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) 140 cm² — Method: the area of a trapezium is the mean of the two parallel sides multiplied by the perpendicular height. Working: (16 + 24) ÷ 2 = 20, then 20 × 7 = 140. Answer: 140 cm². The distractors: 280 cm² comes from multiplying the sum of the parallel sides by the height, (16 + 24) × 7, and forgetting to halve; 168 cm² comes from using only the longer parallel side, 24 × 7, as though the shape were a rectangle; 47 cm comes from adding all three given lengths, 16 + 24 + 7, which gives a length rather than an area.
- (b) 70 cm³ — Area of the triangular cross-section = base × height ÷ 2. Base × height = 3.5 × 4 = 14 cm², and half of that is 14 ÷ 2 = 7 cm². Volume of the prism = cross-sectional area × length = 7 × 10 = 70 cm³. A pupil who forgets to halve when finding the triangle's area gets 3.5 × 4 × 10 = 140 cm³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length stops at 7 cm³. A pupil who ignores the height altogether, multiplying base by length, gets 3.5 × 10 = 35 cm³. The correct volume is 70 cm³.
- (b) 50 cm² — Method: the area of a rectangle is length × width. Working: 2 × 25 = 50. Answer: 50 cm². The distractors: 54 cm comes from working out the perimeter, 2 × (2 + 25), which is a length and not an area; 27 cm comes from adding the two sides, 2 + 25, instead of multiplying them; 25 cm² comes from halving the product, 50 ÷ 2, which is the rule for the area of a triangle rather than of a rectangle.
- (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.
- (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.
- (b) 6 m — Volume of a cuboid = length × width × height, so height = volume ÷ (length × width) = 360 ÷ (12 × 5) = 360 ÷ 60 = 6 m. A pupil who divides the volume by the length only gets 360 ÷ 12 = 30 m. A pupil who divides the volume by the width only gets 360 ÷ 5 = 72 m. A pupil who subtracts length × width from the volume instead of dividing gets 360 − 60 = 300 m. The correct height is 6 m.
- (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.
- (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) £50 — Area = 1/2 × (3.5 + 6.5) × 4 = 1/2 × 10 × 4 = 20 m². Cost = 20 × £2.50 = £50. (£100 comes from forgetting to halve the trapezium area, giving 40 m² instead of 20 m²; £65 comes from using only the longer parallel side, 6.5 × 4 = 26 m², instead of the trapezium formula; £35 comes from adding all three given lengths, 3.5 + 6.5 + 4, and treating that total as the area in square metres.)
- (b) 8 cm — Area = base × height, so height = area ÷ base = 136 ÷ 17 = 8 cm. (16 cm comes from using the triangle formula, thinking area = 1/2 × base × height and so height = 2 × area ÷ base; 2312 cm comes from multiplying the area by the base instead of dividing; 119 cm comes from subtracting the base from the area instead of dividing the area by the base.)
- (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.
- (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.
- (a) 180 cm³ — Method: the volume of a right prism is the area of its cross-section multiplied by its length, and the area of a triangle is half the base multiplied by the perpendicular height. Working: the cross-section has area (6 × 5) ÷ 2 = 15 cm², and 15 × 12 = 180. Answer: 180 cm³. The distractors: 360 cm³ comes from taking the cross-section as 6 × 5 = 30 and never halving it, which measures the rectangle around the triangular face rather than the face itself; 66 cm³ comes from adding the base and the perpendicular height and halving, (6 + 5) ÷ 2 = 5.5, which is the trapezium rule used where the triangle rule is needed, and then multiplying by the 12 cm length; 15 cm³ comes from working out the triangular cross-section correctly and stopping there, so the 12 cm length is never used and an area is handed in as a volume.
- (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.
- (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.)
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