Volume of Composite Solids

Find the volume of a solid made of cuboids, prisms, cylinders and half-cylinders by adding its parts or taking away a hole, and the capacity and the material of hollow boxes and pipes.

Before you start

The volume of any solid with a uniform cross-section is area of cross-section × length. A cuboid has volume length × breadth × height (Volume of Prisms).
A cylinder of radius $r$ and height $h$ has volume $\pi r^{2}h$ ; a half-cylinder of radius $r$ and length $l$ has volume $\frac{1}{2}\pi r^{2}l$ (Volume of Cylinders).
The area of a composite figure, such as a rectangle with a semicircle on one side, is found by adding the areas of its parts (Composite Figures with Circular Parts).
$1 \text{ cm}^{3} = 1$ mL, $1000 \text{ cm}^{3} = 1$ L and $1 \text{ m}^{3} = 1\,000\,000 \text{ cm}^{3}$ (Converting Units of Area and Volume).

Composite Solids: Adding the Parts

A composite solid is a solid made up of two or more simple solids whose volumes you already know how to find: cuboids, prisms, cylinders and half-cylinders. A trophy stand, a two-tier cake, a step block, a letter box and a tunnel are all composite solids.

To find its volume, split it into simple solids that fit together with no gap and no overlap. Each part has its own volume, and together they fill the whole solid, so the volumes add up.

Volume of a composite solid

Volume of a composite solid $=$ sum of the volumes of its parts

Where a part has been taken away (a hole, the inside of a hollow box): volume $=$ volume of the whole solid $-$ volume of the part taken away

How to work

1. Name the simple solids the composite solid is made of.

2. Read off each part's own measurements. Some must be worked out from the others: the height of one part may be the total height less the height of another.

3. Find the volume of each part, keeping full values, then add (or subtract). Round only the final answer.

Example 1: A cylinder on a cuboid

A stand for a trophy is made of a cylinder standing on the middle of a cuboid, as shown.

Find the volume of the stand, (a) in terms of $\pi$ , (b) correct to 3 significant figures, using the $\pi$ button on your calculator.

Solution

The stand is a cuboid with a cylinder on it.

Volume of the cuboid $= 20 \times 20 \times 8 = 3200 \text{ cm}^{3}$

Volume of the cylinder $= \pi r^{2}h = \pi \times 6^{2} \times 15 = 540\pi \text{ cm}^{3}$

(a) Volume of the stand $= (3200 + 540\pi) \text{ cm}^{3}$ . The two terms cannot be put together: 3200 is a number and $540\pi$ is a number times $\pi$ .

(b) Volume $= 3200 + 1696.460\ldots = 4896.460\ldots \approx 4900 \text{ cm}^{3}$

The volume of the stand is $(3200 + 540\pi) \text{ cm}^{3}$ , which is about $4900 \text{ cm}^{3}$ .

Example 2: A step block, found three ways

A concrete block with two steps is 60 cm wide. Its measurements are shown.

Find its volume in $\text{cm}^{3}$ and in $\text{m}^{3}$ .

Solution

The block can be split into two cuboids in more than one way. Each way must give the same volume.

The lower step is $40 - 25 = 15$ cm deep and $30 - 15 = 15$ cm high.

Way 1: a level cut. The cut makes a bottom cuboid P, 40 cm by 15 cm by 60 cm, and a cuboid Q on top of it, 25 cm by 15 cm by 60 cm.

Volume $= 40 \times 15 \times 60 + 25 \times 15 \times 60 = 36\,000 + 22\,500 = 58\,500 \text{ cm}^{3}$

Way 2: an upright cut. The cut makes a tall cuboid P, 25 cm by 30 cm by 60 cm, and a low cuboid Q, 15 cm by 15 cm by 60 cm.

Volume $= 25 \times 30 \times 60 + 15 \times 15 \times 60 = 45\,000 + 13\,500 = 58\,500 \text{ cm}^{3}$

Way 3: one prism. The block has the same L-shaped cross-section all the way across, so it is a prism 60 cm long.

Area of cross-section $= 40 \times 15 + 25 \times 15 = 600 + 375 = 975 \text{ cm}^{2}$

Volume $= 975 \times 60 = 58\,500 \text{ cm}^{3}$

All three ways agree. In cubic metres: $58\,500 \div 1\,000\,000 = 0.0585 \text{ m}^{3}$ .

The volume of the block is $58\,500 \text{ cm}^{3}$ , which is $0.0585 \text{ m}^{3}$ .

Example 3: A two-tier cake

Mrs Goh bakes a two-tier cake. The bottom tier is a cylinder of diameter 24 cm and height 10 cm. The top tier, a cylinder of diameter 16 cm and height 8 cm, stands on the middle of it.

Find the volume of the cake, (a) in terms of $\pi$ , (b) correct to 3 significant figures, using the $\pi$ button on your calculator.

Solution

The radii are $24 \div 2 = 12$ cm and $16 \div 2 = 8$ cm.

Volume of the bottom tier $= \pi \times 12^{2} \times 10 = 1440\pi \text{ cm}^{3}$

Volume of the top tier $= \pi \times 8^{2} \times 8 = 512\pi \text{ cm}^{3}$

(a) Volume of the cake $= 1440\pi + 512\pi = 1952\pi \text{ cm}^{3}$ . Both terms are numbers times $\pi$ , so they add.

(b) $1952\pi = 6132.388\ldots \approx 6130 \text{ cm}^{3}$

The volume of the cake is $1952\pi \text{ cm}^{3}$ , which is about $6130 \text{ cm}^{3}$ .

Tunnels and Arches: A Composite Cross-Section

Some composite solids have the same cross-section all along their length: a tunnel whose end is a rectangle with a semicircle on top, an arch-shaped letter box, a step with a rounded edge. Such a solid is not a prism (its cross-section has a curved side), but it is a solid with a uniform cross-section, so its volume is area of cross-section × length.

The cross-section is a composite figure, and its area is found as you found the areas of composite figures with circular parts. There are then two ways to the volume, and they must agree:

Two ways to the volume of a tunnel-shaped solid

Way 1: volume $=$ (area of the rectangle $+$ area of the semicircle) $\times$ length

Way 2: volume $=$ volume of the cuboid $+$ volume of the half-cylinder

The semicircle sits on the top side of the rectangle, so its diameter is the width of the rectangle.

Example 4: The earth removed for an underpass

An underpass is a tunnel 12 m long. Its cross-section is a rectangle 4 m wide with sides 2.5 m high, and a semicircle on top.

(a) Find the volume of earth removed to make the underpass, correct to 3 significant figures. Use the $\pi$ button on your calculator.

(b) Removing the earth costs $45 for each cubic metre. Find the cost.

Solution

The semicircle's diameter is the width, 4 m, so its radius is $4 \div 2 = 2$ m.

(a) Way 1: area of cross-section × length.

Area of the rectangle $= 4 \times 2.5 = 10 \text{ m}^{2}$

Area of the semicircle $= \frac{1}{2}\pi r^{2} = \frac{1}{2} \times \pi \times 2^{2} = 2\pi \text{ m}^{2}$

Area of cross-section $= (10 + 2\pi) \text{ m}^{2}$

Volume $= (10 + 2\pi) \times 12 = 120 + 24\pi = 195.398\ldots \text{ m}^{3}$

Way 2: a cuboid and a half-cylinder, both 12 m long.

Volume of the cuboid $= 4 \times 2.5 \times 12 = 120 \text{ m}^{3}$

Volume of the half-cylinder $= \frac{1}{2}\pi r^{2}l = \frac{1}{2} \times \pi \times 2^{2} \times 12 = 24\pi \text{ m}^{3}$

Volume $= 120 + 24\pi = 195.398\ldots \approx 195 \text{ m}^{3}$ , as before.

(b) Cost $= 195.398\ldots \times 45 = 8792.920\ldots$ , so the cost is $8792.92 to the nearest cent. (Using the rounded 195 would give $8775, which is $17.92 out: keep the full value.)

About $195 \text{ m}^{3}$ of earth is removed, at a cost of $8792.92.

Taking a Part Away: Holes and Pipes

When a hole is drilled right through a solid, the material taken out is itself a simple solid: a cylinder for a round hole, a cuboid for a square one. Its length is the distance right through the solid. Find the volume of the solid as if it had no hole, then subtract the volume of the hole.

Example 5: A block with a hole

A wooden block is a cuboid 10 cm long, 8 cm wide and 6 cm high. A hole of radius 2 cm is drilled from the middle of its top face straight down through the block.

Find the volume of wood left, correct to 3 significant figures. Use the $\pi$ button on your calculator.

Solution

Volume of the cuboid $= 10 \times 8 \times 6 = 480 \text{ cm}^{3}$

The hole is a cylinder of radius 2 cm, as long as the block is high: 6 cm.

Volume of the hole $= \pi \times 2^{2} \times 6 = 24\pi \text{ cm}^{3}$

Volume of wood $= 480 - 24\pi = 480 - 75.398\ldots = 404.601\ldots \approx 405 \text{ cm}^{3}$

The volume of wood left is about $405 \text{ cm}^{3}$ .

A pipe (or a hollow cylinder) is a cylinder with a cylinder-shaped hole along its middle. Its external radius $R$ is the radius of the outside; its internal radius $r$ is the radius of the hole. The material of the pipe is the large cylinder less the hole, both as long as the pipe.

Volume of material in a pipe

For a pipe of external radius $R$ , internal radius $r$ and length $l$ :

volume of material $= \pi R^{2}l - \pi r^{2}l$

which can be factorised as $\pi l(R^{2} - r^{2})$ . Square each radius first, then subtract: $R^{2} - r^{2}$ is not $(R - r)^{2}$ .

A pipe whose wall is $t$ thick: internal radius $=$ external radius $- t$ , but internal diameter $=$ external diameter $- 2t$ , since the wall is on both sides of the hole.

Example 6: The mass of a metal pipe

A metal pipe is 60 cm long. Its external diameter is 12 cm and its internal diameter is 10 cm. $1 \text{ cm}^{3}$ of the metal has a mass of 7.8 g.

Use the $\pi$ button on your calculator to find (a) the volume of metal in the pipe, (b) the mass of the pipe in kilograms, each correct to 3 significant figures.

Solution

External radius $R = 12 \div 2 = 6$ cm; internal radius $r = 10 \div 2 = 5$ cm; length $l = 60$ cm.

(a) Volume of metal $= \pi R^{2}l - \pi r^{2}l = \pi \times 6^{2} \times 60 - \pi \times 5^{2} \times 60$

$= 2160\pi - 1500\pi = 660\pi = 2073.451\ldots \approx 2070 \text{ cm}^{3}$

(b) Each $\text{cm}^{3}$ has a mass of 7.8 g, so mass $= 2073.451\ldots \times 7.8 = 16\,172.918\ldots$ g

$= 16\,172.918\ldots \div 1000 = 16.172\ldots$ kg $\approx 16.2$ kg

The pipe holds about $2070 \text{ cm}^{3}$ of metal and has a mass of about 16.2 kg.

Hollow Boxes: Walls of a Given Thickness

A wooden box has external measurements (its outside) and internal measurements (the space inside). The wood takes up the difference. Cut the box down through its middle and look at it from the front: there is a wall at each end of the length, so the inside is shorter than the outside by two thicknesses. The same is true of the breadth.

The height depends on the lid. An open box (no lid) has only a base below the space inside, so its inside is lower by one thickness. A closed box has a base and a lid, so its inside is lower by two thicknesses.

Internal measurements of a box

For a box whose walls, base and lid are each $t$ thick:

Internal length $=$ external length $- 2t$ and internal breadth $=$ external breadth $- 2t$

Internal height $=$ external height $- t$ for an open box, and external height $- 2t$ for a closed box

The capacity (how much the box holds) is found from the internal measurements.

Volume of material $=$ external volume $-$ internal volume

Example 7: An open box

An open wooden box (it has no lid) measures 30 cm by 20 cm by 15 cm high on the outside. The wood is 1 cm thick.

Find (a) the capacity of the box in litres, (b) the volume of wood used to make it.

Solution

Internal length $= 30 - 2 \times 1 = 28$ cm; internal breadth $= 20 - 2 \times 1 = 18$ cm.

The box is open, so there is wood only below the space inside: internal height $= 15 - 1 = 14$ cm.

(a) Capacity $= 28 \times 18 \times 14 = 7056 \text{ cm}^{3} = 7056 \div 1000 = 7.056$ L

(b) External volume $= 30 \times 20 \times 15 = 9000 \text{ cm}^{3}$

Volume of wood $= 9000 - 7056 = 1944 \text{ cm}^{3}$

The box holds 7.056 L, and $1944 \text{ cm}^{3}$ of wood is used.

Example 8: A closed box, and its mass

A closed wooden box measures 24 cm long, 16 cm wide and 12 cm high on the outside. The wood is 2 cm thick everywhere. $1 \text{ cm}^{3}$ of the wood has a mass of 0.65 g.

Find (a) the capacity of the box, (b) the volume of wood, (c) the mass of the empty box in kilograms, correct to 3 significant figures.

Solution

Internal length $= 24 - 2 \times 2 = 20$ cm; internal breadth $= 16 - 2 \times 2 = 12$ cm.

The box is closed, with wood below and above the space inside: internal height $= 12 - 2 \times 2 = 8$ cm.

(a) Capacity $= 20 \times 12 \times 8 = 1920 \text{ cm}^{3}$

(b) External volume $= 24 \times 16 \times 12 = 4608 \text{ cm}^{3}$

Volume of wood $= 4608 - 1920 = 2688 \text{ cm}^{3}$

The box holds $1920 \text{ cm}^{3}$ (1.92 L); it is made of $2688 \text{ cm}^{3}$ of wood and has a mass of about 1.75 kg.

Space Left in a Box, and Solids Made of Cubes

A can packed in a box leaves empty space round it. The space left is the volume of the box less the volume of what is in it. When a can fits exactly, it touches the four sides, the bottom and the lid, so the box is as long and as wide as the can's diameter and as tall as the can.

Example 9: The space round a can

A can of radius 4 cm and height 12 cm fits exactly in a box with a square base.

(a) Write down the measurements of the box.

(b) Find the volume of space left in the box, correct to 3 significant figures. Use the $\pi$ button on your calculator.

Solution

(a) The can touches all four sides, so the box is as long and as wide as the can's diameter, $2 \times 4 = 8$ cm; it is as tall as the can, 12 cm. The box is 8 cm by 8 cm by 12 cm.

(b) Volume of the box $= 8 \times 8 \times 12 = 768 \text{ cm}^{3}$

Volume of the can $= \pi \times 4^{2} \times 12 = 192\pi = 603.185\ldots \text{ cm}^{3}$

Space left $= 768 - 192\pi = 164.814\ldots \approx 165 \text{ cm}^{3}$

The box is 8 cm by 8 cm by 12 cm; about $165 \text{ cm}^{3}$ of it, or 21.5%, is empty space.

A solid may be built of small identical cubes. If every length of the solid is a whole number of cube edges, the cubes fill it exactly, with no gaps and no pieces cut, and the number of cubes is the volume of the solid divided by the volume of one cube. If a length is not a whole number of edges, count the cubes along each edge instead, as for boxes in a carton.

Example 10: How many cubes?

The L-shaped solid is made of identical cubes of edge 2 cm glued together.

How many cubes are there?

Solution

Every length (10 cm, 4 cm, 6 cm, and the upright part's $10 - 4 = 6$ cm) is a whole number of 2 cm edges, so the cubes fit exactly.

The solid is a prism 6 cm long. Area of its L-shaped cross-section $= 10 \times 4 + 4 \times 6 = 40 + 24 = 64 \text{ cm}^{2}$

Volume $= 64 \times 6 = 384 \text{ cm}^{3}$ ; volume of one cube $= 2^{3} = 8 \text{ cm}^{3}$

Number of cubes $= 384 \div 8 = 48$

Check by counting. The bottom block, 10 cm by 4 cm by 6 cm, is $5 \times 2 \times 3 = 30$ cubes. The upright block on its left, 4 cm by 6 cm by 6 cm, is $2 \times 3 \times 3 = 18$ cubes. $30 + 18 = 48$ .

The solid is made of 48 cubes.

Common mistakes

Using the whole height for one part. In Example 1 the cylinder is 15 cm high, not $8 + 15 = 23$ cm. Read off each part's own measurements, and work out the ones not given (the lower step in Example 2 is $40 - 25 = 15$ cm deep).
Taking a whole circle, or the wrong radius, for an arch. The top of a tunnel is a semicircle: half of $\pi r^{2}$ . Its diameter is the width of the tunnel, so a tunnel 4 m wide has $r = 2$ m.
Taking the thickness off only once. A box has a wall at each end of its length and of its breadth: take off $2t$ from each. Take $t$ off the height of an open box and $2t$ off the height of a closed one.
Using the external measurements for the capacity. What a box holds is the space inside it: use the internal measurements.
Writing $\pi(R - r)^{2}l$ for a pipe. The material is $\pi R^{2}l - \pi r^{2}l$ . For the pipe of Example 6, $\pi(6 - 5)^{2} \times 60 = 60\pi$ , but the true volume is $660\pi$ .
Mixing units. Change every length to one unit (all metres or all centimetres) before multiplying.
Rounding a part before adding. Keep each part's full value (or keep it in terms of $\pi$ ) and round only the final answer.

Practice

Try each question before you open its solution. Each question says which value of π to use; give answers that are not exact correct to 3 significant figures, and money to the nearest cent.

Question 1

A lamp base is a cylinder standing on the middle of a cuboid. Find its volume, leaving your answer in terms of $\pi$ .

ShowHide solution

Volume of the cuboid $= 12 \times 12 \times 4 = 576 \text{ cm}^{3}$

The cylinder's radius is $6 \div 2 = 3$ cm. Volume of the cylinder $= \pi \times 3^{2} \times 10 = 90\pi \text{ cm}^{3}$

Volume of the lamp base $= (576 + 90\pi) \text{ cm}^{3}$

The volume is $(576 + 90\pi) \text{ cm}^{3}$ .

Question 2

The podium for a school sports day is made of three wooden blocks side by side, each 50 cm wide and 60 cm from front to back. The blocks for first, second and third places are 60 cm, 40 cm and 30 cm high.

Find the volume of the podium in $\text{m}^{3}$ .

ShowHide solution

Work in centimetres, then change the answer. Each block is a cuboid 50 cm wide and 60 cm from front to back.

Volume $= 50 \times 60 \times 60 + 50 \times 60 \times 40 + 50 \times 60 \times 30$

$= 180\,000 + 120\,000 + 90\,000 = 390\,000 \text{ cm}^{3}$

Or as one prism 60 cm long: area of the stepped cross-section $= 50 \times 40 + 50 \times 60 + 50 \times 30 = 6500 \text{ cm}^{2}$ , and $6500 \times 60 = 390\,000 \text{ cm}^{3}$ .

In cubic metres: $390\,000 \div 1\,000\,000 = 0.39 \text{ m}^{3}$ .

The volume of the podium is $0.39 \text{ m}^{3}$ .

Question 3

A letter box at the gate of a house is 40 cm from front to back. Its cross-section is a rectangle 28 cm wide and 30 cm high with a semicircle on top.

Taking $\pi = \frac{22}{7}$ , find the volume of the letter box in $\text{cm}^{3}$ and in litres.

ShowHide solution

The semicircle's diameter is the width, 28 cm, so $r = 14$ cm.

Area of the rectangle $= 28 \times 30 = 840 \text{ cm}^{2}$

Area of the semicircle $= \frac{1}{2} \times \frac{22}{7} \times 14^{2} = \frac{1}{2} \times \frac{22}{7} \times 196 = 308 \text{ cm}^{2}$

Volume $= (840 + 308) \times 40 = 1148 \times 40 = 45\,920 \text{ cm}^{3}$

Check as a cuboid and a half-cylinder: $28 \times 30 \times 40 + \frac{1}{2} \times \frac{22}{7} \times 14^{2} \times 40 = 33\,600 + 12\,320 = 45\,920 \text{ cm}^{3}$ .

In litres: $45\,920 \div 1000 = 45.92$ L.

The volume of the letter box is $45\,920 \text{ cm}^{3}$ , which is 45.92 L.

Question 4

A concrete block is a cuboid 40 cm long, 20 cm wide and 20 cm high, with two holes of radius 5 cm going straight down through it.

Find the volume of concrete in the block, correct to 3 significant figures. Use the $\pi$ button on your calculator.

ShowHide solution

Volume of the cuboid $= 40 \times 20 \times 20 = 16\,000 \text{ cm}^{3}$

Each hole is a cylinder of radius 5 cm and length 20 cm (the height of the block).

Volume of the two holes $= 2 \times \pi \times 5^{2} \times 20 = 1000\pi \text{ cm}^{3}$

Volume of concrete $= 16\,000 - 1000\pi = 16\,000 - 3141.592\ldots = 12\,858.407\ldots \approx 12\,900 \text{ cm}^{3}$

The block contains about $12\,900 \text{ cm}^{3}$ of concrete.

Question 5

A concrete pipe for a monsoon drain is 2 m long. Its external diameter is 1.4 m and its internal diameter is 1.2 m.

Find the volume of concrete in the pipe, (a) in terms of $\pi$ , (b) correct to 3 significant figures, using the $\pi$ button on your calculator.

ShowHide solution

$R = 1.4 \div 2 = 0.7$ m, $r = 1.2 \div 2 = 0.6$ m, $l = 2$ m.

(a) Volume of concrete $= \pi R^{2}l - \pi r^{2}l = \pi \times 0.7^{2} \times 2 - \pi \times 0.6^{2} \times 2$

$= 0.98\pi - 0.72\pi = 0.26\pi \text{ m}^{3}$

(b) $0.26\pi = 0.816\,814\ldots \approx 0.817 \text{ m}^{3}$

The pipe contains $0.26\pi \text{ m}^{3}$ , about $0.817 \text{ m}^{3}$ , of concrete.

Question 6

A metal water pipe is 50 cm long. Its external diameter is 10 cm and its wall is 1 cm thick.

Find (a) its internal diameter, (b) the volume of metal in the pipe, in terms of $\pi$ and correct to 3 significant figures, using the $\pi$ button on your calculator.

ShowHide solution

(a) The wall is on both sides of the hole, so internal diameter $= 10 - 2 \times 1 = 8$ cm.

(b) External radius $R = 10 \div 2 = 5$ cm; internal radius $r = 5 - 1 = 4$ cm (or $8 \div 2 = 4$ cm); $l = 50$ cm.

Volume of metal $= \pi R^{2}l - \pi r^{2}l = \pi \times 5^{2} \times 50 - \pi \times 4^{2} \times 50$

$= 1250\pi - 800\pi = 450\pi = 1413.716\ldots \approx 1410 \text{ cm}^{3}$

Check with the factorised form: $\pi \times 50 \times (25 - 16) = 450\pi$ .

(a) The internal diameter is 8 cm. (b) The pipe holds $450\pi \text{ cm}^{3}$ , about $1410 \text{ cm}^{3}$ , of metal.

Question 7

An open wooden planter box at a void deck measures 60 cm by 25 cm by 20 cm high on the outside. The wood is 2 cm thick.

Find (a) the volume of soil needed to fill the box, in litres, correct to 3 significant figures, (b) the volume of wood in the box.

ShowHide solution

Internal length $= 60 - 2 \times 2 = 56$ cm; internal breadth $= 25 - 2 \times 2 = 21$ cm.

The box is open: internal height $= 20 - 2 = 18$ cm.

(a) Capacity $= 56 \times 21 \times 18 = 21\,168 \text{ cm}^{3} = 21\,168 \div 1000 = 21.168$ L $\approx 21.2$ L

(b) External volume $= 60 \times 25 \times 20 = 30\,000 \text{ cm}^{3}$

Volume of wood $= 30\,000 - 21\,168 = 8832 \text{ cm}^{3}$

About 21.2 L of soil fills the box, which is made of $8832 \text{ cm}^{3}$ of wood.

Question 8

Ravi makes a closed wooden box measuring 40 cm by 30 cm by 25 cm high on the outside, from wood 2.5 cm thick. $1 \text{ cm}^{3}$ of the wood has a mass of 0.7 g.

Find (a) the capacity of the box in litres, (b) the mass of the box in kilograms.

ShowHide solution

Internal length $= 40 - 2 \times 2.5 = 35$ cm; internal breadth $= 30 - 2 \times 2.5 = 25$ cm.

The box is closed: internal height $= 25 - 2 \times 2.5 = 20$ cm.

(a) Capacity $= 35 \times 25 \times 20 = 17\,500 \text{ cm}^{3} = 17.5$ L

(b) External volume $= 40 \times 30 \times 25 = 30\,000 \text{ cm}^{3}$

Volume of wood $= 30\,000 - 17\,500 = 12\,500 \text{ cm}^{3}$

Mass $= 12\,500 \times 0.7 = 8750$ g $= 8.75$ kg

The box holds 17.5 L and has a mass of 8.75 kg.

Question 9

Six cans of radius 4 cm and height 12 cm stand in two rows of three in a box 24 cm by 16 cm by 12 cm. Each can touches the cans next to it and the sides of the box.

Find the volume of empty space in the box, correct to 3 significant figures. Use the $\pi$ button on your calculator.

ShowHide solution

Check that they fit: three diameters make $3 \times 8 = 24$ cm, two make $2 \times 8 = 16$ cm, and the cans are as tall as the box.

Volume of the box $= 24 \times 16 \times 12 = 4608 \text{ cm}^{3}$

Volume of the six cans $= 6 \times \pi \times 4^{2} \times 12 = 1152\pi = 3619.114\ldots \text{ cm}^{3}$

Empty space $= 4608 - 1152\pi = 988.885\ldots \approx 989 \text{ cm}^{3}$

About $989 \text{ cm}^{3}$ of the box is empty.

Question 10

A concrete step is 1 m long. Its cross-section is a rectangle 30 cm wide and 15 cm high, with a quarter circle of radius 15 cm joined to one side to round off the front.

Taking $\pi = 3.142$ , find the volume of the step in $\text{cm}^{3}$ and in $\text{m}^{3}$ , each correct to 3 significant figures.

ShowHide solution

Change the length to centimetres: 1 m $= 100$ cm.

Area of the rectangle $= 30 \times 15 = 450 \text{ cm}^{2}$

Area of the quarter circle $= \frac{1}{4}\pi r^{2} = \frac{1}{4} \times 3.142 \times 15^{2} = \frac{1}{4} \times 3.142 \times 225 = 176.7375 \text{ cm}^{2}$

Area of cross-section $= 450 + 176.7375 = 626.7375 \text{ cm}^{2}$

Volume $= 626.7375 \times 100 = 62\,673.75 \text{ cm}^{3} \approx 62\,700 \text{ cm}^{3}$

In cubic metres: $62\,673.75 \div 1\,000\,000 = 0.062\,673\,75 \approx 0.0627 \text{ m}^{3}$

The volume of the step is about $62\,700 \text{ cm}^{3}$ , or $0.0627 \text{ m}^{3}$ .

Question 11

A wooden cube of edge 9 cm has a square hole of side 3 cm cut through the middle of it, from front to back. Siti wants to build the same shape from small cubes of edge 3 cm.

How many small cubes does she need?

ShowHide solution

Every length (9 cm, 3 cm, and the 3 cm of wood on each side of the hole) is a whole number of 3 cm edges, so the small cubes fit exactly.

Volume of the solid $= 9 \times 9 \times 9 - 3 \times 3 \times 9 = 729 - 81 = 648 \text{ cm}^{3}$

Volume of one small cube $= 3^{3} = 27 \text{ cm}^{3}$

Number of cubes $= 648 \div 27 = 24$

Check: the whole cube would be $3 \times 3 \times 3 = 27$ small cubes, and the hole takes out a row of 3 of them, front to back: $27 - 3 = 24$ .

She needs 24 small cubes.

Question 12

A solid is made of a cylinder of radius 3.5 cm standing on the middle of a cuboid 10 cm by 10 cm by 4 cm. The volume of the solid is $708 \text{ cm}^{3}$ .

(not drawn to scale)

Taking $\pi = \frac{22}{7}$ , find the height $h$ cm of the cylinder.

ShowHide solution

Volume of the cuboid $= 10 \times 10 \times 4 = 400 \text{ cm}^{3}$

Volume of the cylinder $= \frac{22}{7} \times 3.5^{2} \times h = \frac{22}{7} \times 12.25 \times h = 38.5h \text{ cm}^{3}$

The two parts make up the solid, so $400 + 38.5h = 708$ .

$38.5h = 708 - 400 = 308$

$h = 308 \div 38.5 = 8$

Check: $400 + 38.5 \times 8 = 400 + 308 = 708$ .

The cylinder is 8 cm high.

Question 13

Mr Tan buys a concrete flower pot in the shape of an open cylinder. Its external diameter is 60 cm and its external height is 50 cm. The wall and the base are each 5 cm thick. $1 \text{ cm}^{3}$ of the concrete has a mass of 2.4 g.

Use the $\pi$ button on your calculator to find (a) the volume of soil the pot holds, in litres, (b) the mass of the empty pot in kilograms, each correct to 3 significant figures.

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External radius $= 60 \div 2 = 30$ cm. The wall goes all round, so the internal radius is $30 - 5 = 25$ cm (the internal diameter is $60 - 2 \times 5 = 50$ cm).

The pot is open, with concrete only below the soil: internal height $= 50 - 5 = 45$ cm.

(a) Capacity $= \pi \times 25^{2} \times 45 = 28\,125\pi = 88\,357.293\ldots \text{ cm}^{3}$

$= 88\,357.293\ldots \div 1000 = 88.357\ldots$ L $\approx 88.4$ L

(b) External volume $= \pi \times 30^{2} \times 50 = 45\,000\pi \text{ cm}^{3}$

Volume of concrete $= 45\,000\pi - 28\,125\pi = 16\,875\pi = 53\,014.376\ldots \text{ cm}^{3}$

Mass $= 53\,014.376\ldots \times 2.4 = 127\,234.502\ldots$ g $= 127.234\ldots$ kg $\approx 127$ kg

The pot holds about 88.4 L of soil and has a mass of about 127 kg.

Key points

A composite solid is made of simple solids: cuboids, prisms, cylinders, half-cylinders. Its volume is the sum of the volumes of its parts, which fit with no gap and no overlap.
Where a part is taken away (a hole), subtract its volume from the volume of the whole solid.
A solid can often be split in more than one way; every way gives the same volume.
A composite solid with the same cross-section all along it (a tunnel, an arch) has volume $=$ area of cross-section $\times$ length, the area found as for a composite figure.
A tunnel's semicircle has the tunnel's width as its diameter; its half-cylinder has volume $\frac{1}{2}\pi r^{2}l$ .
A pipe of external radius $R$ , internal radius $r$ and length $l$ has $\pi R^{2}l - \pi r^{2}l$ of material; a wall $t$ thick makes $r = R - t$ and the internal diameter the external diameter $- 2t$ .
A box with walls $t$ thick: internal length and breadth are external $- 2t$ ; internal height is external $- t$ if open, external $- 2t$ if closed.
Capacity uses the internal measurements; the volume of material is external volume $-$ internal volume.
Mass $=$ volume $\times$ mass of $1 \text{ cm}^{3}$ ; cost $=$ volume $\times$ cost of one unit of volume.
Small cubes fill a solid exactly when every length is a whole number of cube edges; then the number of cubes is the volume divided by the volume of one cube.
Keep full values (or $\pi$ ) through the working and round only the final answer.