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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- (b) Square — A square has all four sides equal, all four angles equal to 90°, and diagonals that are equal in length and bisect each other at right angles — every part of the description matches, so Square is correct. A rhombus has all four sides equal and diagonals bisecting at right angles, but its interior angles are not generally 90° (only a square, a special rhombus, has that), so it does not fully match. A rectangle has four 90° angles and equal diagonals, but its sides are not all equal in general, so it fails the equal-sides condition. A kite has two pairs of adjacent equal sides rather than all four sides equal, and its diagonals are not generally equal in length, so it fails both conditions.
- (b) 12 cm — Method: the area of a rectangle is one side multiplied by the other, so when the area and one side are known the other side is found by reversing that multiplication — divide the area by the side that is known. Working: 96 ÷ 8 = 12. Answer: 12 cm. The distractors: 88 cm comes from 96 − 8, subtracting the known side as though the area had been made by adding the two sides together; 768 cm comes from 96 × 8, running the area rule forwards on the two numbers given instead of reversing it; 40 cm comes from reading the 96 as a perimeter — halving it to 48 and taking the 8 cm side away — which reverses the perimeter rule rather than the area rule.
- (a) 65° — Method: a line that divides an angle into two equal parts gives each part half of the original angle, so halve 130°. Working: 130 ÷ 2 = 65. Answer: 65°. The distractors: 130° is the whole of angle ABC, written down without halving it; 50° comes from working out 180 − 130, using the angles on a straight line instead of dividing the angle in two; 32.5° comes from dividing by 4 instead of by 2, as though the line split the angle into four equal parts.
- (b) 2 cm — Method: the volume of a cube is edge × edge × edge, so finding the edge from the volume means undoing a cube, not a square or a halving. Working: edge × edge × edge = 8; testing whole numbers, 1 × 1 × 1 = 1 is too small and the next whole number gives 2 × 2 × 2 = 8, which matches. Answer: 2 cm. The distractors: 8 cm comes from writing the volume down as the edge length and changing only the unit; 4 cm comes from halving the volume, 8 ÷ 2, as though a cube were undone by dividing by 2; 2.83 cm is the square root of 8, from taking a square root where a cube root is needed.
- (d) SSS, using shared side QS — PQ equals RQ and PS equals RS are two given pairs of equal sides, and QS is common to both triangles, so QS equals itself and gives a third pair of equal sides. Three pairs of equal sides is exactly the SSS condition, so 'SSS, using shared side QS' is correct. 'SAS, using the angle at Q' is wrong because no angle is given anywhere in this question; angle PQS and angle RQS are not stated to be equal, and assuming they are would be assuming the very thing being proved. 'Only two pairs of sides — not enough' is wrong because it forgets that the shared side QS is itself a third pair of equal sides. 'Cannot prove — no angle given' is wrong because SSS is one of the four basic congruence conditions and specifically requires no angle at all.
- (a) 7 — Rearranging F + V − E = 2 gives F = 2 − V + E = 2 − 10 + 15 = 7. A candidate who works out E − V without the +2, giving 15 − 10, answers 5. A candidate who rearranges with a sign error, working out 2 + V − E = 2 + 10 − 15 = −3 and then drops the negative sign, answers 3. A candidate who adds all three numbers together, V + E + 2 = 10 + 15 + 2, answers 27, having used the wrong operation entirely. The correct number of faces is 7.
- (a) (2, 6) — To translate by the vector (−3, 4), add −3 to the x-coordinate and add 4 to the y-coordinate: (5 − 3, 2 + 4) = (2, 6). A pupil who adds 3 instead of subtracting for the x-coordinate gets (8, 6). A pupil who swaps the x- and y-components of the vector gets (5 + 4, 2 − 3) = (9, −1). A pupil who subtracts both components instead of adding the y-component gets (5 − 3, 2 − 4) = (2, −2). The correct image is (2, 6).
- (a) 9 — A triangular prism has two triangular ends and three rectangular side faces. Each triangular end contributes 3 edges, giving 6 edges for both ends, and three more edges run lengthways to join the two ends together: 6 + 3 = 9 edges. Choosing 6 counts only the vertices (3 on each triangular end). Choosing 5 counts the faces (2 triangular + 3 rectangular) instead of the edges. Choosing 12 is the edge count of a cuboid, not a triangular prism.
- (c) 32 — Method: alternate angles between parallel lines are equal, so 2x + 10 = 74. Working: subtracting 10 from both sides gives 2x = 64; dividing by 2 gives x = 32. Answer: x = 32. A candidate who forgets to subtract 10 first and divides 74 by 2 directly gets 37. A candidate who treats the angles as co-interior instead of alternate, so that the two expressions add to 180° rather than being equal, gets 48 after solving. A candidate who makes a sign error and treats the equation as 2x equalling 10 minus 74 instead of 74 minus 10 gets −32.
- (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.)
- (d) (−5, −4) — Method: every vertex of a translated shape moves by the same vector, so find that vector from the one vertex whose image is given, then apply it to A. Working: C(4, 5) moves to (0, −1), so across 0 − 4 = −4 and up −1 − 5 = −6, giving the vector $\binom{-4}{-6}$. Applying it to A(−1, 2): −1 − 4 = −5 and 2 − 6 = −4. Answer: the image of A is (−5, −4). Working the vector out as object minus image gives 4 to the right and 6 up, which applied to A gives (3, 8). Getting the horizontal movement right but reversing the vertical one gives (−5, 8). Treating (0, −1) as the image of every vertex ignores that a translation carries each vertex to a different place.
- (b) 50 cm² — Method: the area of a rectangle is length × width. Working: 2 × 25 = 50. Answer: 50 cm². The distractors: 54 cm comes from working out the perimeter, 2 × (2 + 25), which is a length and not an area; 27 cm comes from adding the two sides, 2 + 25, instead of multiplying them; 25 cm² comes from halving the product, 50 ÷ 2, which is the rule for the area of a triangle rather than of a rectangle.
- (c) 90° — A tangent to a circle always meets the radius drawn to the point of contact at a right angle, so the angle between the tangent and the radius at P is 90°. 180° confuses the tangent with the diameter through P, as if the radius continued in a straight line into the tangent. 45° halves the true angle by mistake. 60° comes from confusing this fact with the angle of an equilateral triangle.
- (b) 70 cm³ — Area of the triangular cross-section = base × height ÷ 2. Base × height = 3.5 × 4 = 14 cm², and half of that is 14 ÷ 2 = 7 cm². Volume of the prism = cross-sectional area × length = 7 × 10 = 70 cm³. A pupil who forgets to halve when finding the triangle's area gets 3.5 × 4 × 10 = 140 cm³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length stops at 7 cm³. A pupil who ignores the height altogether, multiplying base by length, gets 3.5 × 10 = 35 cm³. The correct volume is 70 cm³.
- (b) 25 minutes — Method: a rate in litres per minute can only be used on a volume measured in litres, so convert the tank first and then divide. Working: 1 m³ = 1000 litres, so the tank holds 0.45 × 1000 = 450 litres, and the time is 450 ÷ 18 = 25. Answer: 25 minutes. Using 1 m³ = 100 litres gives 45 ÷ 18 = 2.5 minutes. Using 1 m³ = 1 000 000 litres, which is the factor that turns cubic metres into cubic centimetres, gives 450 000 ÷ 18 = 25 000 minutes. Multiplying by the rate instead of dividing by it gives 450 × 18 = 8100.
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