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.
Answer key: Geometry and measures worksheet — GCSE Higher
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- (a) 376.8 cm³ — Volume of a cone = (1/3)πr²h. Substitute r = 6 and h = 10: (1/3) × 3.14 × 6² × 10 = (1/3) × 3.14 × 36 × 10 = (1/3) × 1130.4 = 376.8 cm³.
- (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.
- (a) 1/2 — cos 0° = 1 and sin 30° = 1/2, so cos 0° − sin 30° = 1 − 1/2 = 1/2. 1 comes from writing down cos 0° alone and forgetting to subtract sin 30°. 3/2 comes from adding the two values instead of subtracting. −1/2 comes from working out sin 30° − cos 0°, the two terms the wrong way round.
- (b) 8.4 cm — Two sides and the angle between them are known, so use the cosine rule: BC² = AB² + AC² − 2 × AB × AC × cos(BAC). Substituting, BC² = 81 + 36 − 45.64 = 71.36. Taking the square root: BC = √71.36 = 8.4 cm (1 d.p.). Leaving out the factor of 2 in the formula gives BC² = 81 + 36 − 22.82 = 94.18, so BC = 9.7 cm. Adding the cosine term instead of subtracting it gives BC² = 81 + 36 + 45.64 = 162.64, so BC = 12.8 cm. Using sin65° in place of cos65° gives BC² = 81 + 36 − 97.88 = 19.12, so BC = 4.4 cm. Keep the factor of 2, subtract the cosine term, and the correct length is 8.4 cm.
- (d) £982.20 — Method: find the missing side BC with the cosine rule, add it to AB and AC for the perimeter, then multiply by the cost per metre. Working: BC² = 45² + 38² − 2 × 45 × 38 × cos 110° = 4638.71, so BC = 68.108 m, perimeter = 45 + 38 + 68.108 = 151.108 m, and cost = 151.108 × £6.50 = £982.20 — keep the unrounded perimeter, because the rounded 151.1 m would give £982.15. Forgetting the negative sign in the cosine rule gives BC = 48.0 m and a cost of £851.18; costing only the missing side BC and forgetting to include AB and AC gives £442.70; and stopping after finding the perimeter, without multiplying by the cost per metre, gives £151.11. The perimeter is only the halfway point of this question — the cost still has to be worked out.
- (c) ∠XYZ — The angle at a named vertex is written with that vertex's letter in the middle, flanked by its two neighbouring vertices. The angle at Y sits between X and Z, its neighbours in quadrilateral WXYZ, so it is written ∠XYZ. ∠WXY names the angle at X, since X is the middle letter, not Y. ∠YZW names the angle at Z, since Z is the middle letter. ∠ZWX names the angle at W, since W is the middle letter.
- (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.
- (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) 1.5 m — By Pythagoras' theorem, diagonal² = 1.2² + 0.9² = 1.44 + 0.81 = 2.25, so diagonal = √2.25 = 1.5 m. 2.1 m comes from simply adding the two sides (1.2 + 0.9) instead of using Pythagoras' theorem. 0.3 m comes from subtracting the two sides (1.2 − 0.9) instead. 2.25 m comes from correctly finding 1.2² + 0.9² = 2.25 but forgetting to take the square root at the end.
- (b) Neither, because both coordinates differ — A line segment is horizontal only when both points share the same y-coordinate, and vertical only when both points share the same x-coordinate. Here A has x-coordinate −4 and B has x-coordinate 2, which differ, and A has y-coordinate 3 and B has y-coordinate 6, which also differ, so the segment is neither horizontal nor vertical. Every point has a y-coordinate and an x-coordinate, so simply having one is not a reason for the line to be horizontal or vertical — both of those wrong reasons ignore that the coordinates must match, not just exist. The x-coordinates do increase from A to B, but an increasing x-coordinate on its own describes a slope, not a horizontal line.
- (a) 80° — In triangle ABD, the three angles sum to 180°: angle BAD = 180° − 35° − 65° = 80°. ABCD is a cyclic quadrilateral, so the exterior angle at C, angle BCE, is equal to the interior angle at the opposite vertex, angle BAD: angle BCE = 80°. Adding the two given angles in triangle ABD instead of subtracting them from 180°, 35° + 65° = 100°, and then applying the exterior-angle rule correctly still gives the wrong angle BAD and so the wrong exterior angle, 100°. Halving the correct angle BAD, 80° ÷ 2 = 40°, confuses the exterior-angle rule with a different circle theorem in which one angle is half of another. Doubling it instead, 80° × 2 = 160°, makes the same kind of confusion in the other direction.
- (a) 8 — A cylinder has two flat circular faces (the top and the base) and one curved surface, so only its 2 flat faces are wrapped. A cuboid has 6 faces and all of them are flat, so all 6 are wrapped. Adding these: 2 + 6 = 8, so 8 is correct. 9 comes from wrongly counting the cylinder's curved surface as a flat face. 6 comes from counting only the cuboid and forgetting the cylinder's two flat circular faces. 3 comes from counting only the cylinder and including its curved surface in that total.
- (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.
- (c) 18 — Rearranging F + V − E = 2 gives E = F + V − 2. Substitute F = 8 and V = 12: 8 + 12 − 2 = 18 edges. Choosing 20 comes from adding the faces and vertices but forgetting to subtract the 2 (8 + 12 = 20). Choosing 22 comes from adding the 2 instead of subtracting it (8 + 12 + 2 = 22). Choosing 16 comes from subtracting 2 twice by mistake (8 + 12 − 2 − 2 = 16).
- (b) Hexagon — Cutting straight across a prism, at right angles to its length, always gives a cross-section that is the same shape as its end faces. The end faces of a hexagonal prism are hexagons (6-sided), so the cross-section is a hexagon. Choosing Pentagon comes from miscounting the sides of the hexagonal end as five instead of six. Choosing Rectangle comes from cutting along the LENGTH of the prism instead of across it, which gives a rectangular face, not the cross-section asked for. Choosing Triangle comes from confusing a hexagonal prism with a triangular prism.
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