Printable · GCSE Foundation · ages 14-16
Geometry and measures worksheet — GCSE Foundation
Fifteen questions across the geometry and measures statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Geometry and measures worksheet — GCSE Foundation
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- 1.In triangle ABC, angle ABC = 90° and angle BAC = 30°. The hypotenuse AC = 12 cm. Work out the exact length of AB.
- 2.A scale drawing uses a scale of 1 : 60. A path on the drawing is measured as 9.8 cm long. What is the real length of the path, in metres, to 1 decimal place?
- 3.A bead starts at position (2, −1) on a grid, in centimetres. It is moved by the column vector u, with top number 3 and bottom number 5, and then moved by the column vector v, with top number −7 and bottom number 2. Work out the coordinates of the bead's final position.
- 4.The bearing of a harbour B from a ferry's position A is 260°. What is the bearing of A from B?
- 5.A cylinder has a radius of 5 cm. Its volume is 471 cm³. Using π = 3.14, work out the height of the cylinder.
- 6.A circle has a diameter of 24 cm. A sector of this circle has an angle of 90°. Using π = 3.14, work out the area of the sector.
- 7.A photograph is 10 cm wide and 15 cm tall. It is enlarged to make a similar poster that is 40 cm wide. The poster costs £0.80 per centimetre of its height to print, based on its full height. Work out the cost of printing the poster.
- 8.Two similar ponds have perimeters in the ratio 4 : 11. The perimeter of the larger pond is 88 m. Work out the perimeter of the smaller pond.
- 9.In triangle ABC the angle at C is 90°, AC = 8 cm and the angle at A is 30°. Work out the length of BC. Give your answer to 1 decimal place.
- 10.A solid ball has a radius of 7 cm. Work out the surface area of the ball. Use π = 3.14.
- 11.Three angles meet at a single point on one side of a straight line. Two of the angles are 48° and 65°. Work out the third angle.
- 12.Two straight roads cross at a junction. The angle between them, measured with a protractor, is 118°. What is the size of the angle vertically opposite to it?
- 13.A triangular prism has a cross-section that is a triangle with a base of 3.5 cm and a perpendicular height of 4 cm. The prism is 10 cm long. Work out the volume of the prism.
- 14.Two vertical masts stand on level ground, 40 m apart. One mast is 30 m tall and the other is 50 m tall. Work out the straight-line distance between the tops of the two masts. Give your answer to 1 decimal place.
- 15.A garden water trough is a prism whose cross-section is a right-angled triangle with base 40 cm and height 30 cm. The trough is 120 cm long. Work out the volume of water needed to fill the trough completely.
Answer key
- (d) 6√3 cm — Method: AB lies alongside the 30° angle at A and AC is the hypotenuse, so the ratio needed is cosine: cos 30° = AB ÷ AC. Working: the exact value of cos 30° is √3/2, so AB = 12 × √3 ÷ 2, and half of 12 is 6. Answer: AB = 6√3 cm, which is about 10.4 cm. Using sine by mistake gives 12 × 1/2 = 6 cm, which is the length of BC rather than AB. Using tan 30° = 1/√3 gives 12 ÷ √3, which is 4√3 cm. Remembering cos 30° as √3 rather than as √3 halved gives 12√3 cm, longer than the hypotenuse and so impossible.
- (a) 5.9 — The real length is 9.8 × 60 = 588 cm. Converting to metres, by dividing by 100, gives 5.88 m, which rounds to 5.9 m to 1 decimal place. A candidate who rounds 5.88 down instead of up gets 5.8 m. A candidate who forgets to convert from centimetres to metres gets 58.8. A candidate who divides by 60 instead of multiplying gets 0.16, to 2 decimal places. The real length, to 1 decimal place, is 5.9 m.
- (d) (−2, 6) — Method: add the top numbers of both vectors to the starting x-coordinate, and the bottom numbers of both vectors to the starting y-coordinate. Working: x-coordinate 2 + 3 + (−7) = −2; y-coordinate −1 + 5 + 2 = 6. Answer: (−2, 6). A candidate who only applies vector u and forgets v gets (5, 4). A candidate who only applies vector v and forgets u gets (−5, 1). A candidate who works out the combined vector u + v but forgets to add it to the starting point gets (−4, 7).
- (a) 080° — The back bearing differs from the given bearing by 180°. Because 260° is greater than 180°, subtract 180°: 260 − 180 = 80°, so the bearing of A from B is 080°. Choosing 440° adds 180° instead of subtracting it, even though the result would be more than a full turn (260 + 180 = 440). Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 100° comes from measuring the reflex angle the other way round the circle (360 − 260 = 100) instead of applying the 180° back-bearing rule.
- (d) 6 cm — Volume = πr²h, so height = volume ÷ (πr²) = 471 ÷ (3.14 × 25) = 471 ÷ 78.5 = 6 cm. (30 cm comes from dividing by πr instead of πr², missing one factor of the radius; 150 cm comes from dividing by π only, without using r² at all; 24 cm comes from treating the given 5 cm as a diameter and using a radius of 2.5 cm instead.)
- (b) 113.04 cm² — Sector area is (angle ÷ 360) × π × radius², and radius means the RADIUS, not the diameter: here the diameter is 24 cm, so the radius is 12 cm. The fraction is 90 ÷ 360 = 1/4, and 12² = 144, so the area is 0.25 × 3.14 × 144 = 113.04 cm². Using the diameter itself as if it were the radius gives 0.25 × 3.14 × 576 = 452.16 cm². Using the arc-length formula, 2 × π × radius, instead of the area formula gives 0.25 × 2 × 3.14 × 12 = 18.84 cm². Using the radius instead of its square gives 0.25 × 3.14 × 12 = 9.42 cm².
- (d) £48 — Scale factor = new width ÷ original width = 40 ÷ 10 = 4. Poster height = 15 × 4 = 60 cm. Cost = 60 × £0.80 = £48. (£12 comes from forgetting to scale the height at all, and pricing the original 15 cm height; £15.20 comes from adding the scale factor 4 to the height instead of multiplying by it; £3 comes from dividing the height by the scale factor instead of multiplying by it.)
- (a) 32 m — The scale factor from the larger pond to the smaller pond is 4 ÷ 11, so the smaller perimeter is 88 × 4 ÷ 11 = 32 m. The distractor 242 m comes from using the ratio the wrong way round, 88 × 11 ÷ 4 = 242. The distractor 84 m comes from subtracting the smaller ratio number, 88 − 4 = 84, instead of scaling. The distractor 121 m comes from multiplying 11 × 11 = 121, ignoring the given perimeter altogether.
- (d) 4.6 cm — Method: BC is opposite the 30° angle and AC is next to it, so the ratio that links the two is tan θ = opposite ÷ adjacent. Working: tan 30° = BC ÷ 8, so BC = 8 × tan 30° = 4.6188…, which is 4.6 to 1 decimal place. Answer: 4.6 cm. The distractors: 13.9 cm comes from dividing by tan 30° instead of multiplying by it; 4.0 cm comes from using sin 30°, which treats the 8 cm side as the hypotenuse when it is the side next to the 30° angle; 6.9 cm comes from using cos 30° in place of tan 30°, which gives the wrong pair of sides.
- (a) 615.44 cm² — Method: the surface area of a sphere is 4πr². Square the radius, multiply by π, then multiply by 4. Working: r² = 7² = 49, then 3.14 × 49 = 153.86, then 4 × 153.86 = 615.44. Answer: 615.44 cm². The distractors: 153.86 cm² comes from stopping at πr², which is the area of a flat circle of radius 7 cm and leaves out the factor of 4 that a curved surface needs; 87.92 cm² comes from 4 × 3.14 × 7, using the radius itself where the formula asks for its square; 1436.03 cm³ comes from working out the volume of the ball with (4 ÷ 3) × π × r³ instead of its surface area, which is a different measure and carries a cubic unit.
- (c) 67° — Angles on a straight line add up to 180°. Add the two known angles: 48° + 65° = 113°. Subtract from 180°: 180° − 113° = 67°.
- (a) 118° — When two straight lines cross, the angles that are vertically opposite each other are always equal. So the angle vertically opposite 118° is also 118°. A candidate who instead finds the angle next to it on the straight line, using 180° − 118° = 62°, has found the adjacent angle, not the vertically opposite one. A candidate who answers 180° has confused the rule with angles on a straight line. A candidate who doubles the angle, giving 236°, has applied no valid angle rule at all. The vertically opposite angle is 118°.
- (b) 70 cm³ — Area of the triangular cross-section = base × height ÷ 2. Base × height = 3.5 × 4 = 14 cm², and half of that is 14 ÷ 2 = 7 cm². Volume of the prism = cross-sectional area × length = 7 × 10 = 70 cm³. A pupil who forgets to halve when finding the triangle's area gets 3.5 × 4 × 10 = 140 cm³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length stops at 7 cm³. A pupil who ignores the height altogether, multiplying base by length, gets 3.5 × 10 = 35 cm³. The correct volume is 70 cm³.
- (a) 44.7 m — Method: the line joining the two tops is the hypotenuse of a right-angled triangle whose horizontal side is the gap between the masts and whose vertical side is the difference in their heights, so Pythagoras' theorem applies. Working: the difference in heights is 50 − 30 = 20 m, so d² = 40² + 20² = 1600 + 400 = 2000 and d = √2000 = 44.721…, which is 44.7 m to 1 decimal place. Answer: 44.7 m. The distractors: 34.6 m comes from subtracting the squares, √(40² − 20²), instead of adding them; 60.0 m comes from adding the two sides of the triangle, 40 + 20, rather than using Pythagoras' theorem; 50.0 m is the height of the taller mast, copied from the question in place of the distance asked for.
- (b) 72 000 cm³ — Cross-sectional area = 1/2 × 40 × 30 = 600 cm². Volume = cross-sectional area × length = 600 × 120 = 72 000 cm³. (144 000 cm³ comes from forgetting the 1/2 in the triangle's area, using 40 × 30 as the cross-section; 36 000 cm³ comes from halving the correct volume again, as if the 1/2 applied a second time; 720 cm³ comes from adding the cross-sectional area and the length, 600 + 120, instead of multiplying them.)
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