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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- (d) 15 cm — Method: any two sides of a triangle must together be longer than the third, so the third side must be longer than the difference of the two given sides and shorter than their sum. Working: the difference is 20 − 8 = 12 cm and the sum is 20 + 8 = 28 cm, so the third side must be between 12 cm and 28 cm, and 15 cm lies inside that range. Answer: 15 cm. The distractors: 12 cm is exactly the difference, so the three lengths would lie flat along a straight line and never close into a triangle; 5 cm is shorter than the difference — 5 + 8 = 13 cm cannot reach across the 20 cm side — and is chosen by candidates who check no lower limit at all; 30 cm is longer than the sum of the other two, so those two sides could never meet, and it is chosen by candidates who check no upper limit.
- (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) 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.
- (a) a line parallel to both, 3 cm from each — Being equidistant from two parallel lines 6 cm apart means being exactly halfway between them all along their length, tracing out a third line, parallel to both, at 3 cm from each — half of the 6 cm gap. "a line parallel to both, 6 cm from each" repeats the full gap instead of halving it, which puts those points past one of the lines entirely. "a circle of radius 3 cm, centred midway" applies to a locus equidistant from a single fixed POINT, not from two parallel lines running the full length. "the perpendicular bisector of the gap" crosses the gap at right angles and meets each line at only one point — it is not the whole locus, which runs parallel to the lines, not across them.
- (a) $\binom{1}{6.5}$ — After the first flight, the drone is at (3.5 − 6.2, −2 + 4.5) = (−2.7, 2.5). The second vector takes it from (−2.7, 2.5) to (−1.7, 9): subtract the coordinates, (−1.7 − (−2.7), 9 − 2.5) = (1, 6.5). The distractor $\binom{−4.4}{6.5}$ comes from treating the drone's position after the first flight as (2.7, 2.5) instead of (−2.7, 2.5), giving −1.7 − 2.7 = −4.4 for the top number. The distractor $\binom{−5.2}{11}$ comes from finding the vector straight from the start point (3.5, −2) to (−1.7, 9), ignoring the first flight altogether. The distractor $\binom{−1}{−6.5}$ comes from subtracting the wrong way round, (−2.7 − (−1.7), 2.5 − 9), which reverses both signs of the correct vector.
- (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.
- (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.
- (c) 500 cm² — The area of a triangle is half of base × height. First, base × height = 40 × 25 = 1,000. Half of 1,000 is 500 cm². 1,000 cm² forgets to halve and just gives base × height. 65 cm² adds the base and height together instead of multiplying them. 2,000 cm² doubles base × height instead of halving it.
- (a) the lens-shaped region where the two circles overlap — The region within 30 km of the ship is a circle of radius 30 km around the ship; the region within 20 km of the lighthouse is a circle of radius 20 km around the lighthouse. Since 30 + 20 = 50 km is more than the 40 km between them, and 30 − 20 = 10 km is less than 40 km, the two circles genuinely overlap, in a lens-shaped region where both conditions hold at once. "the whole smaller circle, since it fits inside the larger one" would only be true if the 40 km distance between the centres were small enough for one circle to swallow the other, which it is not here. "no overlap — 30 km and 20 km don't add to 40 km" wrongly assumes the two radii must add up to exactly the distance between the centres for an overlap to exist; any total greater than that distance is enough. "a straight strip, exactly halfway between the two" confuses this with an equidistant locus, which is a different problem from the one asked here.
- (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.
- (a) 32 m — The scale factor from the larger pond to the smaller pond is 4 ÷ 11, so the smaller perimeter is 88 × 4 ÷ 11 = 32 m. The distractor 242 m comes from using the ratio the wrong way round, 88 × 11 ÷ 4 = 242. The distractor 84 m comes from subtracting the smaller ratio number, 88 − 4 = 84, instead of scaling. The distractor 121 m comes from multiplying 11 × 11 = 121, ignoring the given perimeter altogether.
- (b) 6 cm — Method: the right angle is at A, so BC is the hypotenuse and AC is the side opposite the angle at B; sin B = opposite ÷ hypotenuse therefore gives AC ÷ BC = 3/5. Working: AC ÷ 10 = 3/5, so AC = 10 × 3 ÷ 5 = 6. Answer: 6 cm. The distractors: 8 cm is AB, the side next to the angle at B, which is what cos B = 4/5 produces — the right method used on the wrong side; 3 cm comes from reading the 3 in the ratio as a length and never scaling it up to the 10 cm hypotenuse; 30 cm comes from multiplying by 3 and forgetting to divide by 5.
- (d) 2 — Method: if f is 3 times e, then each part of f equals 3 times the matching part of e. Working: using the bottom numbers, 6 = 3 × k, so k = 2. Answer: k = 2. A candidate who multiplies instead of dividing, working out 6 × 3, gets 18. A candidate who uses the top numbers' ratio instead, 15 ÷ 5, and gives that ratio as k gets 3. A candidate who adds instead of using the multiple relationship, working out 6 + 3, gets 9.
- (b) AB = DE — RHS needs a right angle, the hypotenuse, and one OTHER side to be equal; the right angles and hypotenuses are already equal, so a matching pair of the remaining sides, AB = DE, completes RHS. Angle A = angle D is an extra ANGLE fact, not the extra SIDE fact that RHS specifically requires. AC being parallel to DF says nothing about either triangle's side lengths, so it cannot complete a congruence condition. Being drawn the same way up is about orientation on the page, not about any measurement, so it proves nothing about congruence.
- (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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