Printable · GCSE Higher · ages 14-16
Geometry and measures worksheet — GCSE Higher
Fifteen questions across the geometry and measures statements at Higher 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 Higher
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- (a) £157.00 — Area covered = (angle ÷ 360) × π × r² = (90 ÷ 360) × 3.14 × 400 = 0.25 × 1256 = 314 km². Charge = 314 × £0.50 = £157.00. (£7.85 comes from forgetting to square the radius, using 0.25 × 3.14 × 20 = 15.7 km² and then charging that; £628.00 comes from finding the area of a full circle, 3.14 × 400 = 1256 km², and forgetting the angle fraction before charging; £15.70 comes from using the arc length formula, 0.25 × 2 × 3.14 × 20 = 31.4, in place of the area, and charging that.)
- (a) 69.3 m — Method: each observer gives a right-angled triangle with the mast as the opposite side, so tan θ = 60 ÷ distance and the distance from the foot of the mast is 60 ÷ tan θ; because both stand on the same side, the gap between them is the difference of those two distances. Working: from Amelia, 60 ÷ tan 30° = 103.92… m; from Noah, 60 ÷ tan 60° = 34.64… m; the gap is 103.92… − 34.64… = 69.28… m, which is 69.3 m to 1 decimal place. Answer: 69.3 m. The distractors: 103.9 m is Amelia's own distance from the foot of the mast, written down before the second distance has been taken away; 34.6 m is Noah's distance from the foot of the mast; 138.6 m comes from adding the two distances, which would be right only if the two observers stood on opposite sides of the mast.
- (d) 4.44 m — The real van's length is 18.5 × 24 = 444 cm, which converts to 4.44 m. '0.77 m' comes from dividing by 24 instead of multiplying, using the ratio the wrong way round (18.5 ÷ 24 = 0.77 cm), and then writing that figure down as metres. '44.4 m' comes from converting 444 cm to metres with the decimal point in the wrong place. '444 m' comes from working out 444 cm correctly but forgetting to convert it into metres at all.
- (b) 13.3 cm — The apex is directly above the centre of the square base, so the height, half the base diagonal, and a slant edge form a right-angled triangle with the slant edge as the hypotenuse. Half the base diagonal is 14 ÷ 2 = 7 cm. Using Pythagoras' theorem, height = √(15² − 7²) = √(225 − 49) = √176 = 13.3 cm (1 d.p.). Using the slant edge itself as the height, without applying Pythagoras' theorem at all, gives 15 cm. Using the full base diagonal (14 cm) instead of half of it gives √(15² − 14²) = √(225 − 196) = √29 = 5.4 cm (1 d.p.), far too short for a pyramid this size. Adding the two squares instead of subtracting them, √(15² + 7²) = √(225 + 49) = √274 = 16.6 cm (1 d.p.), gives a length longer than the slant edge itself, which cannot be the height.
- (b) £709 — Method: first find the area of the triangular platform with Area = (1/2)ab sin C, then multiply by the cost per square metre. Working: Area = 1/2 × 11.5 × 8.2 × sin 72° = 44.8 m² (1 d.p.); cost = area × £15.80, which rounds to £709 to the nearest pound. Answer: £709. Leaving out the 1/2 in the area formula doubles the area, giving a cost of £1417; using cos 72° instead of sin 72° gives a much smaller area and a cost of £230; and squaring the 11.5 m side instead of multiplying the two different given sides together gives a cost of £994. Find the exact area first — don't round it early — then multiply by the cost per square metre and round only the final answer.
- (d) Two angles and a side: sine rule finds other sides. — Method: match the data you are given to the rule that needs it. Working: the sine rule a/sin A = b/sin B = c/sin C needs a complete angle-side pair to set up its ratio, so the statement that two angles and a side (AAS or ASA) let the sine rule find the other sides is the correct one — the third angle comes from the angle sum, and each unknown side is then opposite a known angle. The statement that two sides and the angle between them (SAS) call for the sine rule is wrong: no angle-side pair is complete, so the cosine rule is what works there. The statement that the sine rule finds any angle from three sides (SSS) is wrong for the same reason in reverse — no angle is known at all, so the cosine rule must find the first one. The statement that the cosine rule finds a missing angle directly from two sides and a non-included angle (SSA) is wrong: the cosine rule reports the angle enclosed by the two sides it uses, so with SSA it is the sine rule that reaches the missing angle, and the ambiguous case is then settled from the wording of the question.
- (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) 18 — Method: convert the real length to centimetres, then divide by the scale factor to shrink it down to the model's size. Working: 4.32 m = 432 cm; 432 cm / 24 = 18 cm. A student who answers 432 has converted the units correctly but forgotten to divide by the scale factor at all. A student who answers 10368 has multiplied by the scale factor instead of dividing (432 x 24). A student who answers 1.8 has converted the metres to centimetres by multiplying by 10 instead of 100, getting 43.2 cm, and then divided by 24. Answer: 18 cm.
- (d) 38 m² — The two fences meet at a right angle, so the rope sweeps a quarter of a circle: 90 ÷ 360 = 1/4. Grazing area = (90 ÷ 360) × 3.14 × 7² = 0.25 × 153.86 = 38.465 m², which rounds to 38 m². (5 m² comes from forgetting to square the rope length; 154 m² comes from finding the area of a full circle and forgetting the angle fraction; 11 m² comes from using the arc length formula instead of the sector area formula.)
- (b) 7 cm — The side opposite the 30° angle is found using sin 30° = opposite/hypotenuse, so opposite = 14 × sin 30° = 14 × 1/2 = 7 cm. 7√3 cm comes from using cos 30° = √3/2 instead of sin 30° (mixing up the opposite and adjacent sides). 14/√3 cm comes from using tan 30° = 1/√3 instead of sin 30°. 28 cm comes from dividing 14 by sin 30° instead of multiplying by it.
- (d) 15° — Sector area = (angle ÷ 360) × π × r², so 4.71 = (angle ÷ 360) × 3.14 × 36 = (angle ÷ 360) × 113.04. Dividing gives angle ÷ 360 = 4.71 ÷ 113.04 = 1/24, so angle = 360 ÷ 24 = 15°. (90° comes from forgetting to square the radius, using (angle ÷ 360) × 3.14 × 6 = 18.84 in place of 113.04; 3.75° comes from using the diameter, 12 cm, in place of the radius, giving (angle ÷ 360) × 3.14 × 144 = 452.16; 45° comes from using the arc length formula, (angle ÷ 360) × 2 × 3.14 × 6 = 37.68, instead of the sector area formula.)
- (b) 32 cm² — The area of a trapezium is half of the sum of the parallel sides, multiplied by the height. Add the parallel sides: 6 + 10 = 16. Multiply by the height: 16 × 4 = 64. Half of 64 is 32 cm². 64 cm² forgets to halve and just gives (6+10)×4. 8 cm² averages the two parallel sides, (6+10)÷2 = 8, but forgets to multiply by the height. 20 cm² treats it as a triangle using only the longer parallel side as the base: half of 10 × 4.
- (d) 21.2 m — Method: the cable is the hypotenuse of a right-angled triangle whose vertical side is the drop from the roof to the bracket and whose horizontal side is 15 m, so use Pythagoras' theorem. Working: the drop is 20 − 5 = 15 m, so c² = 15² + 15² = 225 + 225 = 450 and c = √450 = 21.213…, which is 21.2 m to 1 decimal place. Answer: 21.2 m. The distractors: 25.0 m comes from using the whole 20 m height of the roof as the vertical side and forgetting that the bracket is already 5 m up; 30.0 m comes from adding the two sides of the triangle, 15 + 15, instead of using Pythagoras' theorem; 15.0 m is the horizontal distance on its own, which would be the length of the cable only if it ran level.
- (d) £817 — Method: find the area of the plot with 1/2 × a × b × sin C, then multiply the area by the cost of a square metre. Working: the 108° angle is between AB and AC, so the area is 1/2 × 23.5 × 17.2 × sin 108° = 202.1 × 0.95106 = 192.21 m². The cost is 192.21 × 4.25 = 816.89. Answer: the turf costs £817 to the nearest pound. The distractors: £1634 comes from leaving out the factor 1/2, so the area is taken as 384.42 m²; £859 comes from leaving the sine out and using 202.1 m² as the area, which treats the two sides as a base and a perpendicular height; £192 is the area of the plot written down as though it were the cost, stopping one step short of the question.
- (d) 114.6° — Method: rearrange Area = (1/2)ab sin C to make sin C the subject, then use the obtuse branch since the question states the angle is obtuse. Working: sin C = 2 × 45 ÷ (11 × 9) = 0.909, so the acute angle is sin⁻¹(0.909) = 65.4°, and the obtuse angle is 180 − 65.4 = 114.6°. Answer: 114.6°. Giving the acute angle straight from the calculator, 65.4°, ignores that the question asks for the obtuse one; forgetting to double the area before dividing gives sin C = 45 ÷ (11 × 9) = 0.4545, whose obtuse angle is 180 − 27.0 = 153.0°; and that same forgotten-doubling error taken on the acute branch instead gives 27.0°. Always double the area first, and then take 180° minus the calculator's answer whenever the question specifically asks for the obtuse angle.
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