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.
Answer key: Geometry and measures worksheet — GCSE Foundation
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- (a) 5 — Looking straight down on a row of 5 cubes standing side by side, each cube contributes exactly one square to the view from above, since the cubes do not overlap and none is hidden behind another — so the plan shows 5 squares in a row. "1" comes from treating the whole row as a single block instead of counting each cube. "10" comes from doubling the count, perhaps by also counting a front elevation's squares alongside the plan's. "25" comes from squaring the number of cubes (5 × 5) instead of counting them.
- (a) Similar, but AAA alone does not prove congruence — Three equal corresponding angles (AAA) show that the two triangles are similar — the same shape — but says nothing about their size, so it does not prove congruence. They could be congruent, or one could simply be an enlargement of the other; without matching side lengths, congruence is not established, so 'congruent because AAA proves congruence' is wrong. Equal angles do not force equal sides — a triangle can be enlarged to any size while keeping the same angles, so that option is also wrong. Something CAN be said here — that the triangles are similar — so 'no relationship can be determined' is wrong too.
- (d) (6, 3) — For an enlargement centred at the origin, each coordinate is multiplied by the scale factor: (2, 1) → (2 × 3, 1 × 3) = (6, 3). ((5, 4) comes from adding the scale factor to each coordinate instead of multiplying; (6, 1) comes from multiplying only the x-coordinate by 3 and leaving the y-coordinate unchanged; (2, 3) comes from multiplying only the y-coordinate by 3 and leaving the x-coordinate unchanged.)
- (a) 28 cm — Method: the perimeter is the total distance round the outside, and a rhombus has four sides of equal length, so the perimeter is 4 × the side length. Working: 4 × 7 = 28. Answer: 28 cm. The distractors: 14 cm comes from 2 × 7, adding only one pair of sides and forgetting that a rhombus has two pairs; 49 cm comes from working out 7 × 7, which is the calculation for the area of a square rather than a distance round the outside; 11 cm comes from adding the side length to the number of sides, 7 + 4, instead of multiplying them.
- (c) A↔N, B↔L, C↔M (ABC≅NLM) — Matching equal side lengths: AB (8 cm) equals NL (8 cm), BC (10 cm) equals LM (10 cm), and CA (6 cm) equals MN (6 cm). This gives the correspondence A with N, B with L, and C with M, so triangle ABC is congruent to triangle NLM, making 'A↔N, B↔L, C↔M (ABC≅NLM)' correct. 'A↔L, B↔M, C↔N (ABC≅LMN)' simply matches the vertices in the order they are written without checking the side lengths: AB (8 cm) would need to equal LM (10 cm), which is false. 'A↔M, B↔N, C↔L (ABC≅MNL)' also fails this check, since AB (8 cm) would need to equal MN (6 cm), which is false. 'A↔N, B↔M, C↔L (ABC≅NML)' gets A correct but swaps B and C, so AB (8 cm) would need to equal NM (6 cm), which is also false.
- (d) £48 — Scale factor = new width ÷ original width = 40 ÷ 10 = 4. Poster height = 15 × 4 = 60 cm. Cost = 60 × £0.80 = £48. (£12 comes from forgetting to scale the height at all, and pricing the original 15 cm height; £15.20 comes from adding the scale factor 4 to the height instead of multiplying by it; £3 comes from dividing the height by the scale factor instead of multiplying by it.)
- (b) SAS (two sides and the included angle) — PQ = ST and QR = TU are two pairs of matching sides, and the angle between them (angle Q and angle T) is 90° in both triangles — two sides and the angle between them are equal, so the triangles are congruent by SAS, using only the measurements given. RHS needs the hypotenuses to be known equal; here only the two legs are given, so you would first have to work out each hypotenuse by Pythagoras' theorem — extra working the question rules out. SSS has the same problem: the third side of each triangle is not given, only calculable. ASA needs two angles and the side between them, but here it is two sides and the angle between them that are given, not two angles.
- (a) 62° — Angle KPQ and angle MQP are co-interior (allied) angles between parallel lines, so they add up to 180°: angle MQP = 180° − 118° = 62°. 118° comes from treating the angles as equal, as if this were a corresponding or alternate angle pair, instead of using the co-interior rule. 90° comes from wrongly assuming the crossing line meets JK and LM at right angles. 59° comes from halving 118° instead of subtracting it from 180°.
- (d) 0 — A vertex is a sharp corner point where edges meet. A cylinder has two curved, circular edges but no sharp corner points at all, so it has 0 vertices, making that the correct answer. '2' wrongly treats the two circular edges themselves as vertices, but an edge is not the same as a vertex. '4' overcounts by treating each circular edge as if it had two end-vertices, which does not apply to a continuous curved edge. '1' wrongly imagines the curved surface itself forming a single corner point, which it does not.
- (c) 67° — Angles on a straight line add up to 180°. Add the two known angles: 48° + 65° = 113°. Subtract from 180°: 180° − 113° = 67°.
- (a) 75° — The three angles in the triangle formed by the two supports and the ground add up to 180°. Two of the angles are 34° and 71°, so the third angle is 180 − 34 − 71 = 75, which is 75°. 105° comes from adding the two given angles together instead of subtracting them from 180° (34 + 71 = 105). 146° comes from computing 180 − 34 = 146 and forgetting to also subtract 71°. 37° comes from subtracting the two given angles from each other instead of using the triangle's angle sum (71 − 34 = 37).
- (d) A rectangle — Method: to find the plan view, work out the outline traced when looking straight down onto the solid from directly above, not the outline shown in the angled sketch. Working: this solid has two identical flat round ends joined by one curved surface, and it is lying on its side rather than standing upright; viewed from above, the curved surface gives two straight edges running the full length of the solid, the width of the round ends apart, and each flat round end — seen edge-on from directly above — also becomes a straight edge of that same width, with no curve remaining. Four straight edges, with opposite sides equal and meeting at right angles, form a rectangle. Answer: a rectangle. The distractors: a circle comes from picturing the solid as if it were standing upright on one of its flat ends, giving the plan of an upright version instead of working out the plan of the solid as it actually lies. An oval comes from copying the foreshortened shape of a round end as it is drawn in the angled sketch, instead of working out the true shape seen from directly above, which has no such foreshortening. A rectangle with rounded ends comes from carrying the curve of the round ends over into the plan view, when in fact a flat round end viewed edge-on from directly above shows no curve at all, only a straight edge.
- (c) £7.85 — Arc length = (90 ÷ 360) × 2 × 3.14 × 10 = 0.25 × 62.8 = 15.7 cm. Cost = 15.7 × £0.50 = £7.85. (£31.40 comes from finding the full circumference and forgetting the angle fraction; £15.70 comes from using the diameter, 20 cm, in place of the radius; £157.00 comes from multiplying the arc length by the radius instead of by the cost per centimetre.)
- (c) 5 hours — The distance from the origin to (7, 24) is √(7² + 24²) = √(49 + 576) = √625 = 25 km. Travelling at 5 km per hour, the time taken is 25 ÷ 5 = 5 hours. 25 hours mistakes the distance itself for the time, forgetting to divide by the speed. 125 hours comes from multiplying the distance by the speed, 25 × 5, instead of dividing. 0.2 hours comes from dividing the speed by the distance, 5 ÷ 25, the wrong way round.
- (b) (3, −3) — Subtracting column vectors means subtracting the top numbers and subtracting the bottom numbers, in the order given: top = 5 − 2 = 3, bottom = 1 − 4 = −3, giving (3, −3). A candidate who works out v − u instead of u − v, reversing the order, gets (−3, 3). A candidate who makes a sign error on the bottom number, treating 1 − 4 as 3 instead of −3, gets (3, 3). A candidate who adds the vectors instead of subtracting gets (7, 5). Because the question asks for u − v, not v − u, the correct answer is (3, −3).
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