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.
Answer key: Geometry and measures worksheet — GCSE Foundation
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- (c) 27.7 cm — Arc length = (150 ÷ 360) × 2 × 3.14 × 6 = (5 ÷ 12) × 37.68 = 15.7 cm. The perimeter of a sector also includes the two straight radii, so perimeter = 15.7 + 6 + 6 = 27.7 cm. (15.7 cm comes from stopping after the arc length and forgetting the two straight edges; 21.7 cm comes from adding only one radius instead of two; 31.4 cm comes from doubling the arc length instead of adding the two straight edges.)
- (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.
- (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) (−2, 10) — To translate M(3, 8) by $\binom{−5}{2}$, add −5 to the x-coordinate and 2 to the y-coordinate: (3 + (−5), 8 + 2) = (−2, 10). (5, 3) comes from swapping the two components of the vector before applying them. (−2, 8) applies only the x-component and forgets to add the y-component. (3, 10) applies only the y-component and forgets to add the x-component.
- (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.
- (d) 10 cm — For a triangle to exist, any two sides must add up to more than the third side. 9 + 10 = 19 > 15, and 15 − 9 = 6 < 10, so 10 cm satisfies the triangle inequality. The other lengths fail: 6 cm gives 9 + 6 = 15, which is not more than 15; 24 cm and 26 cm are each at least as large as 9 + 15 = 24.
- (c) 36 cm² — The area of a parallelogram is base × height = 9 × 4 = 36 cm². 26 cm² is 2 × (9 + 4), the perimeter you would get by treating the perpendicular height as though it were a side length — and a perimeter is a length, not an area. 18 cm² uses the triangle formula, half of base × height, by mistake. 72 cm² doubles the correct area.
- (d) 4 m — The sloping slide is the hypotenuse of a right-angled triangle with the other two sides 2.4 m and 3.2 m. By Pythagoras' Theorem, hypotenuse² = 2.4² + 3.2² = 5.76 + 10.24 = 16. Square root: √16 = 4 m. Adding the two sides instead of squaring them gives 2.4 + 3.2 = 5.6 m. Truncating each squared side to a whole number before adding (5.76 to 5 and 10.24 to 10) gives √15 ≈ 3.87 m. Forgetting to take the square root and leaving the sum of the squares as the answer gives 5.76 + 10.24 = 16, written as 16 m.
- (c) $\binom{5}{-2}$ — Method: the top number of a column vector is the change in the x-coordinate and the bottom number is the change in the y-coordinate, each worked out as image minus object. Working: across, 7 − 2 = 5; up, 3 − 5 = −2. So the point moves 5 to the right and 2 down. Answer: $\binom{5}{-2}$. Subtracting the other way round, object minus image, gives $\binom{-5}{2}$, which is the journey from B back to A. Recording the vertical change as 2 because the gap between 3 and 5 is 2, without noting the direction, gives $\binom{5}{2}$, a movement 2 upwards. Adding the coordinates instead of subtracting them gives $\binom{9}{8}$.
- (b) 270° — Bearings are measured clockwise from north (000°). Due north is 000°, due east is 090°, due south is 180° and due west is 270°. Choosing 090° gives the bearing for due east, not west. Choosing 180° gives the bearing for due south. Choosing 000° gives the bearing for due north.
- (b) An angle between 90° and 180° — Method: angle names are fixed by two markers, a right angle at 90° and a straight line at 180°. An obtuse angle is one that has opened past the right angle but has not reached the straight line. Working: an obtuse angle is greater than 90° and smaller than 180°. Answer: an angle between 90° and 180°. The distractors: an angle between 0° and 90° is the definition of an acute angle, chosen by a candidate who remembers the wrong side of the right-angle marker; an angle between 180° and 360° is the definition of a reflex angle, chosen by a candidate who knows the angle is past 90° and then takes the largest category; an angle of exactly 90° is the right angle itself, which is the marker an obtuse angle has passed rather than the obtuse angle.
- (c) RHS — Both triangles are right-angled, and have equal hypotenuses (13 cm) and one equal corresponding side (5 cm), so they are congruent by the RHS (right angle, hypotenuse, side) condition — the three facts given are exactly a right angle, a hypotenuse and one other side. The distractor SAS would apply if two sides and the angle between those two sides were matched, but the right angle here lies between the 5 cm side and the third side, not between the 5 cm side and the hypotenuse. The distractor SSS needs all three pairs of sides matched, and only two sides of each triangle are given — the third side would first have to be calculated by Pythagoras, so SSS is not the condition the given facts supply. The distractor AAS would need two angles and a non-included side matched, but only one angle in each triangle is given.
- (a) 250° — The back bearing (the bearing of A from B) differs from the bearing of B from A by exactly 180°. Because the given bearing, 070°, is less than 180°, add 180°: 070 + 180 = 250°, so the bearing of A from B is 250°. Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 110° comes from subtracting 180° from 070° and dropping the negative sign (070 − 180 = −110) instead of adding 180°. Choosing 160° comes from adding only 90° instead of 180° (070 + 90 = 160).
- (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).
- (c) 8 cm — Area of a trapezium = (sum of parallel sides) ÷ 2 × height, so 45 = (a + 10) ÷ 2 × 5. Dividing 45 by 5 gives 9, so (a + 10) ÷ 2 = 9, meaning a + 10 = 18, so a = 18 − 10 = 8 cm. A pupil who correctly finds that the two parallel sides add up to 18 but forgets to subtract the known side of 10 cm gives the sum of both parallel sides, 18 cm, as the answer. A pupil who then adds 10 again by mistake instead of subtracting gets 18 + 10 = 28 cm. A pupil who halves the correct answer by mistake gets 8 ÷ 2 = 4 cm. The correct length of the other parallel side is 8 cm.
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