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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- (b) 31.0 cm² — Method: Area = 1/2ab sin C needs the angle BETWEEN the two given sides, so first find angle BAC using the angle sum of a triangle. Working: angle BAC = 180° − 65° − 65° = 50°, the angle between AB and AC; Area = 1/2 × 9 × 9 × sin 50° = 31.0 cm². Using the given base angle 65° in place of the included angle 50° gives 36.7 cm²; leaving out the 1/2 altogether gives 62.0 cm²; and using cos 50° instead of sin 50° gives 26.0 cm². The formula only works with the angle that sits between the two sides being multiplied — here that means finding the missing angle first.
- (a) 59.0° — Method: rearrange Area = (1/2)ab sin C to make sin C the subject: sin C = 2 × Area ÷ (a × b). Working: sin C = 2 × 24 ÷ (8 × 7) = 0.857, so C = sin⁻¹(0.857) = 59.0° (1 d.p.), which is acute as the question requires. Answer: 59.0°. Taking the obtuse angle instead of the acute one asked for, 180 − 59.0 = 121.0°; forgetting to double the area before dividing gives sin C = 24 ÷ (8 × 7) = 0.4286, whose acute angle is 25.4°; and combining that same forgotten-doubling error with the obtuse branch gives 180 − 25.4 = 154.6°. Always double the area first, and then pick the acute branch, since that is what this question asks for.
- (c) 152.2 cm² — Method: find the area of one of the 5 isosceles triangles with 1/2ab sin C, then multiply by 5 for the whole pentagon. Working: one triangle has area 1/2 × 8 × 8 × sin 72° = 30.4338 cm², so the pentagon's area = 5 × 30.4338 = 152.2 cm² — keep the unrounded triangle area, since 5 × 30.4 would give 152.0. Reporting the area of a single triangle and forgetting to multiply by 5 gives 30.4 cm²; multiplying by 6 instead of 5, as for a hexagon, gives 182.6 cm²; and leaving out the 1/2 from each triangle before multiplying by 5 gives 304.3 cm². A regular pentagon splits into exactly 5 triangles at its centre, each with a 72° angle, since 360° ÷ 5 = 72°.
- (b) 21 — Both legs of the journey are on the same bearing, 070°, so the ship travels in one straight line the whole way and the distances simply add: 12 + 9 = 21 km. A candidate who subtracts instead of adding gets 12 − 9 = 3 km. A candidate who multiplies the two distances gets 12 × 9 = 108. A candidate who assumes the ship changed direction and treats the two legs as the sides of a right-angled triangle works out √(12² + 9²) = √225 = 15 km — but the bearing does not change, so there is no triangle and no hypotenuse to find. Because the ship stays on one straight line, the total distance is 21 km.
- (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.
- (c) 27.7 cm — Arc length = (150 ÷ 360) × 2 × 3.14 × 6 = (5 ÷ 12) × 37.68 = 15.7 cm. The perimeter of a sector also includes the two straight radii, so perimeter = 15.7 + 6 + 6 = 27.7 cm. (15.7 cm comes from stopping after the arc length and forgetting the two straight edges; 21.7 cm comes from adding only one radius instead of two; 31.4 cm comes from doubling the arc length instead of adding the two straight edges.)
- (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.
- (b) 4 m — The two triangles are similar, so height ÷ shadow is the same for both objects. For the post, height ÷ shadow = 1.2 ÷ 1.8. Flagpole height = 6 × (1.2 ÷ 1.8) = 4 m. A student who multiplies the flagpole's shadow by the post's height without applying the ratio gets 6 × 1.2 = 7.2 m. A student who assumes a constant DIFFERENCE between shadow and height instead of a constant ratio finds the post's shadow minus height, 1.8 − 1.2 = 0.6 m, then subtracts that from the flagpole's shadow to get 6 − 0.6 = 5.4 m.
- (d) 13 cm — Use Pythagoras' theorem in three dimensions: for a cuboid with edges a, b and c the space diagonal d satisfies d² = a² + b² + c². Substitute a = 3, b = 4, c = 12: d² = 3² + 4² + 12² = 9 + 16 + 144 = 169. Take the square root: d = √169 = 13 cm. Adding the three edges directly, 3 + 4 + 12 = 19 cm, ignores that Pythagoras' theorem is about squares, not lengths, and gives 19 cm. Stopping after squaring and adding, without taking the square root, leaves 169 cm — the squared length, not the length itself. Using only the 4 cm and 12 cm edges finds the diagonal of one face, √(4² + 12²) = √160 = 12.6 cm (1 d.p.), and leaves out the third dimension entirely.
- (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.)
- (a) 035° — A bearing is measured clockwise from north, so an angle of 35° clockwise from north is a bearing of 035° (written with three figures). Choosing 325° measures the angle anticlockwise instead of clockwise (360 − 35 = 325). Choosing 215° adds 180° to the angle, mixing this up with a back-bearing calculation (35 + 180 = 215). Choosing 350° reorders the digits of 035, writing the ones digit before the tens digit by mistake.
- (a) 20 litres — Volume of water = length × width × depth of water = 40 × 25 × 20 = 20 000 cm³. Since 1000 cm³ = 1 litre, divide by 1000: 20 000 ÷ 1000 = 20 litres. A pupil who uses the full height of the tank, 30 cm, instead of the water depth, 20 cm, gets 40 × 25 × 30 = 30 000 cm³ = 30 litres. A pupil who forgets to convert cm³ to litres at all gives 20 000 litres. A pupil who divides by 1000 twice by mistake gets 20 000 ÷ 1000 ÷ 1000 = 0.02 litres. The correct volume of water is 20 litres.
- (c) 36.7 m — Method: use the cosine rule to find the missing side BC, then add all three sides for the perimeter. Working: BC² = 13² + 10² − 2 × 13 × 10 × cos 72°, so BC = 13.7 m, and perimeter = 13 + 10 + 13.7 = 36.7 m. Forgetting the negative sign in the cosine rule (using + instead of −) gives BC = 18.7 m and a perimeter of 41.7 m; leaving AC out of the total and only adding AB and BC gives 26.7 m; and adding AB twice instead of AB and AC gives 39.7 m. Always add all three named sides once the missing one has been found.
- (a) 9.3 cm — Method: the 40° angle is not between the two known sides, so call the unknown side x and put it into the cosine rule, which turns into a quadratic equation with two positive roots. Working: AC faces angle ABC, so 6² = 7² + x² − 2 × 7 × x × cos 40°, that is 36 = 49 + x² − 10.7246x, which rearranges to x² − 10.7246x + 13 = 0. The discriminant is 10.7246² − 4 × 13 = 115.02 − 52 = 63.02, whose square root is 7.9384, so x = (10.7246 + 7.9384) ÷ 2 = 9.3315 or x = (10.7246 − 7.9384) ÷ 2 = 1.3931. The longer of the two is wanted. Answer: BC = 9.3 cm to 1 decimal place. The distractors: 1.4 cm is the shorter root, taken by a candidate who solves the quadratic correctly but does not read which of the two lengths is wanted; 4.5 cm comes from treating the 40° as the angle between the two given sides and working out 7² + 6² − 2 × 7 × 6 × cos 40° directly, when 40° lies at B and faces AC; 3.6 cm comes from assuming the triangle is right-angled with AB as the hypotenuse and using the square root of 7² − 6².
- (d) 1 kg — First find the total mass in grams: 4 × 250 = 1000 g. Then convert to kilograms by dividing by 1000: 1000 ÷ 1000 = 1 kg. Finding the correct total in grams but forgetting to divide by 1000 gives 1000 kg. Adding the number of packs to the pack mass instead of multiplying, 4 + 250 = 254 g, gives 0.254 kg. Dividing the pack mass by the number of packs instead of multiplying, 250 ÷ 4 = 62.5 g, gives 0.0625 kg.
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