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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- (c) 100√3 cm — The height is opposite the 60° angle, so height = 2 × sin 60° = 2 × √3/2 = √3 m. Converting to centimetres: √3 m = 100√3 cm. '√3 cm' forgets to convert the answer from metres to centimetres. '200√3 cm' comes from mis-recalling sin 60° as √3 instead of √3/2, dropping the denominator of the exact value: 2 × √3 = 2√3 m = 200√3 cm. '50√3 cm' comes from halving the ladder's length before multiplying by sin 60°, instead of using the full 2 m.
- (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.
- (d) 1 — An isosceles trapezium has one line of symmetry, running through the midpoints of the two parallel sides. 0 would be true for a scalene trapezium, whose non-parallel sides are unequal, but this trapezium's non-parallel sides are equal, so it is symmetrical. 2 is the number of lines of symmetry of a rectangle, not a trapezium. 4 comes from wrongly applying the 'number of sides equals number of lines of symmetry' rule, which only holds for REGULAR polygons — a trapezium is not regular.
- (b) x = 5 — Every point on a vertical line has the same x-coordinate, so the equation of a vertical line through (5, 2) is x = 5. "y = 5" mixes up the coordinates, using the x-value of 5 to write a y-equation. "y = 2" is the equation of the horizontal line through (5, 2), not the vertical one. "x = 2" uses the correct letter but the wrong coordinate, the y-value of 2 instead of the x-value of 5.
- (d) 5 — A square-based pyramid has four vertices at the corners of the square base, plus one more vertex at the apex where the four triangular faces meet: 4 + 1 = 5 vertices. Choosing 8 counts the edges instead of the vertices. Choosing 4 counts only the base corners and forgets the apex at the top. Choosing 6 is the vertex count of a triangular prism, not a square-based pyramid.
- (d) SSS — All three pairs of corresponding sides are equal (PQ = ST, QR = TU, PR = SU), so the triangles are congruent by the SSS (side-side-side) condition. The distractor SAS would apply if two sides and the included angle were given equal, not three sides. The distractor ASA would apply if two angles and the included side were given equal. The distractor RHS would apply only for right-angled triangles with the hypotenuse and one other side equal.
- (a) (5, 7) — Method: a midpoint is the mean of the two end points, so for each coordinate (start + end) ÷ 2 = midpoint; rearranging that gives end = 2 × midpoint less the start. Working: for x, (1 + x) ÷ 2 = 3, so 1 + x = 6 and x = 5. For y, (3 + y) ÷ 2 = 5, so 3 + y = 10 and y = 7. B is therefore (5, 7). Answer: (5, 7). The distractors: (2, 2) comes from subtracting A from the midpoint, (3 − 1, 5 − 3), which gives the step from A to the midpoint and stops there instead of taking that same step a second time; (4, 8) comes from adding A to the midpoint, (3 + 1, 5 + 3), without doubling the midpoint first; (6, 10) comes from doubling the midpoint, (2 × 3, 2 × 5), and then forgetting to take A off.
- (d) 77 — Method: angles that meet at a point add up to 360°. Working: 82 + 105 + 96 = 283; 360 − 283 = 77. Answer: 77°. A candidate who gives the sum of the three known angles and forgets to subtract it from 360° gets 283. A candidate who leaves out the 105° angle, working out 360 − 82 − 96, gets 182. A candidate who leaves out the 82° angle, working out 360 − 105 − 96, gets 159.
- (b) £157 — Method: round the hours worked up to the next whole hour, multiply by the hourly rate, then add the call-out fee. Working: 3 hours 30 minutes rounds up to 4 hours; 28 × 4 = 112; 112 + 45 = 157. Answer: £157. A candidate who uses the unrounded time of 3.5 hours, working out 28 × 3.5 = 98 and adding 45, gets £143. A candidate who forgets the call-out fee, giving only 28 × 4, gets £112. A candidate who rounds down to 3 hours instead of up, working out 28 × 3 = 84 and adding 45, gets £129.
- (b) A triangular prism standing on its triangular end — Method: a rectangular elevation with no sloping sides means the solid keeps the same cross-section all the way from the bottom to the top; work out which solid, standing the right way up, has a triangular cross-section that stays that shape as you go higher. Working: a triangular prism standing upright on its triangular end has a triangle as its plan view, and because the cross-section is constant all the way up, both the front and side elevations are plain rectangles. Answer: a triangular prism standing on its triangular end. The distractors: a triangle-based pyramid standing on its triangular base does give a triangle as its plan view, but its cross-section shrinks towards the apex, so its front and side elevations come to a point and are triangles, not rectangles. A cuboid standing on a rectangular face is wrong because its plan view is a rectangle, not a triangle. A triangular prism lying on one of its rectangular faces is wrong because it is then the triangular end that faces the side, so its plan view is a rectangle and one of its elevations is a triangle.
- (b) 2.8 m — Use Pythagoras' Theorem: the plank is the hypotenuse (3.5 m) of a right-angled triangle formed with the wall and the ground (2.1 m). height² = 3.5² − 2.1² = 12.25 − 4.41 = 7.84. height = √7.84 = 2.8 m. A student who subtracts the two given lengths directly instead of using Pythagoras gets 3.5 − 2.1 = 1.4 m. A student who doubles the distance from the wall by mistake gets 2.1 × 2 = 4.2 m.
- (b) 045° — North is 000° and east is 090°; halfway between them, measured clockwise, is 045°. Choosing 090° gives east itself, not the midpoint between north and east. Choosing 135° gives the midpoint between east (090°) and south (180°) instead of between north and east. Choosing 225° is the back bearing of 045° (045° + 180°), not the bearing itself.
- (b) (3, −2) — A 90° clockwise rotation about the origin maps (x, y) to (y, −x): (2, 3) → (3, −2). A pupil who uses the rule for a 90° anticlockwise rotation instead, (x, y) → (−y, x), gets (−3, 2). A pupil who uses the rule for a 180° rotation, (x, y) → (−x, −y), gets (−2, −3). A pupil who swaps the coordinates but forgets to change any sign gets (3, 2). The correct image is (3, −2).
- (a) 5 — For any regular polygon, the order of rotational symmetry is always equal to the number of sides, which is also equal to the number of lines of symmetry. Since this polygon has 5 lines of symmetry, it has 5 sides, so its order of rotational symmetry is 5. 4 comes from subtracting one from the number of sides by mistake. 6 comes from adding one to the number of sides by mistake. 10 comes from doubling the number of lines of symmetry instead of using it directly.
- (d) 12 — Method: rearrange F + V − E = 2 so that E is on its own: E = F + V − 2. Working: E = 6 + 8 − 2 = 12. A student who answers 14 has added F and V but forgotten to subtract 2 at all. A student who answers 16 has added 2 instead of subtracting it. A student who answers 10 has subtracted 2 twice by mistake. Answer: 12 edges.
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