Frustum of a cone

A frustum of a cone is what you get when you slice a cone with a plane parallel to its base, then remove the small cone at the top. The shape that remains is a truncated cone: it has two circular faces (one large, one small), and a slanting curved surface connecting them. A bucket is essentially a frustum; so is a lampshade, a flowerpot, and the head of a drinking glass.

The frustum has two radii, $r_1$ (the bigger circular face) and $r_2$ (the smaller). Its height $h$ is the perpendicular distance between the two parallel circular faces. Its slant height $\ell$ is the length of the slanting side, which by Pythagoras is

$\ell = \sqrt{h^2 + (r_1 - r_2)^2}.$

(Imagine drawing a perpendicular from the smaller circle's edge to the larger circle, forming a right triangle.)

Formulas

Curved surface area:

$\text{CSA}_{\text{frustum}} = \pi (r_1 + r_2) \ell.$

Total surface area (curved surface + the two circular faces):

$\text{TSA}_{\text{frustum}} = \pi(r_1 + r_2)\ell + \pi r_1^2 + \pi r_2^2.$

Volume:

$V_{\text{frustum}} = \frac{1}{3} \pi h \left(r_1^2 + r_2^2 + r_1 r_2\right).$

Derivation of the volume formula (brief)

A cone of radius $r_1$ and height $H$ is sliced by a plane parallel to its base, producing a smaller cone on top of radius $r_2$ and height $H - h$ . By similar triangles, $r_2/r_1 = (H - h)/H$ , so $H = h r_1/(r_1 - r_2)$ and $H - h = h r_2/(r_1 - r_2)$ .

Frustum volume = volume of original cone $-$ volume of small cone $= (1/3)\pi r_1^2 H - (1/3) \pi r_2^2 (H - h)$ .

Substituting and simplifying gives $V = (1/3)\pi h(r_1^2 + r_2^2 + r_1 r_2)$ .

When is a frustum used?

In the board exam, frustum problems are usually about a bucket (whose surface area and capacity are asked). The bucket has:

An open top (radius $r_1$ , the larger).
A closed bottom (radius $r_2$ , the smaller).
A slanting metal surface (CSA).

The surface area of the bucket (excluding the open top) is $\text{CSA} + \pi r_2^2 = \pi(r_1 + r_2)\ell + \pi r_2^2$ .

If both top and bottom are closed (a sealed frustum), use the full TSA.

Worked examples

Example 1. A frustum has $r_1 = 4, r_2 = 1, h = 4$ . Find the slant height, CSA, TSA, and volume.

$\ell = \sqrt{16 + 9} = 5$ .

CSA $= \pi(4 + 1)(5) = 25\pi \approx 78.57$ cm $^2$ .

TSA $= 25\pi + 16\pi + \pi = 42\pi \approx 132$ cm $^2$ .

$V = (1/3)\pi(4)(16 + 1 + 4) = (1/3)\pi(4)(21) = 28\pi \approx 88$ cm $^3$ .

Example 2. A bucket is a frustum with radii $20$ cm and $12$ cm, height $15$ cm. Find the capacity (volume) and the metal sheet area (CSA + bottom disc).

$\ell = \sqrt{225 + 64} = \sqrt{289} = 17$ cm.

$V = (1/3)(22/7)(15)(400 + 144 + 240) = (1/3)(22/7)(15)(784) = 22 \cdot 5 \cdot 784/7 = 22 \cdot 5 \cdot 112 = 12\,320$ cm $^3 = 12.32$ litres.

Sheet area $= \pi(r_1 + r_2)\ell + \pi r_2^2 = (22/7)(32)(17) + (22/7)(144) = 1709.71 + 452.57 = 2162.28$ cm $^2$ .

Example 3. A milk container is a frustum with radii $20$ cm and $8$ cm and slant height $25$ cm. Find the cost of metal sheet at Rs. $1.40$ per cm $^2$ .

Slant height is given ( $25$ cm). Height $h = \sqrt{\ell^2 - (r_1 - r_2)^2} = \sqrt{625 - 144} = \sqrt{481}$ (not needed if only sheet area is asked).

Sheet area = CSA + small disc (the bottom; the top is open) $= \pi(20 + 8)(25) + \pi(64) = 700\pi + 64\pi = 764\pi$ .

With $\pi = 22/7$ : $764 \cdot 22/7 \approx 2400.6$ cm $^2$ .

Cost $\approx 2400.6 \cdot 1.40 =$ Rs. $3360.85$ .

Example 4. A frustum-shaped lampshade has radii $7$ cm and $14$ cm, slant height $12$ cm. Find the CSA.

CSA $= \pi(7 + 14)(12) = 252\pi \approx 791.71$ cm $^2$ .

Example 5. A drinking glass is a frustum with radii $5$ cm and $3$ cm and height $14$ cm. Find its capacity in mL.

$V = (1/3)(22/7)(14)(25 + 9 + 15) = (1/3)(22/7)(14)(49) = 22 \cdot 14 \cdot 49/(3 \cdot 7) = 22 \cdot 2 \cdot 49/3 = 2156/3 \approx 718.67$ cm $^3 = 718.67$ mL.

Example 6. A flowerpot is a frustum with radii $r_1 = 10$ cm, $r_2 = 5$ cm, $h = 12$ cm. Find the volume and the total external surface area (open top, closed bottom).

$\ell = \sqrt{144 + 25} = 13$ .

$V = (1/3)\pi(12)(100 + 25 + 50) = (1/3)\pi(12)(175) = 700\pi \approx 2200$ cm $^3$ .

Surface (CSA + bottom) $= \pi(15)(13) + \pi(25) = 195\pi + 25\pi = 220\pi \approx 691.4$ cm $^2$ .

Try it yourself

A frustum has $r_1 = 9, r_2 = 5, h = 12$ . Find the slant height and CSA.
A bucket holds $12.32$ litres of water, with radii $20$ and $12$ cm. Find its height.
A frustum has radii $6$ and $3$ cm and slant height $5$ cm. Find $h$ and $V$ .
A glass tumbler is a frustum with radii $3.5$ and $2.5$ cm and height $14$ cm. Find the capacity.
A bucket is a frustum with radii $15$ and $9$ cm and height $12$ cm. Find the metal sheet area (open top).
A solid frustum has $r_1 = 14, r_2 = 7, h = 24$ . Find its TSA and volume.
A frustum of a cone has $r_1 = 28, r_2 = 21$ and slant $20$ . Find the height.
A frustum has radii $r_1, r_2$ and is exactly half the volume of the cone that would close at the apex. Find the relation between $r_1$ and $r_2$ .
The slant heights of two frusta of cones (same height $h$ ) are $\ell_1$ and $\ell_2$ . Compare $r_1 + r_2$ for them.
A frustum has volume $1232$ cm $^3$ , $r_1 = 7, r_2 = 3.5$ . Find $h$ .
A flowerpot (frustum) has radii $7$ cm and $5$ cm and slant height $13$ cm. Find the external surface area (closed bottom, open top).
Find the slant height of a frustum with $r_1 = 12, r_2 = 5$ and $h = 24$ .

Pitfalls / Insight

(1) The slant height formula uses $(r_1 - r_2)^2$ , not $r_1^2 - r_2^2$ . Many students get this wrong. The right triangle is formed by the height $h$ and the difference of radii.

(2) The volume formula is not the average of the two cones' volumes. It's $(1/3)\pi h(r_1^2 + r_2^2 + r_1 r_2)$ . The $r_1 r_2$ term is essential.

(3) For a bucket, the open top is excluded from the metal area. For a sealed frustum, both discs are included.

(4) When converting cm $^3$ to litres: $1000$ cm $^3 = 1$ L.

(5) The slant height $\ell$ is not the height $h$ of the frustum. They are related by Pythagoras: $\ell = \sqrt{h^2 + (r_1 - r_2)^2}$ .