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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- (b) 55° — Method: complementary angles are a pair that add up to 90°, so subtract the known angle from 90°. Working: 90 − 35 = 55. Answer: 55°. The distractors: 145° comes from subtracting from 180°, which is the rule for supplementary angles and not for complementary ones; 325° comes from subtracting from 360°, the total of the angles at a point; 90° is the total that the pair must make, written down in place of the missing angle.
- (d) 56.52 cm — Circumference = πd. Substitute d = 18: circumference = 3.14 × 18 = 56.52 cm.
- (b) (8, −6) — Multiplying a column vector by a scalar multiplies both numbers in the vector by that scalar: 2p = (2×4, 2×(−3)) = (8, −6). (8, −3) doubles the top number but leaves the bottom number, −3, unchanged. (4, −6) doubles the bottom number but leaves the top number, 4, unchanged. (8, 6) doubles both numbers correctly in size but loses the negative sign on the bottom number.
- (c) 12 cm — By Pythagoras' Theorem, QR² = PQ² + PR², so PR² = QR² − PQ² = 15² − 9² = 225 − 81 = 144. Square root: √144 = 12 cm.
- (a) Every square is a rectangle — Method: test each statement against the definitions. A rectangle is a quadrilateral with four right angles and opposite sides equal; a square is a quadrilateral with four right angles and all four sides equal; a rhombus is a quadrilateral with all four sides equal. Working: a square has four right angles and its opposite sides are equal, so every square meets the definition of a rectangle and the statement that every square is a rectangle is true. Answer: every square is a rectangle. The distractors: the claim that every rectangle is a square reverses the inclusion, and fails for any rectangle whose length and width differ; the claim that every rhombus is a rectangle treats four equal sides as enough, and drops the right-angle condition — a tilted rhombus has no right angles; the claim that every rectangle is a rhombus reads 'opposite sides equal' as though it meant 'all four sides equal'.
- (b) The radius — Method: each of the four names describes a line in a fixed position relative to the circle, so match the description in the question against those positions. Working: the line described has one end at the centre and its other end on the circle. A chord joins two points that both lie on the circle, so it is ruled out. A diameter joins two points on the circle by passing through the centre, so it is twice as long as the line described. A tangent touches the circle at exactly one point and never reaches the centre. The line with one end at the centre and one end on the circle is a radius. Answer: The radius. The distractors: The diameter comes from remembering only that the line involves the centre, and not noticing that it stops there instead of carrying on to the far side; The chord comes from seeing one end on the circle and assuming both ends are; The tangent comes from matching the phrase 'a point on the circle' to the line that touches at exactly one point, ignoring that a tangent never reaches the centre.
- (d) 6 cm — Volume = πr²h, so height = volume ÷ (πr²) = 471 ÷ (3.14 × 25) = 471 ÷ 78.5 = 6 cm. (30 cm comes from dividing by πr instead of πr², missing one factor of the radius; 150 cm comes from dividing by π only, without using r² at all; 24 cm comes from treating the given 5 cm as a diameter and using a radius of 2.5 cm instead.)
- (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.
- (b) Wrong: 1 litre = 1000 ml, so 3.2 l = 3200 ml. — 1 litre = 1000 ml, so 3.2 litres = 3.2 × 1000 = 3200 ml — Sam is wrong because he multiplied by 100 instead of 1000. Saying Sam is correct accepts the wrong multiplier. Saying Sam is wrong only because 3.2 should be rounded first misses the real error, which is the multiplier, not the starting number. Saying '1 litre is 100 ml' misstates the basic fact and blames the wrong part of Sam's working.
- (d) 50 — The angles in any triangle add up to 180°, so x = 180 − 90 − 40 = 50. "90" comes from subtracting only the 90° angle from 180° and forgetting to also subtract the 40°. "130" comes from adding 90° and 40° together instead of subtracting them from 180°. "45" comes from halving the 90° angle instead of using the angle sum of the triangle at all.
- (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) tan 45° — Method: replace each ratio by its exact value, then compare. Working: a right-angled triangle with a 45° angle is isosceles, so its opposite and adjacent sides are equal and the tangent of 45° is exactly 1. The others are cos 30° = √3/2, about 0.87; sin 45° = √2/2, about 0.71; and cos 60° = 1/2. Answer: tan 45°, the only one of the four that reaches 1. Reading √3/2 as though it were √3, about 1.73, makes cos 30° look the largest, but the division by 2 is part of the value. Ranking by the size of the angle also fails here, because the cosine of an angle falls as the angle grows.
- (c) 13 — Area of the trapezium = 1/2 × (8 + 12) × 5 = 1/2 × 100 = 50 m². Number of bags = 50 ÷ 4 = 12.5, which rounds up to 13 bags since seed is sold only in whole bags. A student who mistakenly uses 2 m² of coverage per bag instead of 4 m² finds 50 ÷ 2 = 25 bags.
- (a) Trapezium — A trapezium is defined as a quadrilateral with exactly one pair of parallel sides. A parallelogram has two pairs, and a kite and rhombus are defined by side lengths, not by having only one pair of parallel sides.
- (d) 60° — Method: the six angles at the centre together make one complete turn of 360°, and because the hexagon is regular they are all equal, so divide 360° by 6. Working: 360 ÷ 6 = 60. Answer: 60°. The distractors: 120° is the interior angle of a regular hexagon, 720 ÷ 6, which is the angle at a vertex and not the angle at the centre; 45° comes from dividing 360 by 8, treating the hexagon as though it had eight sides; 30° comes from halving the angle at the centre, as though each of the six triangles were split again by a line of symmetry.
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