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.All four sides of a rhombus are the same length. One side of a rhombus is 7 cm long. Work out the perimeter of the rhombus.
- 2.A gardener wants to buy edging trim to go all the way around a circular flower bed with a diameter of 8 m. Work out the length of edging trim needed, using π = 3.14. Give your answer to the nearest metre.
- 3.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.
- 4.A solid ball has a radius of 7 cm. Work out the surface area of the ball. Use π = 3.14.
- 5.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.
- 6.A square tile has sides of length 4.5 cm. Work out the perimeter of the tile.
- 7.A circular tabletop has a radius of 15 cm. Work out the area of the tabletop. Use π = 3.14. Give your answer to 1 decimal place.
- 8.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.
- 9.A rectangular photograph is 12 cm long and 5 cm wide. Work out the perimeter of the photograph.
- 10.A circular pizza base has a radius of 12 cm. Work out the area of the pizza base. Use π = 3.14.
- 11.A composite solid is made from a cylinder of radius 3 cm and height 10 cm, with a cone of the same radius and vertical height 4 cm fixed on top, apex upward. Work out the total volume of the solid. Use π = 3.14 and give your answer correct to 1 decimal place.
- 12.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.
- 13.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.
- 14.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.
- 15.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.
Answer key
- (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.
- (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.
- (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.
- (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) 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) 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.
- (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.
- (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.
- (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.
- (d) 452.16 cm² — Area = πr². Substitute r = 12: area = 3.14 × 12² = 3.14 × 144 = 452.16 cm².
- (b) 320.3 cm³ — Cylinder volume = πr²h = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 282.6 cm³. Cone volume = (1/3)πr²h = (1/3) × 3.14 × 9 × 4 = (1/3) × 113.04 = 37.68 cm³. Total = 282.6 + 37.68 = 320.28 cm³, which rounds to 320.3 cm³.
- (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.
- (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.
- (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.
- (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.
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