Printable · GCSE Foundation · ages 14-16
Circles, composite shapes, spheres, pyramids and cones worksheet — GCSE Foundation
Fifteen questions on "circles, composite shapes, spheres, pyramids and cones" — DfE statement G17. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Answer key: Circles, composite shapes, spheres, pyramids and cones worksheet — GCSE Foundation
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- (c) 21 cm — Method: the perimeter is the distance all the way round the edge, so every side is counted once and the three side lengths are added. Working: the two equal sides give 8 + 8 = 16 cm, and the third side adds 5 cm to that. Answer: the perimeter is 21 cm. The distractors: 16 cm comes from adding the two 8 cm sides and handing that total in before the third side has been included; 13 cm comes from adding one 8 cm side to the 5 cm side, as though the badge carried only the two different lengths printed on it rather than three sides; 24 cm comes from taking all three sides to be 8 cm and working out 3 × 8, which would be the perimeter only if the badge were equilateral.
- (a) 151 cm³ — Volume of the cone = 1/3 × 3.14 × 3² × 10 = 94.2 cm³. Volume of the hemisphere = 1/2 × (4/3 × 3.14 × 3³) = 1/2 × 113.04 = 56.52 cm³. Total volume = 94.2 + 56.52 = 150.72 cm³, which rounds to 151 cm³. A student who uses a full sphere instead of a hemisphere gets 94.2 + 113.04 = 207.24 cm³, rounding to 207. A student who uses a cylinder instead of a cone for the base, forgetting the 1/3, gets 3.14 × 3² × 10 = 282.6 cm³, plus the hemisphere's 56.52 cm³, totalling 339.12 cm³, rounding to 339.
- (d) 452.16 cm² — Area = πr². Substitute r = 12: area = 3.14 × 12² = 3.14 × 144 = 452.16 cm².
- (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.
- (c) 34 cm — Method: a rectangle has two lengths and two widths, so the perimeter is 2 × (length + width). Working: 12 + 5 = 17, then 2 × 17 = 34. Answer: 34 cm. The distractors: 17 cm comes from adding one length and one width and stopping, which is only half of the way round; 24 cm comes from doubling the length alone, 2 × 12, and leaving the two widths out; 60 cm² comes from working out 12 × 5, which is the area of the photograph and carries a squared unit because two lengths have been multiplied.
- (c) 80 cm² — Method: a composite shape made of two rectangles that do not overlap has an area equal to the sum of the two rectangle areas, so work out each area and add them. Working: the first rectangle has area 12 × 4 = 48 cm² and the second has area 4 × 8 = 32 cm², so the total is 48 + 32 = 80. Answer: 80 cm². The distractors: 48 cm² comes from working out the larger rectangle only and treating it as the whole shape; 32 cm² comes from working out the smaller rectangle only; 28 cm comes from adding the four given lengths, 12 + 4 + 4 + 8, which is a distance rather than an area and so carries a plain centimetre unit.
- (c) 18 cm — Method: a square has four equal sides, so the perimeter is 4 × the side length. Working: 4 × 4.5 = 4 × 4 + 4 × 0.5 = 16 + 2 = 18. Answer: 18 cm. The distractors: 9 cm comes from 2 × 4.5, adding only one pair of sides; 16 cm comes from rounding the side length down to 4 cm before multiplying, so the 0.5 cm on each side is lost; 20.25 cm² comes from working out 4.5 × 4.5, which is the area of the tile and carries a squared unit.
- (b) 16.747 m³ — Volume of a sphere = (4/3)πr³, so a hemisphere is half that: (2/3)πr³. Substitute r = 2: (2/3) × 3.14 × 2³ = (2/3) × 3.14 × 8 = (2/3) × 25.12 = 16.7467 m³, which rounds to 16.747 m³.
- (a) 62.8 cm — The ribbon goes once around the circular cross-section, so its length equals the circumference: 2πr = 2 × 3.14 × 10 = 62.8 cm.
- (b) 18.84 m² — The pond has radius 2.5 m (half of the 5 m diameter), so the outer edge of the path has radius 2.5 + 1 = 3.5 m. Path area = area of outer circle − area of pond = 3.14 × 3.5² − 3.14 × 2.5² = 38.465 − 19.625 = 18.84 m².
- (a) 8π cm — Circumference = 2πr. Substitute r = 4: circumference = 2 × π × 4 = 8π cm. Using r in place of 2r (halving the formula) gives 4π cm. Using the area formula πr² in place of the circumference formula gives π × 4² = 16π cm. Multiplying 2 × 4 without including π at all gives 8 cm.
- (c) 17.71 m — The perimeter is the two long sides (5 m + 5 m = 10 m) plus the one remaining short side (3 m) plus the curved semicircular edge. The 3 m side is the diameter of the semicircle, so its radius is 3 ÷ 2 = 1.5 m and the curved length is half the circumference: (1/2) × 2 × 3.14 × 1.5 = 4.71 m. Total: 10 + 3 + 4.71 = 17.71 m.
- (d) 56.5 cm — Circumference = 2πr. With r = 9 cm and π = 3.14, circumference = 2 × 3.14 × 9 = 56.52 cm, which rounds to 56.5 cm. A student who uses πr instead of 2πr, forgetting to double, gets 3.14 × 9 = 28.26 cm, rounding to 28.3 cm. A student who uses the area formula πr² instead of the circumference formula gets 3.14 × 81 = 254.34 cm², rounding to 254.3. A student who doubles the correct circumference by mistake gets 56.52 × 2 = 113.04 cm, rounding to 113.0 cm.
- (d) 56.52 cm — Circumference = πd. Substitute d = 18: circumference = 3.14 × 18 = 56.52 cm.
- (a) 28 cm — Method: the perimeter is the total distance round the outside, and a rhombus has four sides of equal length, so the perimeter is 4 × the side length. Working: 4 × 7 = 28. Answer: 28 cm. The distractors: 14 cm comes from 2 × 7, adding only one pair of sides and forgetting that a rhombus has two pairs; 49 cm comes from working out 7 × 7, which is the calculation for the area of a square rather than a distance round the outside; 11 cm comes from adding the side length to the number of sides, 7 + 4, instead of multiplying them.
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