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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- (a) 314 cm² — Method: the area of a circle is πr², and the radius is half the diameter, so halve the 20 cm before squaring. Working: r = 20 ÷ 2 = 10 cm, so the area is 3.14 × 10² = 3.14 × 100 = 314. Answer: 314 cm². The distractors: 1256 cm² comes from putting the diameter straight into πr² without halving it, 3.14 × 20²; 628 cm² comes from halving correctly but then using 2πr², a mixture of the circumference and area formulae; 62.8 cm² comes from working out πd = 3.14 × 20, which is the circumference of the plate rather than its area.
- (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.
- (d) 706.5 cm² — Area of a circle = πr². With r = 15 cm and π = 3.14, area = 3.14 × 15² = 3.14 × 225 = 706.5 cm². A student who uses the circumference formula 2πr instead of the area formula gets 2 × 3.14 × 15 = 94.2 cm². A student who uses πr instead, forgetting to double, gets 3.14 × 15 = 47.1 cm². A student who squares the diameter (30 cm) instead of the radius gets 3.14 × 900 = 2826.0 cm².
- (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.
- (c) One third of the cylinder's volume — Volume of a cylinder = base area × height. Volume of a cone = 1/3 × base area × height. For the same base radius and height, the cone's volume is exactly one third of the cylinder's, so it uses less wax. A student who thinks the cone is half the cylinder's volume has confused it with a different solid's ratio. A student who thinks the two volumes are the same has ignored the 1/3 factor in the cone formula entirely. A student who thinks the cone is two thirds of the cylinder's volume has the right idea that it is a fraction, but the wrong fraction.
- (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.
- (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) 6.7 cm — Volume of the sphere = (4/3)πr³ = (4/3) × 3.14 × 5³ = (4/3) × 3.14 × 125 = 523.33 cm³. Set this equal to the cylinder's volume πr²h: 523.33 = 3.14 × 25 × h = 78.5h. Divide: h = 523.33 ÷ 78.5 = 6.667 cm, which rounds to 6.7 cm. 1.7 cm comes from using the sphere's diameter, 10 cm, as the cylinder's radius: 523.33 ÷ (3.14 × 10²) = 1.7 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.
- (a) 188.4 cm² — Curved surface area of a cone = πrl. With r = 6 cm, l = 10 cm and π = 3.14, curved surface area = 3.14 × 6 × 10 = 188.4 cm². A student who uses the cylinder's curved surface area formula, 2πrl, instead of the cone's gets 2 × 3.14 × 6 × 10 = 376.8 cm². A student who uses the circle-area formula πr² instead of πrl gets 3.14 × 36 = 113.04 cm². A student who multiplies r × l but leaves out π entirely gets 6 × 10 = 60 cm².
- (c) 2π cm — Method: the circumference of a circle is 2πr, where r is the radius, or equivalently πd, where d is the diameter. Working: r = 1, so the circumference is 2 × π × 1 = 2π cm. Answer: 2π cm. The distractors: π cm comes from using the formula πd but substituting the radius in place of the diameter; 4π cm comes from doubling twice — changing the radius into the diameter of 2 cm and then putting that diameter into 2πr as though it were a radius; π cm² is the area of this circle, π × 1², and comes from reaching for the area formula when a distance round the outside was asked for, which is why it carries a squared unit.
- (c) 3π cm — Circumference = πd. With d = 3 cm, circumference = π × 3 = 3π cm. A student who uses the radius (1.5 cm) instead of the diameter in the formula gets 1.5π cm.
- (c) 25 m — The gardener needs the circumference, since the trim goes around the flower bed. Circumference = πd = 3.14 × 8 = 25.12 m, which rounds to 25 m. A student who mistakes the diameter (8 m) for the radius, then doubles it before multiplying by π, gets 3.14 × 16 = 50.24 m, rounding to 50 m. A student who makes the same mistake but uses the area formula instead of the circumference gets 3.14 × 8² = 200.96 m², rounding to 201. A student who correctly halves the diameter to find the radius (4 m) but then uses πr instead of πd gets 3.14 × 4 = 12.56 m, rounding to 13 m.
- (b) 96 cm³ — Method: the volume of a pyramid is one third of the base area multiplied by the vertical height. Work out the area of the square base, multiply by the height, then divide by 3. Working: the base area is 6 × 6 = 36 cm², then 36 × 8 = 288, and 288 ÷ 3 = 96. Answer: 96 cm³. The distractors: 288 cm³ comes from multiplying the base area by the height and forgetting the one third, which is the volume of a cuboid with the same base and height; 144 cm³ comes from halving that 288 instead of taking a third of it; 16 cm³ comes from using the base edge of 6 cm in place of the base area, (6 × 8) ÷ 3.
- (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.
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