Volume of a Cone

Find the volume of a cone with $V = \frac{1}{3}\pi r^{2} h$ , find its height from the slant height by Pythagoras' theorem, and work back from a volume to a radius or a height.

Before you start

The volume of a cylinder of radius $r$ and height $h$ is $\pi r^{2} h$ : its base area $\pi r^{2}$ times its height.
A pyramid holds a third of the prism with the same base and height (Volume of a Pyramid).
$1000 \text{ cm}^{3} = 1$ litre, and $1 \text{ m}^{3} = 1000$ litres.

The Parts of a Cone

A cone has a circular base and a curved surface that rises to a point, the apex (or vertex) $V$ . In the cones of this chapter the apex is straight above the centre $O$ of the base.

Three lengths describe a cone: the radius $r$ of the base; the height $h$ (perpendicular height), from the apex straight down to the centre of the base, inside the cone; and the slant height $l$ , from the apex down the curved surface to the edge of the base.

The height, a radius and the slant height make a right-angled triangle $VOA$ inside the cone, with the right angle at $O$ . The slant height is its hypotenuse, so it is always the longest of the three.

Radius, height and slant height

$l^{2} = r^{2} + h^{2}$

The slant height is used for the curved surface (Surface Area of a Cone); the volume uses the height $h$ .

The Volume of a Cone

A cone is like a pyramid with a circular base. A pyramid holds a third of the prism with the same base and height; in the same way a cone holds a third of the cylinder with the same base and the same height. Fill a hollow cone with water and pour it into a cylinder of the same radius and height: it takes exactly three cones to fill the cylinder. This shows the third; it does not prove it.

Volume of a cone

$V = \frac{1}{3} \pi r^{2} h$

where $r$ is the radius of the base and $h$ the perpendicular height. Use the calculator's $\pi$ unless a question says "take $\pi = 3.142$ " or "leave your answer in terms of $\pi$ ".

Example 1: Radius and height

A cone has a base radius of $3$ cm and a height of $4$ cm. Find its volume (a) in terms of $\pi$ , (b) correct to 3 significant figures.

Solution

$V = \frac{1}{3} \times \pi \times 3^{2} \times 4 = \frac{1}{3} \times 36\pi = 12\pi$

$12\pi = 37.699\ldots = 37.7$ (3 s.f.)

(a) $12\pi \text{ cm}^{3}$ (b) $37.7 \text{ cm}^{3}$

Example 2: From the diameter

A cone has a base of diameter $10$ cm and a height of $9$ cm. Find its volume.

Solution

The formula needs the radius: $r = 10 \div 2 = 5$ cm.

$V = \frac{1}{3} \times \pi \times 5^{2} \times 9 = 75\pi = 235.61\ldots = 236$ (3 s.f.)

The volume is $236 \text{ cm}^{3}$ .

Example 3: Taking π = 3.142

A cone has a radius of $6$ cm and a height of $14$ cm. Taking $\pi = 3.142$ , find its volume.

Solution

$V = \frac{1}{3} \times 3.142 \times 6^{2} \times 14 = \frac{1}{3} \times 3.142 \times 504 = 527.856 = 528$ (3 s.f.)

The volume is $528 \text{ cm}^{3}$ .

The Height from the Slant Height

If a question gives the slant height, find the height first, from the triangle $VOA$ : $h^{2} = l^{2} - r^{2}$ . Never put the slant height into the volume formula.

Example 4: Radius and slant height

A cone has a base radius of $5$ cm and a slant height of $13$ cm. Find (a) its height, (b) its volume.

Solution

(a) In $\triangle VOA$ , the right angle is at $O$ and $VA = 13$ cm is the hypotenuse.

$h^{2} = 13^{2} - 5^{2} = 169 - 25 = 144\text{,}$ so $h = 12$ cm.

(b) $V = \frac{1}{3} \times \pi \times 5^{2} \times 12 = 100\pi = 314.15\ldots = 314$ (3 s.f.)

(a) $12$ cm (b) $314 \text{ cm}^{3}$

Example 5: Height and slant height

A cone has a height of $6$ cm and a slant height of $10$ cm. Find its volume.

Solution

This time the radius is missing: $r^{2} = l^{2} - h^{2} = 10^{2} - 6^{2} = 100 - 36 = 64\text{,}$ so $r = 8$ cm.

$V = \frac{1}{3} \times \pi \times 8^{2} \times 6 = 128\pi = 402.12\ldots = 402$ (3 s.f.)

The volume is $402 \text{ cm}^{3}$ .

Finding the Radius or the Height from the Volume

Put the volume into $V = \frac{1}{3}\pi r^{2} h$ and solve for the unknown (Changing the Subject of a Formula). Rearranged, $h = \frac{3V}{\pi r^{2}}$ and $r = \sqrt{\frac{3V}{\pi h}}$ .

Example 6: The height

A cone has a volume of $300 \text{ cm}^{3}$ and a base radius of $5$ cm. Find its height.

Solution

$300 = \frac{1}{3} \times \pi \times 5^{2} \times h$

$900 = 25\pi h$

$h = \frac{900}{25\pi} = 11.459\ldots = 11.5$ (3 s.f.)

The height is $11.5$ cm.

Example 7: The radius

A cone has a volume of $500 \text{ cm}^{3}$ and a height of $12$ cm. Find its base radius.

Solution

$500 = \frac{1}{3} \times \pi \times r^{2} \times 12 = 4\pi r^{2}$

$r^{2} = \frac{500}{4\pi} = 39.788\ldots$

$r = \sqrt{39.788\ldots} = 6.307\ldots = 6.31$ (3 s.f.)

Take the square root of the full value of $r^{2}$ , not of a rounded one.

The radius is $6.31$ cm.

Cones in the Real World

Example 8: A heap of sand

At a building site, sand is tipped into a heap in the shape of a cone with a base of diameter $3$ m and a height of $1.2$ m. Find the volume of sand.

Solution

$r = 3 \div 2 = 1.5$ m

$V = \frac{1}{3} \times \pi \times 1.5^{2} \times 1.2 = 0.9\pi = 2.827\ldots = 2.83$ (3 s.f.)

The heap holds about $2.83 \text{ m}^{3}$ of sand.

Example 9: Paper cups

A drinks stall uses paper cups in the shape of a cone, point down, with a radius of $4$ cm and a depth of $9$ cm. How many cups can be filled to the brim from a $2$ -litre bottle of soya bean milk?

Solution

One cup: $V = \frac{1}{3} \times \pi \times 4^{2} \times 9 = 48\pi = 150.79\ldots \text{ cm}^{3}$

$2$ litres $= 2000 \text{ cm}^{3}$ . $2000 \div 150.79\ldots = 13.26\ldots$

Only full cups count, so round down.

$13$ cups can be filled.

Common Mistakes

Using the slant height as the height. For a cone of radius $5$ cm and slant height $13$ cm, $\frac{1}{3}\pi \times 25 \times 13$ is wrong: the height is $12$ cm.
Using the diameter as the radius. Halve the diameter first: a diameter of $10$ cm used as $r$ gives four times the volume.
Forgetting the third, which gives the volume of the cylinder.
Squaring the wrong thing. $\frac{1}{3}\pi r^{2} h$ squares only $r$ , not $h$ and not $\pi r$ .
Rounding $r^{2}$ or $h$ before the last step.

Practice Questions

Try each question before you open its solution. Use the calculator's $\pi$ unless the question says otherwise, and give answers correct to 3 significant figures where they are not exact.

Question 1

Find the volume of a cone with a radius of $6$ cm and a height of $10$ cm.

ShowHide solution

$V = \frac{1}{3} \times \pi \times 36 \times 10 = 120\pi = 376.99\ldots = 377$ (3 s.f.)

$377 \text{ cm}^{3}$

Question 2

A cone has a radius of $9$ cm and a height of $8$ cm. Find its volume, leaving your answer in terms of $\pi$ .

ShowHide solution

$V = \frac{1}{3} \times \pi \times 81 \times 8 = 216\pi$

$216\pi \text{ cm}^{3}$

Question 3

A cone has a base of diameter $14$ cm and a height of $12$ cm. Taking $\pi = 3.142$ , find its volume.

ShowHide solution

$r = 7$ cm. $V = \frac{1}{3} \times 3.142 \times 49 \times 12 = 615.832 = 616$ (3 s.f.)

$616 \text{ cm}^{3}$

Question 4

A cone has a radius of $8$ cm and a slant height of $17$ cm. Find (a) its height, (b) its volume.

ShowHide solution

(a) $h^{2} = 17^{2} - 8^{2} = 289 - 64 = 225\text{,}$ so $h = 15$ cm.

(b) $V = \frac{1}{3} \times \pi \times 64 \times 15 = 320\pi = 1005.3\ldots = 1010$ (3 s.f.)

(a) $15$ cm (b) $1010 \text{ cm}^{3}$

Question 5

A cone has a height of $24$ cm and a slant height of $25$ cm. Find its volume.

ShowHide solution

$r^{2} = 25^{2} - 24^{2} = 625 - 576 = 49\text{,}$ so $r = 7$ cm.

$V = \frac{1}{3} \times \pi \times 49 \times 24 = 392\pi = 1231.5\ldots = 1230$ (3 s.f.)

$1230 \text{ cm}^{3}$

Question 6

A cone with a radius of $3.5$ cm has a volume of $154 \text{ cm}^{3}$ . Find its height.

ShowHide solution

$154 = \frac{1}{3} \times \pi \times 3.5^{2} \times h\text{,}$ so $462 = 12.25\pi h$ .

$h = \frac{462}{12.25\pi} = 12.004\ldots = 12.0$ (3 s.f.)

$12.0$ cm

Question 7

A cone-shaped container $15$ cm deep holds exactly $1$ litre. Find the radius of its top.

ShowHide solution

$1$ litre $= 1000 \text{ cm}^{3}$ . $1000 = \frac{1}{3} \times \pi \times r^{2} \times 15 = 5\pi r^{2}$

$r^{2} = \frac{1000}{5\pi} = 63.66\ldots\text{,}$ so $r = 7.978\ldots = 7.98$ (3 s.f.)

$7.98$ cm

Question 8

An ice-cream cone has a radius of $2.5$ cm and a depth of $11$ cm. It is filled with ice cream level with its top. Find the volume of ice cream.

ShowHide solution

$V = \frac{1}{3} \times \pi \times 2.5^{2} \times 11 = \frac{1}{3} \times \pi \times 68.75 = 71.99\ldots = 72.0$ (3 s.f.)

$72.0 \text{ cm}^{3}$

Question 9

A water tank is a cone standing on its point, with a top of radius $1.2$ m and a depth of $2$ m. Find its capacity in litres.

ShowHide solution

$V = \frac{1}{3} \times \pi \times 1.2^{2} \times 2 = 0.96\pi = 3.0159\ldots \text{ m}^{3}$

$1 \text{ m}^{3} = 1000$ litres, so $3.0159\ldots \times 1000 = 3015.9\ldots = 3020$ litres (3 s.f.)

About $3020$ litres

Question 10

A heap of gravel is a cone with a base of diameter $4$ m and a slant height of $2.5$ m. A lorry carries $5 \text{ m}^{3}$ of gravel at a time. How many trips are needed to carry the heap away?

ShowHide solution

$r = 2$ m. $h^{2} = 2.5^{2} - 2^{2} = 6.25 - 4 = 2.25\text{,}$ so $h = 1.5$ m.

$V = \frac{1}{3} \times \pi \times 4 \times 1.5 = 2\pi = 6.283\ldots \text{ m}^{3}$

$6.283\ldots \div 5 = 1.25\ldots\text{:}$ one trip is not enough, so round up.

$2$ trips

Question 11

A cylindrical can of radius $4$ cm and height $6$ cm is full of water. The water is poured into an empty cone, point down, of radius $6$ cm and depth $12$ cm. (a) Does the water overflow? (b) What fraction of the cone is filled?

ShowHide solution

Can: $\pi \times 4^{2} \times 6 = 96\pi \text{ cm}^{3}\text{.}$ Cone: $\frac{1}{3} \times \pi \times 6^{2} \times 12 = 144\pi \text{ cm}^{3}\text{.}$

$96\pi$ is less than $144\pi$ , so it does not overflow. Fraction: $\frac{96\pi}{144\pi} = \frac{2}{3}\text{.}$

(a) No (b) $\frac{2}{3}$ of its volume

Question 12

Cone A has a radius of $6$ cm and a height of $4$ cm. Cone B has a radius of $4$ cm and the same volume as cone A. Find the height of cone B.

ShowHide solution

$\frac{1}{3} \pi \times 6^{2} \times 4 = \frac{1}{3} \pi \times 4^{2} \times h$

Cancel $\frac{1}{3}\pi$ : $144 = 16h$ , so $h = 9$ .

$9$ cm

Key Points

A cone has a radius $r$ , a height $h$ (from the apex straight down to the centre of the base) and a slant height $l$ (down the curved surface).
$l^{2} = r^{2} + h^{2}$ : the slant height is the hypotenuse of the triangle $VOA$ .
Volume of a cone $= \frac{1}{3}\pi r^{2} h$ , a third of the cylinder with the same base and height.
Use the radius, not the diameter, and the height, not the slant height.
To find $r$ or $h$ from a volume, solve the equation and round only at the end.