Frustums of a Cone

Find the volume and the surface area of a frustum, a cone with its top cut off, as a large cone less a small one, and find the two cones from the frustum by similar figures.

Before you start

A cone of radius $r$ , height $h$ and slant height $l$ has volume $V = \frac{1}{3}\pi r^{2} h$ and curved surface area $\pi r l$ , where $l^{2} = r^{2} + h^{2}$ (Volume of a Cone, Surface Area of a Cone).
Two figures are similar when one is an enlargement of the other: their matching angles are equal and their matching lengths are in the same ratio (Similar Figures). Only ratios of lengths are needed in this lesson, not of areas or volumes.
An equation with a fraction on each side is solved by multiplying both sides by both denominators: from $\frac{2}{3} = \frac{x - 5}{x}$ , $2x = 3(x - 5) = 3x - 15\text{,}$ so $x = 15$ .

What a Frustum Is

Cut a cone with a flat cut parallel to its base, and take away the small cone at the top. The piece left is a frustum of the cone: a bucket, a lampshade, a paper cup, a flowerpot. It has two circular faces of different radii, one above the other, and a curved surface between them.

A frustum is a large cone less a small cone: work out both cones and subtract. In this lesson $R$ and $H$ are the radius and the height of the large cone, and $r$ and $h$ those of the small cone cut off it (above it, as here, or below it, for a bucket standing on its smaller circle). The frustum's own height is then $H - h$ .

Frustum of a cone

Volume of the frustum $=$ volume of the large cone $-$ volume of the small cone $= \frac{1}{3}\pi R^{2} H - \frac{1}{3}\pi r^{2} h$

Curved surface area of the frustum $= \pi R L - \pi r l$ , with $L$ and $l$ the slant heights of the two cones.

For the total surface area, add the circles the object has: both, $\pi R^{2} + \pi r^{2}$ , for a closed frustum; only the bottom for a bucket; none for a lampshade.

Example 1: The small cone given

A cone of radius $6$ cm and height $12$ cm has a small cone of radius $2$ cm and height $4$ cm cut off its top, as shown. Find the volume of the frustum left.

Solution

Large cone: $\frac{1}{3} \times \pi \times 6^{2} \times 12 = 144\pi$

Small cone: $\frac{1}{3} \times \pi \times 2^{2} \times 4 = \frac{16}{3}\pi$

Frustum: $144\pi - \frac{16}{3}\pi = \frac{416}{3}\pi = 435.63\ldots = 436$ (3 s.f.)

The volume of the frustum is $436 \text{ cm}^{3}$ .

Finding the Cones by Similar Figures

Often a question gives only part of the picture: the large cone and where it is cut, or only the frustum's two radii and its height. The missing lengths come from one fact: the large cone is an enlargement of the small one, so the two cones are similar.

Cut both cones down the middle, through the axis: each cut shows an isosceles triangle. Take half of each cut, a right-angled triangle with the cone's height and radius as its two shorter sides. The large cone is an enlargement of the small one from the apex, so the half of its cut is an enlargement of the half of the small cone's cut, from the same apex. An enlargement keeps every angle and multiplies every length by the same scale factor, so the two right-angled triangles are similar (both conditions of Similar Figures hold), and the small cone's height and radius are in the same ratio to the large cone's height and radius.

Similar cones

$\frac{r}{R} = \frac{h}{H}$ , and the height of the frustum is $H - h$ .

Example 2: The radius of the small cone

A cone of radius $8$ cm and height $12$ cm is cut parallel to its base, $9$ cm above the base. Find (a) the radius of the small cone cut off, (b) the volume of the frustum.

Solution

(a) The small cone is $12 - 9 = 3$ cm high. The cones are similar, so $\frac{r}{8} = \frac{3}{12}$ , and $r = 8 \times \frac{3}{12} = 2$ cm.

(b) Large cone: $\frac{1}{3} \times \pi \times 8^{2} \times 12 = 256\pi\text{.}$ Small cone: $\frac{1}{3} \times \pi \times 2^{2} \times 3 = 4\pi\text{.}$

Frustum: $256\pi - 4\pi = 252\pi = 791.68\ldots = 792$ (3 s.f.)

(a) $2$ cm (b) $792 \text{ cm}^{3}$

Example 3: A bucket, by similar cones

A bucket is a frustum of a cone, $20$ cm deep, with a top of radius $15$ cm and a bottom of radius $10$ cm. How many litres does it hold when full?

Solution

Complete the cone below the bucket. Let the large cone (radius $15$ cm) have height $H$ cm; the small cone cut off (radius $10$ cm) then has height $H - 20$ cm.

The cones are similar, so $\frac{10}{15} = \frac{H - 20}{H}$ . Then $10H = 15H - 300$ , so $5H = 300$ and $H = 60$ ; the small cone is $40$ cm high.

Volume $= \frac{1}{3}\pi \times 15^{2} \times 60 - \frac{1}{3}\pi \times 10^{2} \times 40 = \frac{1}{3}\pi (13\,500 - 4000) = \frac{9500}{3}\pi = 9948.3\ldots \text{ cm}^{3}$

$9948.3\ldots \div 1000 = 9.948\ldots = 9.95$ litres (3 s.f.)

The bucket holds $9.95$ litres.

The Surface Area of a Frustum

The curved surface of a frustum is the curved surface of the large cone less that of the small cone: $\pi R L - \pi r l$ . Find the heights of both cones first, then each slant height by Pythagoras' theorem. Then add the circular faces the object has.

Example 4: A lampshade

A lampshade, open at both ends, is a frustum $8$ cm high with circles of radius $12$ cm and $6$ cm. Find the area of material in it.

Solution

Complete the cone: $\frac{6}{12} = \frac{H - 8}{H}$ , so $6H = 12H - 96$ , $H = 16$ cm; the small cone is $8$ cm high.

Here the small cone happens to be as high as the frustum, because the smaller radius is half the larger one. That is not so in general: under the bucket of Example 3 it was $40$ cm, not $20$ cm.

The slant heights are those of the two cones: $L$ runs down the whole side of the large cone, from its apex to the bottom circle, and $l$ down the side of the small cone cut off. Neither is the frustum's own sloping side.

Slant heights: large cone $L = \sqrt{12^{2} + 16^{2}} = \sqrt{400} = 20$ cm; small cone $l = \sqrt{6^{2} + 8^{2}} = \sqrt{100} = 10$ cm.

Curved surface of the frustum $= \pi \times 12 \times 20 - \pi \times 6 \times 10 = 240\pi - 60\pi = 180\pi = 565.48\ldots = 565$ (3 s.f.)

Both ends are open, so no circles are added.

About $565 \text{ cm}^{2}$ of material

Example 5: A basin with a bottom

A plastic basin, open at the top, is a frustum $8$ cm deep, with a top of radius $18$ cm and a bottom of radius $12$ cm. Find the area of plastic in it.

Solution

Complete the cone below the basin: $\frac{12}{18} = \frac{H - 8}{H}$ , so $12H = 18H - 144$ , $6H = 144$ and $H = 24$ cm; the small cone is $16$ cm high.

Slant heights: $L = \sqrt{18^{2} + 24^{2}} = \sqrt{900} = 30$ cm; $l = \sqrt{12^{2} + 16^{2}} = \sqrt{400} = 20$ cm.

Curved surface: $\pi \times 18 \times 30 - \pi \times 12 \times 20 = 540\pi - 240\pi = 300\pi$

The basin has a bottom but no top: add the bottom circle, $\pi \times 12^{2} = 144\pi$ .

Total: $300\pi + 144\pi = 444\pi = 1394.86\ldots = 1390$ (3 s.f.)

About $1390 \text{ cm}^{2}$ of plastic

Common Mistakes

Using the frustum's height for the cone. In $\frac{1}{3}\pi R^{2} H$ , $H$ is the height of the whole cone, not of the frustum.
Taking the small cone as high as the frustum. That is so only when the smaller radius is half the larger one; otherwise find $h$ from $\frac{r}{R} = \frac{h}{H}$ .
Averaging the radii. A cylinder of radius $12.5$ cm and height $20$ cm holds about $9820 \text{ cm}^{3}$ : close to the bucket's $9950 \text{ cm}^{3}$ , but wrong. Take the large cone less the small one.
Adding a circle the object does not have. A lampshade has no circular faces, a bucket only its bottom.
Rounding in the middle. Keep each height and slant height exact or unrounded in the calculator, subtract, and round once.

Practice Questions

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

Question 1

A cone of radius $9$ cm and height $15$ cm has its top $5$ cm cut off parallel to its base; the small cone removed has radius $3$ cm. Find the volume of the frustum.

ShowHide solution

$\frac{1}{3}\pi \times 81 \times 15 - \frac{1}{3}\pi \times 9 \times 5 = 405\pi - 15\pi = 390\pi = 1225.2\ldots = 1230$ (3 s.f.)

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

Question 2

A cone of radius $10$ cm and height $24$ cm is cut parallel to its base, $18$ cm above the base. Find (a) the radius of the small cone cut off, (b) the volume of the frustum.

ShowHide solution

(a) The small cone is $24 - 18 = 6$ cm high. $\frac{r}{10} = \frac{6}{24}$ , so $r = 10 \times \frac{6}{24} = 2.5$ cm.

(b) $\frac{1}{3}\pi \times 10^{2} \times 24 - \frac{1}{3}\pi \times 2.5^{2} \times 6 = 800\pi - 12.5\pi = 787.5\pi = 2474.0\ldots = 2470$ (3 s.f.)

(a) $2.5$ cm (b) $2470 \text{ cm}^{3}$

Question 3

A frustum $6$ cm high has a bottom of radius $9$ cm and a top of radius $6$ cm. Find (a) the height of the cone it was cut from, (b) its volume.

ShowHide solution

(a) $\frac{6}{9} = \frac{H - 6}{H}$ , so $6H = 9H - 54$ , $3H = 54$ and $H = 18$ cm. The small cone is $12$ cm high.

(b) $\frac{1}{3}\pi \times 81 \times 18 - \frac{1}{3}\pi \times 36 \times 12 = 486\pi - 144\pi = 342\pi = 1074.4\ldots = 1070$ (3 s.f.)

(a) $18$ cm (b) $1070 \text{ cm}^{3}$

Question 4

A paper cup is a frustum $8$ cm deep, with a top of radius $4$ cm and a bottom of radius $3$ cm. How many cups can be filled completely from a $1.5$ -litre bottle of barley water?

ShowHide solution

Complete the cone: $\frac{3}{4} = \frac{H - 8}{H}$ , so $3H = 4H - 32$ , $H = 32$ cm; the small cone is $24$ cm high.

One cup: $\frac{1}{3}\pi (16 \times 32 - 9 \times 24) = \frac{296}{3}\pi = 309.97\ldots \text{ cm}^{3}$

$1500 \div 309.97\ldots = 4.83\ldots\text{:}$ only whole cups count.

$4$ cups

Question 5

A lampshade, open at both ends, is a frustum $12$ cm high with circles of radius $15$ cm and $10$ cm. Find the area of fabric needed.

ShowHide solution

Complete the cone: $\frac{10}{15} = \frac{H - 12}{H}$ , so $10H = 15H - 180$ , $5H = 180$ and $H = 36$ cm; the small cone is $24$ cm high.

$L = \sqrt{15^{2} + 36^{2}} = \sqrt{1521} = 39$ cm and $l = \sqrt{10^{2} + 24^{2}} = \sqrt{676} = 26$ cm.

$\pi \times 15 \times 39 - \pi \times 10 \times 26 = 585\pi - 260\pi = 325\pi = 1021.0\ldots = 1020$ (3 s.f.)

No circles: the lampshade is open at both ends.

$1020 \text{ cm}^{2}$

Question 6

A solid block of wood is a frustum $8$ cm high, with circles of radius $15$ cm and $9$ cm. Every face of it is to be varnished. Find the total area to be varnished.

ShowHide solution

Complete the cone: $\frac{9}{15} = \frac{H - 8}{H}$ , so $9H = 15H - 120$ , $6H = 120$ and $H = 20$ cm; the small cone is $12$ cm high.

$L = \sqrt{15^{2} + 20^{2}} = \sqrt{625} = 25$ cm and $l = \sqrt{9^{2} + 12^{2}} = \sqrt{225} = 15$ cm.

Curved surface: $\pi \times 15 \times 25 - \pi \times 9 \times 15 = 375\pi - 135\pi = 240\pi$

The block is solid, closed at both ends: add both circles, $\pi \times 15^{2} + \pi \times 9^{2} = 225\pi + 81\pi = 306\pi\text{.}$

Total: $240\pi + 306\pi = 546\pi = 1715.3\ldots = 1720$ (3 s.f.)

$1720 \text{ cm}^{2}$

Question 7

A concrete stool at a void deck is a solid frustum $45$ cm high, with a top of radius $15$ cm and a bottom of radius $20$ cm. Its curved surface and its top are to be painted; the bottom stands on the floor. Find the area to be painted.

ShowHide solution

Complete the cone: $\frac{15}{20} = \frac{H - 45}{H}$ , so $15H = 20H - 900$ , $5H = 900$ and $H = 180$ cm; the small cone is $135$ cm high.

$L = \sqrt{20^{2} + 180^{2}} = \sqrt{32\,800} = 181.10\ldots$ cm and $l = \sqrt{15^{2} + 135^{2}} = \sqrt{18\,450} = 135.83\ldots$ cm.

Curved surface: $\pi \times 20 \times 181.10\ldots - \pi \times 15 \times 135.83\ldots = 4978.4\ldots$

Top: $\pi \times 15^{2} = 225\pi = 706.85\ldots$

Total: $4978.4\ldots + 706.85\ldots = 5685.3\ldots = 5690$ (3 s.f.)

$5690 \text{ cm}^{2}$

Question 8

Jun Kai finds the volume of a frustum $10$ cm high, with radii $6$ cm and $4$ cm, as $\frac{1}{3}\pi \times 6^{2} \times 10 - \frac{1}{3}\pi \times 4^{2} \times 10\text{.}$ Explain his mistake and find the correct volume.

ShowHide solution

He used the frustum's height, $10$ cm, for both cones. Each cone needs its own height.

Complete the cone: $\frac{4}{6} = \frac{H - 10}{H}$ , so $4H = 6H - 60$ , $H = 30$ cm; the small cone is $20$ cm high.

$\frac{1}{3}\pi \times 36 \times 30 - \frac{1}{3}\pi \times 16 \times 20 = 360\pi - \frac{320}{3}\pi = \frac{760}{3}\pi = 795.87\ldots = 796$ (3 s.f.)

He used the frustum's height for both cones; the volume is $796 \text{ cm}^{3}$ .

Question 9

Siew Ling's mop pail is a frustum $24$ cm deep, with a top of radius $14$ cm and a bottom of radius $10$ cm. Can it hold $10$ litres of water?

ShowHide solution

Complete the cone: $\frac{10}{14} = \frac{H - 24}{H}$ , so $10H = 14H - 336$ , $4H = 336$ and $H = 84$ cm; the small cone is $60$ cm high.

$\frac{1}{3}\pi (14^{2} \times 84 - 10^{2} \times 60) = \frac{1}{3}\pi (16\,464 - 6000) = 3488\pi = 10\,957.8\ldots \text{ cm}^{3}$

$10\,957.8\ldots \div 1000 = 10.95\ldots$ litres, more than $10$ litres.

Yes: it holds about $11.0$ litres.

Key Points

A frustum of a cone is what is left when a cone is cut parallel to its base and the small cone at the top is taken away.
Volume of a frustum $= \frac{1}{3}\pi R^{2} H - \frac{1}{3}\pi r^{2} h\text{:}$ the large cone less the small cone.
Curved surface area of a frustum $= \pi R L - \pi r l$ ; then add the circular faces the object has.
The two cones are similar: $\frac{r}{R} = \frac{h}{H}$ , and the frustum is $H - h$ high.
Keep each part exact (in terms of $\pi$ or unrounded), subtract, and round once.