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) 32 cm² — The area of a trapezium is half of the sum of the parallel sides, multiplied by the height. Add the parallel sides: 6 + 10 = 16. Multiply by the height: 16 × 4 = 64. Half of 64 is 32 cm². 64 cm² forgets to halve and just gives (6+10)×4. 8 cm² averages the two parallel sides, (6+10)÷2 = 8, but forgets to multiply by the height. 20 cm² treats it as a triangle using only the longer parallel side as the base: half of 10 × 4.
- (d) 4 — Method: use the definition of a rectangle — a quadrilateral in which every interior angle is a right angle. Working: a quadrilateral has four interior angles, and in a rectangle all four of them measure 90°, which is consistent with the 360° angle sum because 4 × 90° = 360°. Answer: 4. The distractors: 2 comes from applying the rule that opposite angles are equal and concluding that only one pair of corners is square; 1 comes from thinking that a single square corner is enough to make a shape a rectangle; 0 comes from confusing a rectangle with a general parallelogram, which has no right angles unless it is a rectangle.
- (d) SSS — All three pairs of corresponding sides are equal (PQ = ST, QR = TU, PR = SU), so the triangles are congruent by the SSS (side-side-side) condition. The distractor SAS would apply if two sides and the included angle were given equal, not three sides. The distractor ASA would apply if two angles and the included side were given equal. The distractor RHS would apply only for right-angled triangles with the hypotenuse and one other side equal.
- (c) 14.7 cm — The scale factor from the smaller tile to the larger tile is 7 ÷ 4 = 1.75, so the larger side is 8.4 × 1.75 = 14.7 cm. '4.8 cm' comes from scaling by 4 ÷ 7 instead, using the ratio the wrong way round. '11.4 cm' comes from adding the difference between the ratio numbers, 7 − 4 = 3, onto 8.4, instead of scaling. '58.8 cm' comes from multiplying by 7 on its own, using a ratio number as the scale factor instead of working out 7 ÷ 4 = 1.75 first.
- (d) 1 kg — First find the total mass in grams: 4 × 250 = 1000 g. Then convert to kilograms by dividing by 1000: 1000 ÷ 1000 = 1 kg. Finding the correct total in grams but forgetting to divide by 1000 gives 1000 kg. Adding the number of packs to the pack mass instead of multiplying, 4 + 250 = 254 g, gives 0.254 kg. Dividing the pack mass by the number of packs instead of multiplying, 250 ÷ 4 = 62.5 g, gives 0.0625 kg.
- (a) 376.8 cm³ — Volume of a cone = (1/3)πr²h. Substitute r = 6 and h = 10: (1/3) × 3.14 × 6² × 10 = (1/3) × 3.14 × 36 × 10 = (1/3) × 1130.4 = 376.8 cm³.
- (b) (5, 1) — First scale a by 2: 2a = (2×3, 2×(−2)) = (6, −4). Then add b component by component: (6+(−1), −4+5) = (5, 1). (2, 3) is a + b without doubling a first. (4, 6) doubles both a and b instead of only a. (7, −9) subtracts b from 2a instead of adding it.
- (b) The arcs only meet exactly on AB, not above or below it. — With a radius of exactly half of AB, the arc centred at A and the arc centred at B each reach precisely to the midpoint of AB, so they meet only at that one point, on the line AB itself — there are no two intersection points above and below the line to join, so the perpendicular bisector cannot be drawn. (The claim that this radius is always too short to draw any arc is wrong, since a radius equal to half of AB is a perfectly valid, positive length for a compass arc; the claim that the radius must equal the full length of AB is wrong — a longer radius than half of AB would also work, it does not have to equal AB exactly; the claim that the arcs would not cross line AB at all is wrong, since they do cross it — that is exactly the problem, they meet only on it.)
- (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) √3/2 — cos 30° is one of the exact values you must know: cos 30° = √3/2. The value 1/2 is the exact value of sin 30° (and of cos 60°), not cos 30°. The value √2/2 is the exact value of cos 45° (and of sin 45°). The value 1 is the exact value of cos 0°.
- (d) 12 cm — Since DE is parallel to BC, triangle ADE is similar to triangle ABC (the two triangles share angle A, and the parallel lines make the angles at D and E equal to the angles at B and C). The whole side AB = AD + DB = 3 + 6 = 9 cm. The scale factor from the small triangle to the large triangle is AB ÷ AD = 9 ÷ 3 = 3, so BC = DE × 3 = 4 × 3 = 12 cm. Scaling by DB ÷ AD instead of AB ÷ AD gives 4 × (6 ÷ 3) = 8 cm. Adding DB directly onto DE, instead of scaling, gives 4 + 6 = 10 cm. Treating the triangles as congruent rather than similar, so assuming corresponding sides are simply equal, gives BC = DE = 4 cm.
- (c) 36 cm² — Method: the area of a square is its side length multiplied by itself. Working: 6 × 6 = 36. Answer: 36 cm². The distractors: 24 cm comes from working out the perimeter, 4 × 6, which is a length and not an area; 12 cm comes from doubling the side, 6 × 2, instead of squaring it; 18 cm² comes from halving the product, (6 × 6) ÷ 2, using the rule for the area of a triangle.
- (d) Cylinder — Method: check how many faces are flat and how many are curved, and how many flat faces there are. Working: this solid has a curved surface joining two flat circular faces of equal size, which is exactly a cylinder. A student who answers cone has forgotten a cone has only one flat circular face, not two. A student who answers sphere has forgotten a sphere has no flat faces at all. A student who answers triangular prism has spotted that the solid has two identical end faces but has taken those ends to be triangles rather than circles, and a prism's side faces are flat rectangles, not one curved surface. Answer: cylinder.
- (c) 1.44 m³ — The cross-section is a triangle, so its area = base × height ÷ 2. Base × height = 1.2 × 0.8 = 0.96 m², and half of that is 0.96 ÷ 2 = 0.48 m². The volume of the prism = cross-sectional area × length = 0.48 × 3 = 1.44 m³. A pupil who forgets to halve when finding the triangle's area gets 1.2 × 0.8 × 3 = 2.88 m³. A pupil who ignores the height altogether, treating the cross-section as base × length, gets 1.2 × 3 = 3.6 m³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length of the bed stops at 0.48 m³. The correct volume of soil is 1.44 m³.
- (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.
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