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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Answer key: Geometry and measures worksheet — GCSE Foundation
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- (d) 6.4 cm — The scale factor from the larger sign to the smaller sign is 4 ÷ 9, so the smaller side is 14.4 × 4 ÷ 9 = 6.4 cm. The distractor 32.4 cm comes from using the ratio the wrong way round, 14.4 × 9 ÷ 4 = 32.4. The distractor 9.4 cm comes from subtracting the difference between the ratio numbers (9 − 4 = 5) from the given length, 14.4 − 5 = 9.4. The distractor 3.6 cm comes from dividing the given length by 4 only, 14.4 ÷ 4 = 3.6, without also using the other ratio number.
- (c) 25 m — The gardener needs the circumference, since the trim goes around the flower bed. Circumference = πd = 3.14 × 8 = 25.12 m, which rounds to 25 m. A student who mistakes the diameter (8 m) for the radius, then doubles it before multiplying by π, gets 3.14 × 16 = 50.24 m, rounding to 50 m. A student who makes the same mistake but uses the area formula instead of the circumference gets 3.14 × 8² = 200.96 m², rounding to 201. A student who correctly halves the diameter to find the radius (4 m) but then uses πr instead of πd gets 3.14 × 4 = 12.56 m, rounding to 13 m.
- (a) 8π cm — Circumference = 2πr. Substitute r = 4: circumference = 2 × π × 4 = 8π cm. Using r in place of 2r (halving the formula) gives 4π cm. Using the area formula πr² in place of the circumference formula gives π × 4² = 16π cm. Multiplying 2 × 4 without including π at all gives 8 cm.
- (b) 16.747 m³ — Volume of a sphere = (4/3)πr³, so a hemisphere is half that: (2/3)πr³. Substitute r = 2: (2/3) × 3.14 × 2³ = (2/3) × 3.14 × 8 = (2/3) × 25.12 = 16.7467 m³, which rounds to 16.747 m³.
- (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.
- (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.)
- (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.
- (c) AB means the segment or its length, shown by context — Method: recall the standard convention for a two-letter label such as AB. Working: AB names the segment AND its length; the sentence around it shows which is meant — 'draw AB' means the segment, 'AB = 5 cm' means the length. Options: 'a different symbol is needed' and 'must be written as |AB|' both describe stricter rules than the convention GCSE actually uses; 'AB can only mean the segment' ignores that the same label is also used for the length. Answer: AB means the segment or its length, shown by context.
- (b) 25 minutes — Method: a rate in litres per minute can only be used on a volume measured in litres, so convert the tank first and then divide. Working: 1 m³ = 1000 litres, so the tank holds 0.45 × 1000 = 450 litres, and the time is 450 ÷ 18 = 25. Answer: 25 minutes. Using 1 m³ = 100 litres gives 45 ÷ 18 = 2.5 minutes. Using 1 m³ = 1 000 000 litres, which is the factor that turns cubic metres into cubic centimetres, gives 450 000 ÷ 18 = 25 000 minutes. Multiplying by the rate instead of dividing by it gives 450 × 18 = 8100.
- (b) 320° — A bearing is always measured clockwise from north. An angle measured anticlockwise must be converted by subtracting it from 360°: 360 − 40 = 320°, so the bearing is 320°. Choosing 040° treats the anticlockwise angle as if it were already a clockwise bearing, without converting it. Choosing 220° adds 180° to the angle, mixing this up with a back-bearing calculation (40 + 180 = 220). Choosing 400° adds the angle to 360° instead of subtracting it (40 + 360 = 400), giving a bearing greater than a full turn.
- (a) 113.04 cm³ — Method: the volume of a sphere is (4 ÷ 3) × π × r³. Cube the radius, multiply by π, then multiply by 4 and divide by 3. Working: r³ = 3³ = 27, then 3.14 × 27 = 84.78, then 84.78 × 4 = 339.12 and 339.12 ÷ 3 = 113.04. Answer: 113.04 cm³. The distractors: 84.78 cm³ comes from stopping at πr³ and leaving out the four thirds; 37.68 cm³ comes from squaring the radius instead of cubing it, (4 ÷ 3) × 3.14 × 9; 28.26 cm³ comes from using πr², the area of a circle, and labelling it as a volume.
- (d) £50 — Area = 1/2 × (3.5 + 6.5) × 4 = 1/2 × 10 × 4 = 20 m². Cost = 20 × £2.50 = £50. (£100 comes from forgetting to halve the trapezium area, giving 40 m² instead of 20 m²; £65 comes from using only the longer parallel side, 6.5 × 4 = 26 m², instead of the trapezium formula; £35 comes from adding all three given lengths, 3.5 + 6.5 + 4, and treating that total as the area in square metres.)
- (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).
- (b) 8.64 m — The real length is 240 times the model length: 3.6 × 240 = 864 cm. Converting to metres, 864 cm ÷ 100 = 8.64 m. The distractor 864 m comes from finding the length correctly in centimetres but forgetting to convert to metres. The distractor 0.0864 m comes from dividing by 100 a second time, converting 864 cm to metres twice over (864 ÷ 10 000) instead of once. The distractor 86.4 m comes from converting centimetres to metres by dividing by 10 instead of 100.
- (b) 10.5 — Method: multiply the drawing length by the scale factor to get the real length. Working: 7 cm × 1.5 = 10.5 m. A student who answers 8.5 has added the scale factor to the drawing length instead of multiplying (7 + 1.5). A student who answers 14 has rounded the scale factor up to 2 before multiplying. A student who answers 3.5 has divided the drawing length by 2 instead of multiplying it by 1.5. Answer: 10.5 m.
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