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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- (b) 22 — Method: add the two drawn lengths together first, then apply the scale to the total. Working: 3 cm + 2.5 cm = 5.5 cm; 5.5 cm × 4 = 22 m. A student who answers 10 has only converted one of the two sections (2.5 cm × 4) and forgotten the other. A student who answers 5.5 has added the two drawn lengths but forgotten to apply the scale at all. A student who answers 44 has doubled the correct answer, effectively applying the scale twice. Answer: 22 m.
- (b) 8 cm — Area = base × height, so height = area ÷ base = 136 ÷ 17 = 8 cm. (16 cm comes from using the triangle formula, thinking area = 1/2 × base × height and so height = 2 × area ÷ base; 2312 cm comes from multiplying the area by the base instead of dividing; 119 cm comes from subtracting the base from the area instead of dividing the area by the base.)
- (b) Rotate 180° about the origin, then translate by (6, 0). — Rotating 180° about the origin sends (x, y) to (−x, −y); applied to T's vertices (1, 1), (3, 1) and (1, 4) this gives (−1, −1), (−3, −1) and (−1, −4). Translating this image by the vector (6, 0) adds 6 to every x-coordinate, giving (5, −1), (3, −1) and (5, −4), which matches T′ exactly. Reflecting in the x-axis first changes the sign of the y-coordinate only, and translating that image by (6, 0) gives (7, −1), (9, −1) and (7, −4) — the wrong triangle. Using the correct rotation but translating by (4, 0) instead of (6, 0) gives (3, −1), (1, −1) and (3, −4), shifted 2 units too far left. Reflecting in the y-axis first changes the sign of the x-coordinate only, so translating that image by (6, 0) leaves every y-coordinate positive, giving (5, 1), (3, 1) and (5, 4) — the correct x-coordinates but the wrong sign throughout on y.
- (d) 7.7 m — The horizontal distance is adjacent to the 50° angle and the zip wire is the hypotenuse, so horizontal distance = 12 × cos 50° = 12 × 0.6428... = 7.71...≈ 7.7 m. "9.2 m" uses the sine ratio instead of cosine, 12 × sin 50° = 9.19...≈ 9.2 m, which actually finds the vertical drop of the zip wire, not the horizontal distance. "15.7 m" comes from dividing by the sine ratio instead of multiplying by the cosine ratio, 12 ÷ sin 50° = 15.66...≈ 15.7 m, both the wrong operation and the wrong ratio. "12.0 m" simply uses the length of the zip wire itself as the horizontal distance, ignoring the angle of 50° altogether.
- (a) 7.2 cm — The distance around the mat is the circumference, π × diameter = 3.14 × 20 = 62.8 cm. Yasmin started with 70 cm, so the ribbon left over is 70 − 62.8 = 7.2 cm. 62.8 cm is the circumference itself, the amount of ribbon used, not what is left over. 50 cm comes from subtracting the diameter instead of the circumference: 70 − 20 = 50. 20 cm is simply the diameter of the mat and involves no calculation with the ribbon length at all.
- (b) It is the longest of the three chords — Method: in any circle the length of a chord is decided by how far the chord lies from the centre, because a chord passing nearer the centre cuts further across the circle. Working: the chord through O lies at a distance of zero from the centre, and no chord can lie closer than that, so no chord of the circle can be longer than it; a chord through the centre is a diameter. The other two chords lie at some distance greater than zero, so each of them falls short of that maximum. Answer: It is the longest of the three chords. The distractors: It is the shortest of the three chords comes from reversing the rule and picturing a chord near the centre as a short line tucked inside; It is the same length as the other two chords comes from carrying the fact that all radii of a circle are equal across to chords, which are not all equal; It is half the length of each of the other two chords comes from confusing a chord through the centre with a radius, which really is half a diameter.
- (a) 24 — The display is 4 tins wide, 3 tins high and 2 tins deep, and every position in that block is filled, so the total number of tins is the product of all three measurements: 4 × 3 × 2 = 24. "12" comes from multiplying only the width and height shown in the front elevation (4 × 3), forgetting the depth entirely. "9" comes from adding the three measurements (4 + 3 + 2) instead of multiplying them. "6" comes from multiplying only the height and depth (3 × 2), forgetting the width shown by the front elevation.
- (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.
- (c) 9.9 cm — Method: the 7 cm side is opposite the 45° angle and the hypotenuse is wanted, so use sin θ = opposite ÷ hypotenuse and rearrange it for the hypotenuse. Working: sin 45° = 7 ÷ h, so h = 7 ÷ sin 45° = 9.899…, which is 9.9 to 1 decimal place. Answer: 9.9 cm. The distractors: 5.0 cm comes from multiplying by sin 45° instead of dividing by it; 14.0 cm comes from doubling the 7 cm side, which is the rule for a side opposite 30° and not one opposite 45°; 7.0 cm comes from reading the two equal sides of a 45° right-angled triangle as including the hypotenuse, when the equal pair is the two shorter sides.
- (c) 38° — The interior angles of a hexagon sum to (6 − 2) × 180° = 720°. 130° + 142° + 125° + 150° + 135° = 682°. x = 720° − 682° = 38°. A student who uses 6 × 180° instead of (6 − 2) × 180°, forgetting to subtract the 2, gets 1080° − 682° = 398°. A student who mis-adds the five given angles as 680° instead of 682° gets x = 720° − 680° = 40°. A student who answers with one of the given angles instead of solving for x might write 142°.
- (b) a rectangle — Lying on its side, the cylinder's curved surface touches the table along a straight line, and the two flat circular ends face sideways rather than up or down; viewed from directly above, the outline traced is a rectangle — as long as the cylinder and as wide as its diameter. "a circle" would be correct if the cylinder stood upright on one of its circular ends instead of lying on its side. "a triangle" belongs to a cone lying or standing so that it narrows to a point in that view, which a cylinder never does. "an oval" is a common guess from picturing the round ends, but from directly above those ends are edge-on and contribute to the rectangle's short sides, not a curved outline.
- (c) 15 cm — Method: going from the smaller tile to the larger one, every length is multiplied by the scale factor 5 ÷ 3. Working: 9 ÷ 3 = 3, and 3 × 5 = 15. Answer: 15 cm. The distractors: 11 cm comes from adding the difference between the parts of the ratio, 5 − 3 = 2, to the 9 cm side, treating a ratio as a gap rather than a multiplier; 45 cm comes from multiplying by 5 and forgetting to divide by 3; 27 cm comes from multiplying by 3, which is the part of the ratio belonging to the smaller tile.
- (b) 340 — The real length of the path is 8.5 × 2000 = 17000 cm, which converts to 170 m by dividing by 100. The cyclist rides the path there and back, so the total distance is 170 × 2 = 340 m. A candidate who forgets the return journey gives only the one-way distance, 170 m. A candidate who halves the one-way distance instead of doubling it, reading “there and back” as splitting the journey, gets 85 m. A candidate who doubles the scale factor to 4000 by mistake, getting a one-way length of 340 m, and then doubles that for the return journey, gets 680 m. The total distance the cyclist travels is 340 m.
- (d) 7 cm — Area of a parallelogram = base × height, so height = area ÷ base = 84 ÷ 12 = 7 cm. A pupil who multiplies instead of dividing gets 84 × 12 = 1008 cm. A pupil who divides the base by the area instead of the area by the base gets 12 ÷ 84 ≈ 0.14 cm. A pupil who mistakenly halves the area first, as if this were a triangle, gets (84 ÷ 2) ÷ 12 = 3.5 cm. The correct height is 7 cm.
- (d) 10.3 m — By Pythagoras' theorem, the hypotenuse = √(5² + 9²) = √(25 + 81) = √106 = 10.29...≈ 10.3 m. "106 m" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "14 m" comes from adding the two shorter sides, 5 + 9, instead of using Pythagoras' theorem at all. "10.2 m" comes from rounding 10.29...m down to 10.2 instead of correctly rounding it up to 10.3.
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