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) 3200 — Method: to change cubic metres to litres, multiply by 1000. Working: 3.2 × 1000 = 3200. Answer: 3200 litres. A candidate who multiplies by 100 instead of 1000 gets 320. A candidate who multiplies by 10000 instead of 1000 gets 32000. A candidate who does not convert at all gives 3.2.
- (b) A right-angled triangle — The right angle at Y makes this a right-angled triangle, so that description is correct. Since XY is twice YZ, those two sides cannot be equal. The third side XZ is opposite the right angle, so it is the hypotenuse and is longer than either XY or YZ, so it cannot equal either of them. No two sides are equal, which rules out 'an isosceles triangle' and 'a right-angled isosceles triangle', both of which wrongly assume two equal sides. 'An equilateral triangle' would need all three sides equal, which contradicts XY being twice YZ, so it is wrong too.
- (b) 21 — Both legs of the journey are on the same bearing, 070°, so the ship travels in one straight line the whole way and the distances simply add: 12 + 9 = 21 km. A candidate who subtracts instead of adding gets 12 − 9 = 3 km. A candidate who multiplies the two distances gets 12 × 9 = 108. A candidate who assumes the ship changed direction and treats the two legs as the sides of a right-angled triangle works out √(12² + 9²) = √225 = 15 km — but the bearing does not change, so there is no triangle and no hypotenuse to find. Because the ship stays on one straight line, the total distance is 21 km.
- (c) AB ∥ CD only; EF not confirmed — By convention, lines marked with the same number of arrows are parallel to each other, but lines marked with a different number of arrows belong to a different, unrelated family of parallel lines. AB and CD both have a single arrow, so AB is parallel to CD. EF has a double arrow, showing it is not part of the same family as AB and CD; it may be parallel to some other line marked with a double arrow, but nothing here confirms it is parallel to AB or CD, so 'AB ∥ CD only; EF not confirmed' is correct. 'AB, CD and EF are all parallel' and 'EF is parallel to AB' both wrongly treat every arrow mark as showing the same relationship. 'None of the lines are parallel' wrongly assumes a different arrow count rules out any parallel relationship at all, when it actually just signals a different pairing.
- (c) ∠XYZ — The angle at a named vertex is written with that vertex's letter in the middle, flanked by its two neighbouring vertices. The angle at Y sits between X and Z, its neighbours in quadrilateral WXYZ, so it is written ∠XYZ. ∠WXY names the angle at X, since X is the middle letter, not Y. ∠YZW names the angle at Z, since Z is the middle letter. ∠ZWX names the angle at W, since W is the middle letter.
- (c) 18 — Method: divide the real length by 5 to find how many 'units' of 5 m it contains, then multiply by 2 cm for each unit. Working: 45 ÷ 5 = 9, so the real bridge is 9 lots of 5 m; each lot is represented by 2 cm on the model, so the model length is 9 × 2 = 18 cm. Options: 9 comes from stopping after the division, without multiplying by the 2 cm per unit; 90 comes from multiplying the real length by 2 directly, without dividing by 5 first; 4.5 comes from dividing by 5 and then dividing by 2 again, instead of multiplying by 2. Answer: 18.
- (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.
- (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) No — the angle is not between the two sides — Method: check whether the given angle sits between the two given sides. Working: the 35° angle is marked away from the corner where the 40 cm and 65 cm edges meet, so it is not the included angle — this is SSA, which is not one of the four basic congruence conditions, so it does not guarantee congruence. Options: 'Yes — SAS' wrongly treats any two sides plus any angle as SAS, without checking the angle's position; 'Yes — SSS' wrongly counts the angle as if it were a third side; 'No — only two sides measured' is not the real reason, since SAS itself only needs two sides, so this reasoning is beside the point. Answer: no, the angle is not between the two sides.
- (c) AB, AC and angle BAC: Area = 1/2 × AB × AC × sin(BAC) — Method: Area = 1/2ab sin C only works when the angle used is the one included between the two sides being multiplied. Working: AB and AC meet at A, and angle BAC is the angle at A between them, so the statement pairing AB, AC and angle BAC is the correct one. The statement that three sides with no angle can still go into 1/2 AB × AC × sin(BAC) is wrong: with no angle known, sin(BAC) cannot be evaluated, so a different method must find an angle first. The statement pairing AB and BC with sin(BAC) is wrong: AB and BC meet at B, so the angle between them is angle ABC, not angle BAC — it names the wrong angle for the sides it uses. The statement pairing AB and AC with sin(ABC) is wrong for the same reason: AB and AC meet at A, so their included angle is angle BAC, and angle ABC is not between them at all.
- (b) 12 cm — Method: in similar shapes every length is multiplied by the same scale factor, and a ratio of 1 : 3 means that factor is 3 going from the smaller shape to the larger one. Working: 4 × 3 = 12. Answer: 12 cm. The distractors: 7 cm comes from adding 3 to the side instead of multiplying by it, which is what happens when a ratio is read as a difference; 36 cm comes from multiplying by 3² = 9, the factor that scales areas, and applying it to a length; 4 cm comes from treating the two shapes as congruent, so that corresponding sides stay equal — similar shapes have equal angles, but their sides are in proportion.
- (b) 7 cm — The side opposite the 30° angle is found using sin 30° = opposite/hypotenuse, so opposite = 14 × sin 30° = 14 × 1/2 = 7 cm. 7√3 cm comes from using cos 30° = √3/2 instead of sin 30° (mixing up the opposite and adjacent sides). 14/√3 cm comes from using tan 30° = 1/√3 instead of sin 30°. 28 cm comes from dividing 14 by sin 30° instead of multiplying by it.
- (b) 10 — Corresponding angles are equal, so 5x + 4 = 3x + 24. Subtracting 3x from both sides gives 2x + 4 = 24, then subtracting 4 gives 2x = 20, so x = 10. 14 comes from adding the constants, 4 + 24, instead of subtracting them when rearranging. 19 comes from treating the angles as co-interior (summing to 180°): 5x + 4 + 3x + 24 = 180 gives 8x = 152, so x = 19. 20 correctly reaches 2x = 20 but stops without dividing by 2.
- (d) (−1, 3) — Method: two 90° rotations about the SAME centre, applied one after another, combine into a single 180° rotation about that same centre: use the shortcut (x, y) → (2a − x, 2b − y) for a half-turn about (a, b). Working: with centre (2, 3), doubling each coordinate gives 2 × 2 = 4 and 2 × 3 = 6, so the rule is (x, y) → (4 − x, 6 − y). Applying it to (5, 3) gives 4 − 5 = −1 and 6 − 3 = 3, so the coin ends at (−1, 3). Answer: (−1, 3). Rotate about the centre (2, 3) stated in the game, not about the origin, and remember the button is pressed TWICE: stopping after one press, or rotating about the wrong centre, both leave the coin somewhere else.
- (a) 1 : 8 — Method: in similar solids the ratio of the volumes is the cube of the ratio of corresponding lengths. Working: the heights are in the ratio 1 : 2, so the volumes are in the ratio 1³ : 2³ = 1 : 8. Answer: 1 : 8. The distractors: 1 : 2 leaves the height ratio unchanged, as though a volume scaled in the same way as a length; 1 : 4 comes from squaring, 1² : 2², which is the rule for surface areas and not for volumes; 1 : 6 comes from multiplying each part by 3 instead of raising each part to the power 3.
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