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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- (d) 49° — Method: first combine the two given angles at the centre to find the whole central angle FOM, then apply the angle-at-the-centre theorem, which halves the central angle to give the angle at the circumference on the major arc. Working: angle FOM = angle FOP + angle POM = 55 + 43 = 98 degrees. Since S is on the major arc FM, angle FSM = 98 ÷ 2 = 49 degrees. Answer: 49°. Add BOTH given angles to find the whole angle FOM before you halve anything: halving only one of them, doubling the total instead of halving it, or subtracting from 180° all give the wrong angle for the programme.
- (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 — Method: check which condition matches two sides and the angle between them, since that is what has been measured for each bed. Working: the 60° angle is marked at the corner where the 5 m and 7 m edges meet, so it is the included angle — this is two Sides and the included Angle, SAS. Options: SSS would need a third side measured, but only two edges are known; ASA would need two angles and the side between them, but only one angle is measured; RHS needs a right angle, and 60° is not a right angle. Answer: SAS.
- (a) 3 cm² — Area scale factor = (linear scale factor)² = (1/4)² = 1/16. Area of T = 48 × 1/16 = 3 cm². (12 cm² comes from multiplying by the linear scale factor 1/4 directly, without squaring it; 24 cm² comes from taking the square root of the scale factor instead of squaring it; 768 cm² comes from squaring the reciprocal of the scale factor, 4, instead of the scale factor itself.)
- (a) 49 cm² — Method: two sides and the angle between them are given, so use the cosine rule a² = b² + c² − 2bc cos A with BC facing the 60° angle. By hand, cos 60° = 0.5. Working: BC² = 8² + 5² − 2 × 8 × 5 × cos 60° = 64 + 25 − 80 × 0.5 = 89 − 40 = 49. Answer: BC² = 49 cm². The distractors: 129 cm² comes from adding the final term instead of subtracting it, 89 + 40; 89 cm² comes from leaving the cosine term out altogether and treating the 60° as though it were a right angle, so that Pythagoras applies; 69 cm² comes from forgetting the factor 2 in 2bc cos A and subtracting only 8 × 5 × 0.5 = 20.
- (a) (9, −5) — Method: multiply every part of q by 2, then subtract the matching part from p. Working: 2q = (−4, 6); p − 2q gives top 5 − (−4) = 9 and bottom 1 − 6 = −5. Answer: p − 2q = (9, −5). A candidate who forgets to double q first, working out p − q instead, gets (7, −2). A candidate who doubles p instead of q, working out 2p − q, gets (12, −1). A candidate who adds 2q instead of subtracting it gets (1, 7).
- (b) 62° — Because AT is a diameter, the radius OT lies along AT, and the tangent PT meets every radius at 90°, by the tangent–radius theorem. So PT meets AT at T at a right angle: angle ATP = 90°. The angles of triangle APT sum to 180°, so angle APT = 180° − 90° − 28° = 62°. Copying the given angle PAT straight across, as though the triangle were isosceles, gives 28°. Adding the two known angles instead of subtracting them from 180° gives 90° + 28° = 118°. Naming the right angle itself, angle ATP, instead of the angle that was actually asked for gives 90°. Subtract both known angles from 180°, and 62° is angle APT.
- (c) 62.9 cm² — Method: no two sides are given, so first find AC with the sine rule, then find the area using BC, AC and the angle between them, angle ACB. Working: angle BAC = 180° − 58° − 47° = 75°; by the sine rule, AC = 14 × sin 58° / sin 75° = 12.2915 cm; then area = 1/2 × 14 × 12.2915 × sin 47° = 62.9 cm² — keep the unrounded AC, since rounding it to 12.3 cm shifts the area to 63.0 cm². Pairing 14 with sin 75° and dividing by sin 58° instead (the ratio the wrong way round) gives AC = 15.9 cm and an area of 81.6 cm²; using angle BAC = 75° as the included angle instead of angle ACB gives 83.1 cm²; and assuming the triangle is isosceles with AC = BC = 14 cm, skipping the sine rule step entirely, gives 71.7 cm². The angle used in the area formula must be the one between the two sides being multiplied — here that is angle ACB, between BC and AC.
- (a) 188.4 cm² — Curved surface area of a cone = πrl. With r = 6 cm, l = 10 cm and π = 3.14, curved surface area = 3.14 × 6 × 10 = 188.4 cm². A student who uses the cylinder's curved surface area formula, 2πrl, instead of the cone's gets 2 × 3.14 × 6 × 10 = 376.8 cm². A student who uses the circle-area formula πr² instead of πrl gets 3.14 × 36 = 113.04 cm². A student who multiplies r × l but leaves out π entirely gets 6 × 10 = 60 cm².
- (d) £100.48 — Area of the rug = (angle ÷ 360) × π × r² = (180 ÷ 360) × 3.14 × 16 = 0.5 × 50.24 = 25.12 m². Cost = 25.12 × £4 = £100.48. (£200.96 comes from finding the area of a full circle, 3.14 × 16 = 50.24 m², and forgetting the angle fraction before costing it; £25.12 comes from forgetting to square the radius, using 0.5 × 3.14 × 4 = 6.28 m², and then costing that; £50.24 comes from using the arc length formula, 0.5 × 2 × 3.14 × 4 = 12.56, in place of the area, and costing that.)
- (b) (0, 2) — Method: to rotate about a point that is not the origin, first subtract the centre's coordinates, apply the rotation rule to the shifted point, then add the centre's coordinates back on; only after that do you apply the translation, in the order the question states them. Working: shifting P relative to the centre gives (6 − 1, 4 − 2) = (5, 2); rotating 90° clockwise sends (x, y) to (y, −x), giving (2, −5); adding the centre back on gives (2 + 1, −5 + 2) = (3, −3); applying the translation (−3, 5) gives (3 − 3, −3 + 5) = (0, 2). Answer: (0, 2). Applying the translation BEFORE the rotation, reversing the order the question gives them in, gives (8, 0); stopping after the rotation and forgetting the translation altogether gives (3, −3); and rotating anticlockwise instead of clockwise, using (x, y) → (−y, x), gives (−4, 12). Always carry out the two transformations in the order stated — rotate about the given centre first, then translate — and check each step before moving to the next.
- (b) No, it needs one more square — A closed cube has exactly 6 faces, so its net must be made of exactly 6 identical squares, arranged so each one unfolds to a separate face with none overlapping. This net has only 5 squares, so it is one square short and cannot be folded into a closed cube. Choosing 'Yes, it folds into a cube' ignores that a cube needs 6 faces, not 5. Choosing 'No, it has one square too many' miscounts in the wrong direction — 5 is one too FEW, not one too many. Choosing 'Yes, but only if two squares overlap' is not a valid net: a net's faces must not overlap when folded.
- (c) (3, 2) — For an enlargement centred on the origin, multiply every coordinate by the scale factor: (9 × 1/3, 6 × 1/3) = (3, 2). A pupil who multiplies by 3 instead of by 1/3 gets (27, 18). A pupil who subtracts a third of each coordinate instead of scaling by a third gets (9 − 3, 6 − 2) = (6, 4). A pupil who applies the scale factor to the x-coordinate only gets (3, 6). The correct image is (3, 2).
- (d) SAS — two sides, included angle — Each section has two known sides, 3.6 m and 2.4 m, with the 70° angle between them, matching in both sections; this is exactly the SAS condition, so 'SAS — two sides, included angle' is correct. 'SSS — but only two sides given' is wrong because SSS requires three pairs of equal sides, but only two sides are given for each triangle here. 'ASA — angle between two sides' is wrong because ASA requires two angles with a side between them, but only one angle, 70°, is given, not two. 'Cannot prove — only one angle' is wrong because SAS is specifically designed to prove congruence from exactly two sides and the one angle between them, so no further angle is needed.
- (a) 100° — In a kite, the pair of angles between the unequal sides are equal to each other. Angle X and angle Z are both between one side from the WX/WZ pair and one side from the XY/ZY pair, so angle Z = angle X = 100°.
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