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) 9 cm — The scale factor from triangle ABC to triangle PQR is PQ ÷ AB = 12 ÷ 8 = 1.5. So PR = AC × 1.5 = 6 × 1.5 = 9 cm. (4 cm comes from using the scale factor upside down, AB ÷ PQ; 10 cm comes from adding the difference between AB and PQ, 4 cm, to AC instead of scaling; 16 cm comes from multiplying PQ by the ratio AB : AC instead of scaling AC by the correct scale factor.)
- (b) £33.60 — The area of the parallelogram flower bed is base × height = 3.5 × 2 = 7 m². The cost is 7 × £4.80 = £33.60. £16.80 comes from using the triangle formula instead of the parallelogram formula: 3.5 × 2 = 7, and half of 7 is 3.5 m², then 3.5 × £4.80 = £16.80. £26.40 comes from adding the base and height, 3.5+2 = 5.5, instead of multiplying them, then multiplying by £4.80. £7.00 correctly finds the area, 7 m², but forgets to multiply it by the cost per m².
- (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.
- (a) 3 squares — Method: the plan view shows only the floor positions that have at least one cube standing on them; height does not add extra squares to the plan. Working: the base row occupies three floor positions in a line. The two extra cubes stand on top of two of those same three positions, so they do not create any new floor position. Answer: 3 squares. The distractors: 5 squares comes from adding the total number of cubes used (3 + 2 = 5) instead of counting distinct floor positions. 2 squares comes from counting only the raised two-cube section and ignoring the single cube at the other end of the row. 4 squares comes from counting one of the shared positions twice.
- (b) 115° — Angle APE and angle BPQ are vertically opposite, so angle BPQ = 65°, equal to angle APE. Angle BPQ and angle PQD are co-interior (allied) angles between the parallel lines, and co-interior angles always add up to 180°, so angle PQD = 180 − 65 = 115°. "65°" comes from treating co-interior angles as equal to each other, the way alternate angles are, instead of adding to 180°. "295°" comes from applying the angles-round-a-point fact (360° − 65°) directly to the original 65°, skipping the correct co-interior step. "25°" comes from misremembering co-interior angles as adding to 90° instead of 180° (90 − 65 = 25).
- (a) 20 litres — Volume of water = length × width × depth of water = 40 × 25 × 20 = 20 000 cm³. Since 1000 cm³ = 1 litre, divide by 1000: 20 000 ÷ 1000 = 20 litres. A pupil who uses the full height of the tank, 30 cm, instead of the water depth, 20 cm, gets 40 × 25 × 30 = 30 000 cm³ = 30 litres. A pupil who forgets to convert cm³ to litres at all gives 20 000 litres. A pupil who divides by 1000 twice by mistake gets 20 000 ÷ 1000 ÷ 1000 = 0.02 litres. The correct volume of water is 20 litres.
- (c) 1 — cos 45° = √2/2 and sin 45° = √2/2. Squaring each gives (√2/2)² = 2/4 = 1/2, so (cos 45°)² + (sin 45°)² = 1/2 + 1/2 = 1. The distractor √2 comes from adding cos 45° + sin 45° directly without squaring first (√2/2 + √2/2 = √2). The distractor 2 comes from squaring the top of the fraction, (√2)² = 2, but then dividing by 2 instead of 4 for each term, giving 1 + 1 = 2. The distractor 1/2 comes from squaring only cos 45° and forgetting to add the sin 45° term.
- (d) 1.5 m — By Pythagoras' theorem, diagonal² = 1.2² + 0.9² = 1.44 + 0.81 = 2.25, so diagonal = √2.25 = 1.5 m. 2.1 m comes from simply adding the two sides (1.2 + 0.9) instead of using Pythagoras' theorem. 0.3 m comes from subtracting the two sides (1.2 − 0.9) instead. 2.25 m comes from correctly finding 1.2² + 0.9² = 2.25 but forgetting to take the square root at the end.
- (b) 54 cm² — Area of a parallelogram = base × perpendicular height = 9 × 6 = 54 cm². (27 cm² comes from halving the product as if it were a triangle, 1/2 × 9 × 6; 15 cm² comes from adding the base and height instead of multiplying, 9 + 6; 30 cm² comes from doubling the sum of the base and height, as if finding a perimeter, 2 × (9 + 6).)
- (b) a rectangle — Lying on its side, the cylinder's curved surface touches the table along a straight line, and the two flat circular ends face sideways rather than up or down; viewed from directly above, the outline traced is a rectangle — as long as the cylinder and as wide as its diameter. "a circle" would be correct if the cylinder stood upright on one of its circular ends instead of lying on its side. "a triangle" belongs to a cone lying or standing so that it narrows to a point in that view, which a cylinder never does. "an oval" is a common guess from picturing the round ends, but from directly above those ends are edge-on and contribute to the rectangle's short sides, not a curved outline.
- (a) 110° — Method: in a parallelogram the two angles at the ends of one side are co-interior angles between a pair of parallel sides, so they add up to 180°. Working: angle A + angle B = 180°, so angle B = 180° − 70° = 110°. Answer: 110°. The distractors: 70° comes from applying the rule for opposite angles of a parallelogram, which are equal, to two angles that are next to each other instead; 20° comes from treating the two angles as complementary and working out 90° − 70°; 290° comes from using the 360° angle sum of a quadrilateral and taking away only the one angle that is given, 360° − 70°.
- (d) 240 cm² — Method: the area of a parallelogram is base × perpendicular height, and the perpendicular height is not the sloping side, so it must be found first from the right-angled triangle. Working: the sloping side is the hypotenuse, so the height squared is 13² − 5² = 169 − 25 = 144, giving a height of √144 = 12 cm; then 20 × 12 = 240. Answer: 240 cm². The distractors: 260 cm² comes from using the 13 cm sloping side as the height, 20 × 13, without going through the right-angled triangle at all; 120 cm² comes from finding the height of 12 cm correctly and then halving the product, (20 × 12) ÷ 2, which is the rule for a triangle and not for a parallelogram; 100 cm² comes from using the 5 cm along the base as the height, 20 × 5.
- (c) 3028 cm² — Split the window into a rectangle and a semicircle. Rectangle area = 40 × 60 = 2400 cm². The semicircle has radius 40 ÷ 2 = 20 cm, so its area = 0.5 × 3.14 × 20² = 628 cm². Total area = 2400 + 628 = 3028 cm². A student who uses a full circle instead of a semicircle on top of the rectangle gets 3.14 × 20² = 1256 cm² for the circle, plus 2400 cm² for the rectangle, totalling 3656 cm². A student who uses the rectangle's full width (40 cm) as the radius instead of halving it gets 0.5 × 3.14 × 40² = 2512 cm² for the semicircle, plus 2400 cm² for the rectangle, totalling 4912 cm².
- (c) 1/2 — Method: cos 60° comes from half of an equilateral triangle. Cut an equilateral triangle of side 2 down its middle: each half is right-angled, with hypotenuse 2, base 1 and height √3. Working: cosine is the adjacent side over the hypotenuse, the side alongside the 60° angle is the base of length 1, and the hypotenuse is 2, so cos 60° = 1 ÷ 2. Answer: cos 60° = 1/2. A candidate who assumes the cosine falls steadily from cos 0° = 1 to cos 90° = 0 puts 60° two thirds of the way along and writes 1/3; the cosine does not change at a constant rate. Reading the start of the cosine row gives cos 0° = 1, and reading its end gives cos 90° = 0.
- (c) 1 — Method: a right angle measures 90°, the three angles of any triangle add up to 180°, and a right-angled triangle is defined as a triangle that contains a right angle. Working: taking one right angle out of the total leaves 180° − 90° = 90° to be shared between the other two angles, so both of those must be acute; a second right angle would use the whole of that remaining 90° and leave nothing at all for the third angle, which is impossible. The definition therefore fixes the count at exactly one. Answer: 1. The distractors: 2 comes from counting the two sides that form the right angle instead of counting the angles themselves; 3 comes from reading the name as a description of the whole triangle, so that all three of its angles are taken to be right angles, which would need an angle sum of 3 × 90° = 270°; 0 comes from over-applying the angle sum — a candidate who works out that 90° + 90° = 180° leaves nothing for a third angle can conclude from that alone that no triangle may contain a right angle at all.
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