Printable · GCSE Foundation · ages 14-16
Area of 2D shapes and volume of prisms worksheet — GCSE Foundation
Fifteen questions on "area of 2d shapes and volume of prisms" — DfE statement G16. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Answer key: Area of 2D shapes and volume of prisms worksheet — GCSE Foundation
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- (b) 12 cm — Method: the area of a rectangle is one side multiplied by the other, so when the area and one side are known the other side is found by reversing that multiplication — divide the area by the side that is known. Working: 96 ÷ 8 = 12. Answer: 12 cm. The distractors: 88 cm comes from 96 − 8, subtracting the known side as though the area had been made by adding the two sides together; 768 cm comes from 96 × 8, running the area rule forwards on the two numbers given instead of reversing it; 40 cm comes from reading the 96 as a perimeter — halving it to 48 and taking the 8 cm side away — which reverses the perimeter rule rather than the area rule.
- (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.
- (d) 300 cm³ — Method: the volume of a cuboid is length × width × height. Working: 2 × 10 = 20, then 20 × 15 = 300. Answer: 300 cm³. The distractors: 27 cm³ comes from adding the three edges, 2 + 10 + 15, instead of multiplying them; 400 cm² comes from working out the surface area, 2 × (2 × 10 + 2 × 15 + 10 × 15) = 400, which answers a different question and carries a different unit; 150 cm³ comes from multiplying 10 × 15 and leaving the 2 cm edge out of the calculation altogether.
- (a) 96 cm² — Method: a cube has six identical square faces, so the total surface area is six times the area of one face. Working: one face has area 4 × 4 = 16 cm², and 6 × 16 = 96. Answer: 96 cm². The distractors: 16 cm² is the area of a single face, from stopping before multiplying by the six faces; 64 cm³ comes from working out the volume, 4 × 4 × 4, which is a different measure and carries a different unit; 24 cm² comes from multiplying the six faces by the edge length, 6 × 4, instead of by the area of a face.
- (c) 27 cm² — Area of a rectangle = length × width = 4.5 × 6 = 27 cm². A pupil who adds the two sides instead of multiplying gets 4.5 + 6 = 10.5 cm². A pupil who works out the perimeter instead of the area gets 2 × (4.5 + 6) = 21 cm². A pupil who rounds 4.5 up to 5 before multiplying gets 5 × 6 = 30 cm². The correct area is 27 cm².
- (d) 7 cm — Area of a parallelogram = base × height, so height = area ÷ base = 84 ÷ 12 = 7 cm. A pupil who multiplies instead of dividing gets 84 × 12 = 1008 cm. A pupil who divides the base by the area instead of the area by the base gets 12 ÷ 84 ≈ 0.14 cm. A pupil who mistakenly halves the area first, as if this were a triangle, gets (84 ÷ 2) ÷ 12 = 3.5 cm. The correct height is 7 cm.
- (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.
- (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.
- (a) 32 cm² — Area of a trapezium = 1/2 × (sum of parallel sides) × height = 1/2 × (5 + 11) × 4 = 1/2 × 16 × 4 = 32 cm². (64 cm² comes from forgetting to halve the product; 44 cm² comes from using only the longer parallel side, 11 × 4, instead of both parallel sides; 20 cm² comes from adding all three given lengths together, 5 + 11 + 4, instead of using the area formula.)
- (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.
- (b) 28 cm² — Area of a trapezium = (sum of parallel sides) ÷ 2 × height. Sum of parallel sides = 5.6 + 8.4 = 14 cm. Half of that is 14 ÷ 2 = 7 cm. Area = 7 × 4 = 28 cm². A pupil who forgets to halve gets 14 × 4 = 56 cm². A pupil who uses only the longer parallel side, as if this were a rectangle, gets 8.4 × 4 = 33.6 cm². A pupil who subtracts the parallel sides instead of adding them gets (8.4 − 5.6) ÷ 2 × 4 = 1.4 × 4 = 5.6 cm². The correct area is 28 cm².
- (b) 72 000 cm³ — Cross-sectional area = 1/2 × 40 × 30 = 600 cm². Volume = cross-sectional area × length = 600 × 120 = 72 000 cm³. (144 000 cm³ comes from forgetting the 1/2 in the triangle's area, using 40 × 30 as the cross-section; 36 000 cm³ comes from halving the correct volume again, as if the 1/2 applied a second time; 720 cm³ comes from adding the cross-sectional area and the length, 600 + 120, instead of multiplying them.)
- (b) 60 cm² — Method: the area of a triangle is half the base multiplied by the perpendicular height, and here the perpendicular height is the 12 cm, not the sloping side. Working: 10 × 12 = 120, then 120 ÷ 2 = 60. Answer: 60 cm². The distractors: 65 cm² comes from using the sloping side of 13 cm as the height, (10 × 13) ÷ 2; 120 cm² comes from using the right two lengths but forgetting to halve, 10 × 12; 78 cm² comes from taking 13 cm and 12 cm as the base and the height and ignoring BC altogether, (13 × 12) ÷ 2.
- (d) 283 cm³ — Volume of a cylinder = πr²h. Using π = 3.14: V = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 282.6, which rounds to 283 cm³. A pupil who uses the diameter, 6 cm, instead of the radius gets 3.14 × 6² × 10 = 1130.4 ≈ 1130 cm³. A pupil who forgets to square the radius gets 3.14 × 3 × 10 = 94.2 ≈ 94 cm³. A pupil who uses the circumference formula 2πr instead of πr² gets 2 × 3.14 × 3 × 10 = 188.4 ≈ 188 cm³. The correct volume is 283 cm³.
- (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).)
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