11\)	✓ 1
A1 for 423 cm³ (423.3...)	✓ 1
Part (b)
M1 for \(2 \times \pi \times 3.5 \times 11\)	✓ 1
A1 for 242 cm² (241.9...)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(12^2 + 5^2\) (= 169)	✓ 1
A1 for 13 km	✓ 1
Part (b)
M1 for a correct ratio for the angle east of north, e.g. \(\tan x = \frac{5}{12}\)	✓ 1
A1 for 22.6° (22.62...)	✓ 1
B1 for three-figure bearing written as 022.6° (or 022.6)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(360 \div 5\)	✓ 1
A1 \(72^\circ\)	✓ 1
Part (b)
B1 \(180 - 72 = 108^\circ\) (ECF from the exterior angle in the previous part)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(120^2 + 50^2\)	✓ 1
M1 \(\sqrt{14400+2500}=\sqrt{16900}\)	✓ 1
A1 130 km	✓ 1
Part (b)
M1 total distance \(120+50+130=300\) km, time \(=300\div600\) hours (ECF from the previous part)	✓ 1
A1 30 minutes	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\pi\times1.5^2\times2.4\)	✓ 1
A1 17.0 m³	✓ 1
Part (b)
M1 convert volume to litres: \(16.96\ldots\times1000\)	✓ 1
M1 divide by 40	✓ 1
A1 424 minutes	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\pi\times36\)	✓ 1
A1 113.1 cm	✓ 1
Part (b)
M1 \(\pi\times18^2\)	✓ 1
A1 1018 cm²	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(12\times8.5\)	✓ 1
A1 102 m\(^{2}\)	✓ 1
Part (b)
B1 41 m	✓ 1
Part (c)
M1 \(102\times14.50\) (ECF from (a))	✓ 1
A1 £1479	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{9^{2}+12^{2}}\)	✓ 1
A1 15 km	✓ 1
Part (b)
M1 \(15\div20\) (ECF from (a))	✓ 1
A1 45 minutes	✓ 1

Marking point	Mark	Notes
Part (a)
B1 \(360\div8=45^{\circ}\)	✓ 1
Part (b)
B1 \(180-45=135^{\circ}\)	✓ 1
Part (c)
M1 \((8-2)\times180\) or \(8\times135\)	✓ 1
A1 \(1080^{\circ}\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\pi\times4^{2}\times10\)	✓ 1
A1 503 cm\(^{3}\)	✓ 1
Part (b)
M1 volume \(\times2.7\)	✓ 1
A1 1360 g	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{36^2+77^2}\)	✓ 1
A1 85 cm	✓ 1
Part (b)
M1 36 + 77 + 85 (ECF from (a))	✓ 1
A1 198 cm	✓ 1

Marking point	Mark	Notes
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

Marking point	Mark	Notes
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

Marking point	Mark	Notes
Part (a)
M1 for rectangle area \(1.8 \times 2.2\) (= 3.96)	✓ 1
M1 for semicircle area \(\frac{1}{2}\pi \times 0.9^2\) (= 1.272...)	✓ 1
A1 for 5.23 m² (5.232...)	✓ 1
Part (b)
M1 for \(2 \times 2.2 + 1.8\) (= 6.2)	✓ 1
M1 for semicircle arc \(\frac{1}{2} \times \pi \times 1.8\) (= 2.827...)	✓ 1
A1 for 9.03 m (9.027...)	✓ 1
Part (c)
M1 for \(9.0274 \times 1.40\) (or their length from (b) × 1.40)	✓ 1
A1 for £12.64 (12.638...; 9.03 × 1.40 also gives 12.64)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(28 \tan 61°\)	✓ 1
A1 for 50.5 m (50.51...)	✓ 1
Part (b)
M1 for height of the tip \(28 \tan 70°\) (= 76.9...)	✓ 1
M1 for subtracting the hub height, \(28\tan 70° - 28 \tan 61°\)	✓ 1
A1 for 26.4 m (26.41...)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(7.5^2 + 10^2\) (= 156.25)	✓ 1
A1 for 12.5 cm	✓ 1
Part (b)
M1 for \(\frac{1}{2} \times 7.5 \times 10\) (= 37.5)	✓ 1
A1 for 1125 cm³	✓ 1
Part (c)
M1 for area of both triangles \(2 \times 37.5\) (= 75)	✓ 1
M1 for rectangles \((7.5 + 10 + 12.5) \times 30\) (= 900)	✓ 1
A1 for 975 cm²	✓ 1
Part (d)
M1 for \(975 \div 100 \times 0.35\)	✓ 1
A1 for £3.41 (3.4125)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\sin12^\circ = \dfrac{h}{8.5}\)	✓ 1
M1 \(h = 8.5\sin12^\circ\)	✓ 1
A1 1.77 km	✓ 1
Part (b)
M1 \(8.5\cos12^\circ\)	✓ 1
A1 8.31 km	✓ 1

Marking point	Mark	Notes
Part (a)
B1 \(180-156=24^\circ\)	✓ 1
Part (b)
M1 \(360\div24\)	✓ 1
A1 15 (ECF from the exterior angle)	✓ 1
Part (c)
M1 \((15-2)\times180\)	✓ 1
A1 \(2340^\circ\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{6.5^2-2.5^2}\)	✓ 1
A1 6 m	✓ 1
Part (b)
M1 \(\sin\theta=\dfrac{2.5}{6.5}\) (or \(\tan\theta=\dfrac{2.5}{6}\))	✓ 1
M1 \(\theta=\sin^{-1}(0.3846\ldots)\)	✓ 1
A1 \(22.6^\circ\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\tan18^{\circ}=\dfrac{45}{d}\) (angle of depression equals angle of elevation from ship)	✓ 1
M1 \(d=\dfrac{45}{\tan18^{\circ}}\)	✓ 1
A1 138 m	✓ 1
Part (b)
M1 \(\tan\theta=\dfrac{45}{200}\)	✓ 1
M1 \(\theta=\tan^{-1}\!\left(\dfrac{45}{200}\right)\)	✓ 1
A1 \(12.7^{\circ}\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\pi\times10\)	✓ 1
A1 31.4 m	✓ 1
Part (b)
M1 outer radius 6 m and inner radius 5 m	✓ 1
M1 \(\pi\times6^{2}-\pi\times5^{2}\)	✓ 1
A1 34.6 m\(^{2}\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \((180-48)\div2\)	✓ 1
A1 \(66^{\circ}\) with reason: base angles of an isosceles triangle are equal and angles in a triangle sum to \(180^{\circ}\)	✓ 1
Part (b)
M1 \(180-66\) (ECF from (a)) or \(48+66\)	✓ 1
A1 \(114^{\circ}\) with reason: angles on a straight line sum to \(180^{\circ}\) (or exterior angle equals sum of the two opposite interior angles)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\tfrac{1}{2}(9+13)\times6\)	✓ 1
A1 66 m\(^{2}\)	✓ 1
Part (b)
B1 \(66\times0.25=16.5\) m\(^{3}\) (ECF from (a))	✓ 1
Part (c)
M1 \(16.5\times38\)	✓ 1
A1 £627	✓ 1