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 solid cone has a base radius of 6 cm and a vertical height of 10 cm. Work out the volume of the cone. Use π = 3.14 and give your answer correct to 1 decimal place.
- 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 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.
- 4.A rectangular garden bed measures 5 m by 3 m. A semicircular flower border is attached along one of the 3 m-wide ends, on the outside of the rectangle, so that the 3 m side is the diameter of the semicircle. Work out the total perimeter of the combined shape (the two long sides, the one remaining short side, and the curved edge). Use π = 3.14.
- 5.A rectangular photograph is 12 cm long and 5 cm wide. Work out the perimeter of the photograph.
- 6.A solid pyramid has a square base of side 10 cm and four identical triangular faces. Its apex is directly above the centre of the base, at a vertical height of 12 cm, and the slant height of each triangular face — the distance from the apex to the midpoint of a base edge — is 13 cm. Work out the total surface area of the pyramid.
- 7.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.
- 8.A circular plate has a diameter of 20 cm. Work out the area of the plate. Use π = 3.14.
- 9.A square tile has sides of length 4.5 cm. Work out the perimeter of the tile.
- 10.An ice cream is made from a cone of radius 3 cm and height 10 cm, topped with a hemisphere of the same radius sitting exactly on top of the cone. Work out the total volume of the ice cream. Use π = 3.14. Give your answer to the nearest whole number. Volume of a cone = 1/3 × πr²h. Volume of a sphere = 4/3 × πr³.
- 11.A solid ball has a radius of 7 cm. Work out the surface area of the ball. Use π = 3.14.
- 12.A party hat is in the shape of a cone with a base radius of 6 cm and a slant height of 10 cm. Work out the curved surface area of the party hat. Use π = 3.14 and the formula curved surface area = πrl.
- 13.A circle has a radius of 1 cm. Work out the circumference of the circle. Give your answer in terms of π.
- 14.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.
- 15.A solid rubber ball has a radius of 3 cm. Work out the volume of the ball. Use π = 3.14.
Answer key
- (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) 56.52 cm — Circumference = πd. Substitute d = 18: circumference = 3.14 × 18 = 56.52 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.
- (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.
- (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.
- (b) 360 cm² — Method: the total surface area is the square base plus the four triangular faces, and the height used for a triangular face is the slant height of 13 cm, not the vertical height of 12 cm. Working: the base is 10 × 10 = 100 cm²; one triangular face is (10 × 13) ÷ 2 = 65 cm², so four faces give 4 × 65 = 260 cm²; the total is 100 + 260 = 360. Answer: 360 cm². The distractors: 260 cm² comes from adding the four triangular faces and leaving out the base; 340 cm² comes from using the vertical height of 12 cm as the height of each triangle, 100 + 4 × 60; 620 cm² comes from working out each face as 10 × 13 without halving, 100 + 4 × 130.
- (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) 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.
- (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.
- (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.
- (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) 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.
- (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) 113.04 cm³ — Method: the volume of a sphere is (4 ÷ 3) × π × r³. Cube the radius, multiply by π, then multiply by 4 and divide by 3. Working: r³ = 3³ = 27, then 3.14 × 27 = 84.78, then 84.78 × 4 = 339.12 and 339.12 ÷ 3 = 113.04. Answer: 113.04 cm³. The distractors: 84.78 cm³ comes from stopping at πr³ and leaving out the four thirds; 37.68 cm³ comes from squaring the radius instead of cubing it, (4 ÷ 3) × 3.14 × 9; 28.26 cm³ comes from using πr², the area of a circle, and labelling it as a volume.
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