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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- (a) 33 — Method: total crates = (number of floor positions that actually have crates on them) × (the stack height). Working: there are 4 × 3 = 12 floor positions in the whole arrangement, but one corner position is left empty, leaving 11 filled positions; each filled position is stacked 3 crates high, so 11 × 3 = 33. Answer: 33. The distractors: 36 comes from forgetting to remove the empty corner and using all 12 positions (12 × 3). 35 comes from removing only one crate for the empty corner instead of the full stack of 3 (36 − 1). 11 comes from counting the filled floor positions and stopping there, forgetting that each one carries a stack 3 crates high.
- (c) (6, 2) — Method: add the vector's components to the point's coordinates, x-component to x, y-component to y. Working: A′ = (2 + 4, 3 + (−1)) = (6, 2). Options: (6, 4) comes from adding the size of the y-component, 1, instead of subtracting it; (1, 7) comes from swapping the components, adding the y-component (−1) to the x-coordinate and the x-component (4) to the y-coordinate; (−2, 4) comes from reversing the sign of both components, using the vector (−4, 1) instead of (4, −1). Answer: (6, 2).
- (d) positive — Moving from A to B, the x-coordinate increases from 1 to 4 and the y-coordinate also increases from 1 to 7, so the line rises as it goes from left to right, which means the gradient is positive. "negative" would need y to decrease as x increases, which is not the case here. "zero" would need the y-coordinate to stay the same, but it changes from 1 to 7. "cannot be determined" is wrong because the coordinates of both points are known, so the direction of the line can always be found.
- (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) 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.
- (b) $\binom{-5}{5}$ — Method: add the two flights to get the single vector of the whole journey out, then reverse that vector to get the journey home. Working: across, 6 − 1 = 5; up, 4 − 9 = −5. So the drone finishes at (7, −8), which is 5 to the right of its start and 5 below it. The way home is therefore 5 to the left and 5 up. Answer: $\binom{-5}{5}$. Giving the combined journey itself, $\binom{5}{-5}$, describes the flight out rather than the flight home. Subtracting the second vector instead of adding it makes the journey out 7 across and 13 up, and reversing that gives $\binom{-7}{-13}$. Reversing the horizontal movement but leaving the vertical one alone gives $\binom{-5}{-5}$.
- (c) 15 — The exterior angle is 180° − 156° = 24°, and the number of sides of a regular polygon is 360° divided by the exterior angle, so 360 ÷ 24 = 15. 24° is the exterior angle itself, stopping one step before the final division. 17 comes from finding 15 correctly and then adding 2, muddling the exterior angle rule with the (n − 2) that appears in the interior angle sum formula. 7.5 comes from dividing 180 by the exterior angle instead of 360, using the angles on a straight line rather than the total of the exterior angles of a polygon.
- (d) 180 — Method: bearings are measured clockwise from north and written using three figures. Working: south is a half turn (180°) clockwise from north. A student who answers 090 has confused south with east. A student who answers 270 has confused south with west. A student who answers 018 has written the correct digits in the wrong order. Answer: 180.
- (a) (5, 4) — Method: find the vector from the centre to the point, multiply it by the scale factor, then add the result back to the centre. Working: the vector from (2, 4) to (8, 4) is (6, 0); multiplying by 1/2 gives (3, 0); adding this to the centre (2, 4) gives (5, 4). Options: (4, 2) comes from multiplying the original coordinates by 1/2 directly, ignoring the centre of enlargement; (14, 4) comes from using a scale factor of 2 instead of 1/2, giving (2, 4) + 2×(6, 0) = (14, 4); (8, 2) comes from halving only the y-coordinate and leaving the x-coordinate unchanged. Answer: (5, 4).
- (d) tangent — A line that touches a circle at exactly one point, without crossing into the circle, is called a tangent. A chord is a straight line joining two points ON the circle, so it touches at two points, not one. A radius runs from the centre to the circle's edge, not along the outside of it. A diameter is a chord that passes through the centre, also touching the circle at two points.
- (b) angle B = angle E — AB and BC meet at vertex B, so the angle INCLUDED between them is angle B; making angle B = angle E completes SAS. Angle A sits between AB and AC, not between AB and BC, so it is not the included angle needed for SAS here. Angle C sits between BC and CA, not between AB and BC, so it is not included either. AC = DF would give a third pair of equal sides, which proves congruence by SSS instead of SAS.
- (b) 25 cm² — Method: a square has four equal sides, so divide the perimeter by 4 to recover the side length, then square that side to get the area. Working: 20 ÷ 4 = 5 cm, then 5 × 5 = 25. Answer: 25 cm². The distractors: 400 cm² comes from squaring the perimeter itself, 20 × 20, treating the 20 cm as though it were the side length; 100 cm² comes from dividing the perimeter by 2 rather than by 4, giving a side of 10 cm, and squaring that; 5 cm is the side length, from stopping as soon as the perimeter has been divided by 4 and never squaring it, which also leaves a length where an area was asked for.
- (b) (22, −13) — The vector from the ship to the lighthouse is (12, −4) − (2, 5) = (10, −9). Sailing along this vector twice from the start gives (2, 5) + 2 × (10, −9) = (2 + 20, 5 − 18) = (22, −13). '(12, −4)' stops after the ship reaches the lighthouse and ignores the second identical leg. '(−18, 23)' comes from finding the vector the wrong way round, as (2, 5) − (12, −4) = (−10, 9), and then doubling that. '(2, 5)' comes from adding the vector and then subtracting it again, wrongly cancelling the two legs instead of adding them.
- (c) 67° — Angles on a straight line add up to 180°. Add the two known angles: 48° + 65° = 113°. Subtract from 180°: 180° − 113° = 67°.
- (a) 75° — The three angles in the triangle formed by the two supports and the ground add up to 180°. Two of the angles are 34° and 71°, so the third angle is 180 − 34 − 71 = 75, which is 75°. 105° comes from adding the two given angles together instead of subtracting them from 180° (34 + 71 = 105). 146° comes from computing 180 − 34 = 146 and forgetting to also subtract 71°. 37° comes from subtracting the two given angles from each other instead of using the triangle's angle sum (71 − 34 = 37).
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