GCSE Maths · Volume, Surface Area and 3D Shapes · Higher · Lesson 19.10

GCSE Maths: Composite solids, frustums and similar solids (Higher)

Part of the GCSE Maths volume, surface area and 3d shapes chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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19.10

Composite solids, frustums and similar solids

Edexcel 1MA1: G16, G17, R12
Grade 6-9

Many real objects are made by joining simple solids: an ice-cream is a cone with a hemisphere on top, a grain silo is a cylinder with a cone roof, a capsule is a cylinder with a hemisphere at each end. These are composite solids. To find their volume or surface area, split them into the standard solids you know, work out each part, and combine.

Volume. Add the volumes of the parts. If a part is removed (a hole), subtract its volume. Surface area. Add only the surfaces you can touch from the outside. Where two parts are joined, the faces that meet are hidden inside, so they are not counted. For a cylinder with a hemisphere on top, the top circle of the cylinder is hidden by the hemisphere, so the surface area is: base circle + curved surface of the cylinder + curved surface of the hemisphere.

Three composite solids: a cylinder with a hemisphere on top, a cylinder with a cone on top, and an ice-cream shape (a cone with a hemisphere on top).

Frustums. A frustum is what is left of a cone when a smaller cone is cut off its top by a plane parallel to the base (a flower pot and a bucket are frustums). To find its volume, imagine the whole cone, then subtract the small cone that was removed:

volume of a frustum = volume of large cone − volume of small cone

The small cone is a smaller copy (an enlargement) of the large cone, so its measurements are in the same ratio: radius ÷ height is the same for both. Use this to find the missing height.

Left: a frustum. Right: the small cone cut off the top, put back on, to complete the large cone.

Similar solids (see Chapter 21). Two solids are similar if one is an enlargement of the other. If the lengths are multiplied by the scale factor k, then every area is multiplied by k² and every volume (and the mass of a solid of the same material) by k³. A model that is half the size (k = ½) has ¼ of the surface area and ⅛ of the volume.

Key ideas

Composite solid: split into standard solids, then add volumes (subtract any hole).
For surface area count only the outside surfaces: faces that are joined are hidden and are not counted.
Frustum volume = large cone − small cone; find the missing height with similar triangles (radius ÷ height is the same for both cones).
Similar solids with length scale factor k: area scale factor k², volume scale factor k³.
Work backwards by writing the formula with the numbers, cancelling π, then undoing each step.

Worked example 19.10.1

The solid is a cylinder with a hemisphere on top. Work out its volume, in terms of π.

Method. Split the solid into a cylinder and a hemisphere. Work out the two volumes and add them.

Identify the parts. The solid is a cylinder (radius 3 cm, height 7 cm) with a hemisphere of radius 3 cm on top.
Volume of the cylinder. Volume of a cylinder = πr²h.
π × 3² × 7 = π × 9 × 7 = 63π
Volume of the hemisphere. Half a sphere: ½ × &frac43;πr³ = ⅔πr³.
⅔ × π × 3³ = ⅔ × 27π = 18π
Add the parts. The whole solid is made of these parts, so add them. Keep the π terms together.
63π + 18π = 81π
Write the final answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
81π cm³

Check. Estimate with π ≈ 3: the total is about 243, close to the answer 254. ✓

Worked example 19.10.2

Work out the total surface area of the solid in the last example. Give your answer in terms of π.

Method. Add only the outside surfaces: the base circle, the curved surface of the cylinder and the curved surface of the hemisphere. The circle where they join is inside the solid.

Decide which surfaces are on the outside. The top circle of the cylinder is covered by the hemisphere, so it is not counted. The bottom circle is outside.
Surfaces: base circle + cylinder curved surface + hemisphere curved surface
Area of the base circle. Area of a circle = πr².
π × 3² = 9π
Area of the curved surface of the cylinder. A rectangle of width 2πr and height h: area = 2πrh.
2 × π × 3 × 7 = 42π
Area of the curved surface of the hemisphere. Half of the sphere surface 4πr², which is 2πr².
2 × π × 3² = 18π
Add the parts. The whole solid is made of these parts, so add them. Keep the π terms together.
9π + 42π + 18π = 69π
Write the final answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
69π cm²

Check. Estimate with π ≈ 3: the total is about 207, close to the answer 217. ✓

Worked example 19.10.3

An ice-cream is a cone with a hemisphere on top. The cone has radius 3 cm and vertical height 4 cm. Work out the volume of the ice-cream, to 3 significant figures.

Method. Split the ice-cream into a cone and a hemisphere with the same radius. Add the volumes.

Identify the parts. A cone of radius 3 cm and vertical height 4 cm, with a hemisphere of radius 3 cm on top.
Volume of the cone. Volume of a cone = ⅓πr²h (h is the vertical height).
⅓ × π × 3² × 4 = ⅓ × 36π = 12π
Volume of the hemisphere. Half a sphere: ½ × &frac43;πr³ = ⅔πr³.
⅔ × π × 3³ = ⅔ × 27π = 18π
Add the parts. The whole solid is made of these parts, so add them. Keep the π terms together.
12π + 18π = 30π
Use your calculator. Type the calculation in using the π button on your calculator (not 3.14). Write down the whole display before you round.
12π + 18π = 94.2478…
Round the answer. The question wants 3 significant figures. Look at the digit just after the last one you are keeping: 5 or more rounds up, 4 or less leaves it.
94.2478… = 94.2 cm³

Check. Estimate with π ≈ 3: the total is about 90, close to the answer 94. ✓

Worked example 19.10.4

A cuboid 10 cm by 6 cm by 8 cm has a cylindrical hole of radius 2 cm drilled right through it, from the front to the back. Work out the volume of the remaining solid, to 3 significant figures.

Method. The solid is a cuboid with a cylinder removed. Work out both volumes and subtract.

Identify the parts. The hole goes right through the 6 cm depth of the cuboid, so it is a cylinder of radius 2 cm and height (length) 6 cm.
Volume of the cuboid. Volume = length × width × height.
10 × 6 × 8 = 480
Volume of the cylindrical hole. Volume of a cylinder = πr²h.
π × 2² × 6 = π × 4 × 6 = 24π
Subtract the parts. We take away the part that is missing, so subtract. Keep the π terms together.
480 − 24π = 480 − 24π
Use your calculator. Type the calculation in using the π button on your calculator (not 3.14). Write down the whole display before you round.
480 − 24π = 404.6018…
Round the answer. The question wants 3 significant figures. Look at the digit just after the last one you are keeping: 5 or more rounds up, 4 or less leaves it.
404.6018… = 405 cm³

Check. Estimate with π ≈ 3: the total is about 408, close to the answer 405. ✓

Worked example 19.10.5

Work out the volume of this frustum. Give your answer in terms of π.

Key words. A frustum is what is left of a cone when the top is cut off by a plane parallel to the base. Similar shapes have the same shape but different sizes: all their lengths are in the same ratio.

Method. Frustum volume = volume of the large cone − volume of the small cone that was cut off.

Imagine the missing top of the cone. A frustum is a cone with a smaller cone cut off its top. If we put the small cone back we get a whole large cone.
Large cone: radius 6, height 4 + x. Small cone: radius 3, height x.
Use similar triangles to find x. The two cones have the same shape, so (radius ÷ height) is the same for both.
3 ÷ x = 6 ÷ (4 + x)
Cross-multiply and solve. Multiply both sides to remove the fractions.
3(4 + x) = 6x
12 + 3x = 6x
12 = 3x
x = 4
Heights of the two cones. The large cone is the frustum height plus x.
Large cone height = 4 + 4 = 8 cm
Small cone height = 4 cm
Volume of the large cone. V = ⅓πr²h.
⅓ × π × 6² × 8 = 96π
Volume of the small cone. Same formula with the smaller measurements.
⅓ × π × 3² × 4 = 12π
Subtract. The frustum is what is left when the small cone is removed.
96π − 12π = 84π
Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
V = 84π cm³

Check. Formula check: ⅓πh(R² + Rr + r²) = ⅓ × π × 4 × (36 + 18 + 9) = ⅓ × 252π = 84π. ✓

Watch out. Subtract the volumes (not the heights or radii): large cone − small cone.

Worked example 19.10.6

A solid is a cylinder of radius 3 cm and height h cm with a hemisphere of radius 3 cm on top. Its total volume is 117π cm³. Work out h.

Method. Write the volume of the solid as an expression in h, put it equal to the total, and solve.

Volume of the cylinder. V = πr²h with r = 3.
π × 3² × h = 9πh
Volume of the hemisphere. ⅔πr³ with r = 3.
⅔ × π × 27 = 18π
Add and put equal to the total. The total volume is 117π.
9πh + 18π = 117π
Divide every term by π. The π cancels.
9h + 18 = 117
Solve. Subtract 18, then divide by 9.
9h = 99, so h = 11 cm

Check. Cylinder 9π × 11 = 99π plus hemisphere 18π = 117π. ✓

Watch out. Write the unknown height in the cylinder part only: the hemisphere has fixed size.

Worked example 19.10.7

Two similar cones have heights 4 cm and 6 cm. The volume of the smaller cone is 40 cm³. Work out the volume of the larger cone.

Key words. Two solids are similar if one is an enlargement of the other: every length is multiplied by the same scale factor. See Chapter 21 for more on similar shapes.

Method. For similar solids with length scale factor k: areas scale by k² and volumes by k³.

Find the length scale factor. Divide a length on the solid we are going to by the matching length on the solid we are going from.
6 ÷ 4 = 1.5
Find the volume scale factor. Areas are multiplied by the scale factor squared, and volumes (and masses of the same material) by the scale factor cubed. So we raise 1.5 to the power 3.
1.5³ = 1.5 × 1.5 × 1.5 = 3.375
Multiply the volume. Multiply the known amount by the scale factor.
40 × 3.375 = 135
Write the unit.
135 cm³

Check. Reverse it: 135 ÷ 3.375 = 40. ✓

Watch out. Do not multiply the volume by just the length scale factor 1.5: you must use 1.5³.

Worked example 19.10.8

Two solids are similar. The volumes are 8 cm³ and 27 cm³. Work out the ratio of their lengths and the ratio of their surface areas.

Method. The ratio of volumes is the ratio of lengths cubed. So take the cube root of each part to get the ratio of lengths, then square it for the ratio of areas.

Simplify the volume ratio. Divide by the highest common factor, 1.
8 : 27 = 8 : 27
Cube root of each part. Undo the cubing to go back to lengths.
∛8 : ∛27 = 2 : 3
Square the length ratio for areas. Areas go with the length ratio squared.
2² : 3² = 4 : 9

Check. Cube the lengths again: 2³ : 3³ = 8 : 27. ✓

Watch out. For volumes use the cube root, not the square root.

Watch out

For a composite solid do not add the surface areas of the separate solids: the faces where they are joined are hidden. For similar solids, do not multiply an area or volume by the length scale factor alone: use k² for area and k³ for volume.

Exercise 19L Composite solids and working backwards

Non-calculator

FThe solid is a cylinder with a hemisphere on top. The cylinder has radius 3 cm and height 7 cm. Work out, in terms of π, (a) the volume and (b) the total surface area.

(a)Volume

(b)Total surface area

Done

FA solid is made from a cylinder and a cone. The cylinder has radius 3 cm and height 6 cm. The cone has the same radius and vertical height 4 cm. Work out, in terms of π, (a) the volume and (b) the total surface area.

(a)Volume

(b)Total surface area

Done

FAn ice-cream is a cone with a hemisphere on top. The cone has radius 3 cm and vertical height 4 cm. Work out, in terms of π, (a) the volume of the ice-cream, (b) the area of its curved surface (the area that is outside).

(a)Volume

(b)Outside surface area

Done

PSA cuboid 10 cm by 6 cm by 8 cm has a cylindrical hole of radius 2 cm drilled right through from front to back. Work out the volume of the remaining solid, (a) in terms of π and (b) correct to 3 significant figures.

(a)In terms of π

(b)To 3 significant figures

Done

PSA medicine capsule is a cylinder of radius 2 cm and length 6 cm with a hemisphere of radius 2 cm attached to each end. Work out the total volume, correct to 3 significant figures.

Done

PSA hollow hemispherical bowl has outer radius 6 cm and inner radius 3 cm. Work out the volume of material used to make the bowl, in terms of π.

Done

PSA solid is a cylinder of radius 3 cm and height h cm with a hemisphere of radius 3 cm on top. Its total volume is 117π cm³. Work out h.

Done

RA solid is a cone on top of a cylinder. Both have radius 6 cm. The cylinder has height 10 cm. The total volume is 396π cm³. Work out the vertical height of the cone.

Done

RA cylinder of radius 3 cm and height 7 cm has a hemisphere of radius 3 cm stuck on top. Without using the formula for the whole solid, explain why its surface area (69π cm²) is less than the sum of the surface areas of the separate solid cylinder and solid hemisphere.

Done

Exercise 19M Frustums and similar solids

Non-calculator

FWork out the volume of each frustum. Give your answers in terms of π.

(a)

(b)

Done

PSA bucket is the shape of a frustum. The base has radius 10 cm, the top has radius 15 cm and the height is 20 cm. Work out the capacity of the bucket, (a) in cm³ to 3 significant figures and (b) in litres to 3 significant figures.

(a)Capacity in cm³

(b)Capacity in litres

Done

PSTwo similar solids have lengths in the ratio shown. Work out each missing quantity.

(a)Two similar cones have heights 4 cm and 6 cm. The smaller has volume 40 cm³. Find the volume of the larger.

(b)Two similar bottles have surface areas 100 cm² (height 5 cm) and a height of 8 cm. Find the surface area of the larger bottle.

(c)Two similar statues, made of the same metal, are 10 cm and 15 cm tall. The smaller has mass 250 g. Find the mass of the larger.

(d)Two similar jugs have heights 20 cm and 30 cm. The smaller holds 400 ml. Find the capacity of the larger.

(e)Two similar solids have heights 6 cm and 9 cm. The larger has volume 540 cm³. Find the volume of the smaller.

Done

FThe lengths of two similar solids are in the ratio 6 : 15. Write down the ratio of their volumes.

Done

PSTwo similar solids have volumes 27 cm³ and 64 cm³. Work out (a) the ratio of their lengths and (b) the ratio of their surface areas.

(a)Ratio of lengths

(b)Ratio of surface areas

Done

RThe larger of two similar solids has 8 times the volume of the smaller. How many times larger is its surface area?

Done

RA cone is cut into two pieces by a plane parallel to its base, at one third of the height from the top (apex). The small cone that is cut off is similar to the whole cone. Show that the frustum is 26/27 of the volume of the whole cone.

Done

PSA scale model of a ship is made at a scale of 1 : 50. The model has a hull surface area of 0.4 m² and holds 0.02 m³ of water. Work out (a) the surface area and (b) the volume of the real ship, in m² and m³.

(a)Surface area of the real ship

(b)Volume of the real ship

Done

RA frustum is made by removing the top of a cone. The large cone has radius 12 cm and height 18 cm. The small cone removed has radius 4 cm. Work out the volume of the frustum, in terms of π.

Done