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.
Circles, composite shapes, spheres, pyramids and cones worksheet — GCSE Foundation
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- 1.A circular plate has a diameter of 20 cm. Work out the area of the plate. Use π = 3.14.
- 2.A circular clock face has a diameter of 18 cm. Work out the circumference of the clock face. Use π = 3.14.
- 3.A solid metal sphere has a radius of 5 cm. It is melted down and recast into a solid cylinder with the same radius, 5 cm. Work out the height of the cylinder, so that its volume equals the volume of the sphere. Use π = 3.14 and give your answer correct to 1 decimal place.
- 4.A solid ball has a radius of 7 cm. Work out the surface area of the ball. Use π = 3.14.
- 5.A circular coaster has a radius of 4 cm. Work out the circumference of the coaster. Give your answer in terms of π.
- 6.A window is made from a rectangle with a semicircle on top. The rectangle is 40 cm wide and 60 cm tall. The semicircle has the same width as the rectangle. Work out the total area of the window. Use π = 3.14. Give your answer to the nearest whole number.
- 7.A circular clock face has a radius of 9 cm. Work out the circumference of the clock face. Use π = 3.14. Give your answer to 1 decimal place.
- 8.A cylindrical water tank has a solid hemispherical dome on top, both with radius 2 m. Work out the volume of the hemispherical dome alone. Use π = 3.14 and give your answer correct to 3 decimal places.
- 9.A circular pizza base has a radius of 12 cm. Work out the area of the pizza base. Use π = 3.14.
- 10.A cylindrical candle and a cone-shaped candle have the same base radius and the same height. State the cone's volume as a fraction of the cylinder's volume.
- 11.A solid pyramid has a square base of side 6 cm. Its apex is directly above the centre of the base, at a vertical height of 8 cm. Work out the volume of the pyramid.
- 12.A circle has a radius of 1 cm. Work out the circumference of the circle. Give your answer in terms of π.
- 13.A square tile has sides of length 4.5 cm. Work out the perimeter of the tile.
- 14.A circular garden pond has a diameter of 5 m. A gardener is laying gravel around it in a ring-shaped path 1 m wide, measured outward from the edge of the pond. Work out the area of the gravel path. Use π = 3.14.
- 15.A birthday cake is a cylinder of radius 10 cm and height 8 cm, with a thin ribbon fixed exactly once around the curved side, at half the height, and joined with no overlap. Work out the length of ribbon needed. Use π = 3.14.
Answer key
- (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.
- (d) 56.52 cm — Circumference = πd. Substitute d = 18: circumference = 3.14 × 18 = 56.52 cm.
- (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) 615.44 cm² — Method: the surface area of a sphere is 4πr². Square the radius, multiply by π, then multiply by 4. Working: r² = 7² = 49, then 3.14 × 49 = 153.86, then 4 × 153.86 = 615.44. Answer: 615.44 cm². The distractors: 153.86 cm² comes from stopping at πr², which is the area of a flat circle of radius 7 cm and leaves out the factor of 4 that a curved surface needs; 87.92 cm² comes from 4 × 3.14 × 7, using the radius itself where the formula asks for its square; 1436.03 cm³ comes from working out the volume of the ball with (4 ÷ 3) × π × r³ instead of its surface area, which is a different measure and carries a cubic unit.
- (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) 3028 cm² — Split the window into a rectangle and a semicircle. Rectangle area = 40 × 60 = 2400 cm². The semicircle has radius 40 ÷ 2 = 20 cm, so its area = 0.5 × 3.14 × 20² = 628 cm². Total area = 2400 + 628 = 3028 cm². A student who uses a full circle instead of a semicircle on top of the rectangle gets 3.14 × 20² = 1256 cm² for the circle, plus 2400 cm² for the rectangle, totalling 3656 cm². A student who uses the rectangle's full width (40 cm) as the radius instead of halving it gets 0.5 × 3.14 × 40² = 2512 cm² for the semicircle, plus 2400 cm² for the rectangle, totalling 4912 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.
- (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.
- (d) 452.16 cm² — Area = πr². Substitute r = 12: area = 3.14 × 12² = 3.14 × 144 = 452.16 cm².
- (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.
- (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) 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) 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) 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) 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.
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