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.
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Answer key: Geometry and measures worksheet — GCSE Higher
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- (c) 1 — cos 45° = √2/2 and sin 45° = √2/2. Squaring each gives (√2/2)² = 2/4 = 1/2, so (cos 45°)² + (sin 45°)² = 1/2 + 1/2 = 1. The distractor √2 comes from adding cos 45° + sin 45° directly without squaring first (√2/2 + √2/2 = √2). The distractor 2 comes from squaring the top of the fraction, (√2)² = 2, but then dividing by 2 instead of 4 for each term, giving 1 + 1 = 2. The distractor 1/2 comes from squaring only cos 45° and forgetting to add the sin 45° term.
- (a) 10.4 cm — The scale factor from the smaller triangle to the larger triangle is 8 ÷ 5 = 1.6, so the larger side is 6.5 × 1.6 = 10.4 cm. The distractor 4.0625 cm comes from using the ratio the wrong way round, 6.5 × 5 ÷ 8 = 4.0625. The distractor 9.5 cm comes from adding the difference between the ratio numbers (8 − 5 = 3) to the given length, 6.5 + 3 = 9.5. The distractor 13 cm comes from doubling the given length, treating the scale factor as 2 instead of 1.6.
- (d) 310 — Method: since both turns are clockwise, add both angles to the starting bearing. Working: 245° + 50° = 295°; 295° + 15° = 310°. A student who answers 295 has only added the first turn and forgotten the second one. A student who answers 180 has subtracted both turns instead of adding them. A student who answers 320 has added the two turns as 75° instead of 65° by misreading the second turn. Answer: 310°.
- (c) exterior angle = sum of the two interior opposite angles — Angle ACD is the exterior angle at C, and the exterior angle of a triangle is always equal to the sum of the two interior angles at the other two vertices — here, angle BAC and angle ABC — so angle ACD = 48° + 67° = 115°. "angles in a triangle add up to 180°" is a true fact, but on its own it only gives the INTERIOR angle at C (180 − 48 − 67 = 65°), not the exterior angle 115° — reaching 115° that way still needs a further step. "angles on a straight line add up to 180°" is the fact that links the interior and exterior angles at C to each other, not the fact that gives 115° directly from the two OTHER angles. "exterior angle is always double the smallest angle" is not a genuine angle fact: double the smallest angle here is 48° × 2 = 96°, not 115°.
- (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) 13 cm — Use Pythagoras' theorem in three dimensions: for a cuboid with edges a, b and c the space diagonal d satisfies d² = a² + b² + c². Substitute a = 3, b = 4, c = 12: d² = 3² + 4² + 12² = 9 + 16 + 144 = 169. Take the square root: d = √169 = 13 cm. Adding the three edges directly, 3 + 4 + 12 = 19 cm, ignores that Pythagoras' theorem is about squares, not lengths, and gives 19 cm. Stopping after squaring and adding, without taking the square root, leaves 169 cm — the squared length, not the length itself. Using only the 4 cm and 12 cm edges finds the diagonal of one face, √(4² + 12²) = √160 = 12.6 cm (1 d.p.), and leaves out the third dimension entirely.
- (d) (2, −1) — Method: for an enlargement, image = centre + k × (point − centre), so the centre satisfies centre = (image − k × point) ÷ (1 − k). Working: with k = 5, point (4, 1) and image (12, 9): 5 × (4, 1) = (20, 5); (12, 9) − (20, 5) = (−8, 4); dividing by 1 − 5 = −4 gives (2, −1). Answer: (2, −1), the centre of the enlargement, is the only invariant point since the scale factor is not 1. Subtracting the point itself instead of k times the point, (12, 9) − (4, 1) = (8, 8), then dividing by −4 gives (−2, −2); dividing by k − 1 = 4 instead of 1 − k = −4 gives (−2, 1); and simply taking the midpoint of the point and its image ignores the scale factor altogether and gives (8, 5). The centre of an enlargement is never just the midpoint between a point and its image unless the scale factor happens to be −1 — always use the full centre formula and keep the scale factor k in it.
- (d) y = 2 — Two reflections in perpendicular lines that cross at a point combine to a 180° rotation about that point. The line x = 3 is vertical, so the second line must be horizontal, and it must pass through the centre of rotation (3, 2) — that line is y = 2. Taking the y-coordinate of the centre but writing it against the wrong letter gives y = 3. Assuming the second line must also be vertical, like the first one, and just swapping in the other coordinate gives x = 2. Reaching for the standard mirror line y = x without checking that it actually passes through (3, 2) gives y = x — it does not pass through that point at all. The line that is both perpendicular to x = 3 and through (3, 2) is y = 2.
- (a) 8,000,000 cm³ — Method: change the edge length into centimetres first and then cube it, because 1 m = 100 cm and a volume needs that conversion applied to all three dimensions. Working: 2 m = 2 × 100 = 200 cm, so the volume is 200 × 200 × 200. 200 × 200 = 40,000 and 40,000 × 200 = 8,000,000. Answer: 8,000,000 cm³. The distractors: 8,000 cm³ comes from converting 2 m to 20 cm and cubing that; 80,000 cm³ comes from cubing in metres to get 8 m³ and then multiplying by 10,000, the conversion factor for an area rather than the 1,000,000 a volume needs; 8 cm³ comes from cubing the 2 without converting at all and simply writing cm³ because the question asked for that unit.
- (a) 9 — A cube has 9 planes of symmetry in total: 3 that pass through the middles of pairs of opposite faces, and 6 more that pass through pairs of opposite edges diagonally. A candidate who counts only the 3 face-to-face planes — which is correct for a cuboid with three different edge lengths, but forgets that a cube's equal edges create 6 more diagonal planes — answers 3. A candidate who counts only the 6 diagonal planes and forgets the 3 face-to-face ones answers 6. A candidate who confuses the number of planes of symmetry with the number of edges on a cube answers 12. The correct total for a cube is 9.
- (b) 20 cm — Method: corresponding sides of similar triangles are in the same ratio, and the longest side of one triangle corresponds to the longest side of the other; a ratio of 2 : 5 means each length is multiplied by 5 ÷ 2 = 2.5 going from the smaller triangle to the larger one. Working: the longest side of the smaller triangle is 8 cm, so the matching side of the larger triangle is 8 × 2.5 = 20. Answer: 20 cm. The distractors: 10 cm comes from scaling the shortest side, 4 cm, instead of the longest; 40 cm comes from multiplying by 5 and forgetting to divide by 2; 3.2 cm comes from multiplying by 2 ÷ 5 instead of 5 ÷ 2, which scales from the larger triangle down to the smaller one.
- (c) 9 : 16 — Method: in similar solids the ratio of the surface areas is the square of the ratio of corresponding lengths. Working: the radii are in the ratio 3 : 4, so the surface areas are in the ratio 3² : 4² = 9 : 16. Answer: 9 : 16. The distractors: 27 : 64 is 3³ : 4³, the ratio of the volumes, which cubes the length ratio instead of squaring it; 3 : 4 leaves the length ratio untouched, as though surface area scaled in the same way as a length; 81 : 256 comes from squaring a second time, applying the rule to the ratio 9 : 16 rather than to the radii.
- (b) (5, 1) — First scale a by 2: 2a = (2×3, 2×(−2)) = (6, −4). Then add b component by component: (6+(−1), −4+5) = (5, 1). (2, 3) is a + b without doubling a first. (4, 6) doubles both a and b instead of only a. (7, −9) subtracts b from 2a instead of adding it.
- (a) SSS – all three corresponding sides are equal — All three pairs of corresponding sides are stated as equal — LM = XY, MN = YZ and LN = XZ — with no angle mentioned. This matches the SSS condition, so triangle LMN is congruent to triangle XYZ.
- (b) (1/2)b − (1/2)a — Method: MN runs from M to N, so MN = ON − OM, with OM = (1/2)a and ON = (1/2)b. Working: MN = (1/2)b − (1/2)a. Answer: MN = (1/2)b − (1/2)a. Subtracting the other way round gives (1/2)a − (1/2)b, the reverse vector from N to M; subtracting the wrong way round AND forgetting to halve gives a − b, which is BA, not MN; and adding the two halved vectors instead of subtracting them gives (1/2)a + (1/2)b, which is the position vector of the midpoint of AB. Always subtract the START point's vector from the END point's vector, and halve OA and OB before you combine them, not after.
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