A solid is in the shape of a frustum of a cone. The diameters of the two circular ends are $ 60 \mathrm{~cm} $ and $ 36 \mathrm{~cm} $ and the height is $ 9 \mathrm{~cm} $. Find the area of its whole surface and the volume.

Given:

A solid is in the shape of a frustum of a cone. The diameters of the two circular ends are \( 60 \mathrm{~cm} \) and \( 36 \mathrm{~cm} \) and the height is \( 9 \mathrm{~cm} \).

To do:

We have to find the area of its whole surface and the volume.

Solution:

Upper diameter of the frustum $= 60\ cm$

This implies,

Upper radius $r_1 = \frac{60}{2}$

$= 30\ cm$

Lower diameter of the frustum $= 36\ cm$

This implies,

Lower radius $r_2 = \frac{36}{2}$

$= 18\ cm$

Height of the frustum $h = 9\ cm$

Therefore,

Slant height of the frustum $l=\sqrt{h^{2}+(r_{1}-r_{2})^{2}}$

$=\sqrt{(9)^{2}+(30-18)^{2}}$

$=\sqrt{81+144}$

$=\sqrt{225}$

$=15 \mathrm{~cm}$

Total surface area of the frustum $=\pi[(r_{1}+r_{2}) l+r_{1}^{2}+r_{2}^{2}]$

$=\pi[(30+18) \times 15+(30)^{2}+(18)^{2}]$

$=\pi[48 \times 15+900+324]$

$=\pi[720+900+324]$

$=1944 \pi \mathrm{cm}^{2}$

Volume of the frustum $=\frac{\pi}{3}[r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}] \times h$

$=\frac{\pi}{3}[(30)^{2}+30 \times 18+(18)^{2}] \times 9$

$=\frac{\pi}{3}[900+540+324] \times 9$

$=3 \pi \times 1764$

$=5292 \pi \mathrm{cm}^{3}$

The area of its whole surface is $1944 \pi\ cm^2$ and the volume is $5292 \pi\ cm^3$.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

30 Views

Kickstart Your Career

Get certified by completing the course

Get Started