The radii of the circular bases of a frustum of a right circular cone are $ 12 \mathrm{~cm} $ and $ 3 \mathrm{~cm} $ and the height is $ 12 \mathrm{~cm} $. Find the total surface area and the volume of the frustum.

Given:

The radii of the circular bases of a frustum of a right circular cone are \( 12 \mathrm{~cm} \) and \( 3 \mathrm{~cm} \) and the height is \( 12 \mathrm{~cm} \).

To do:

We have to find the total surface area and the volume of the frustum.

Solution:

Height of the frustum $h = 12\ cm$

Radius of the top of the frustum $r_1 = 12\ cm$

Radius of the bottom of the frustum $r_2 = 3\ cm$

Therefore,

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

$=\sqrt{(12)^{2}+(12-3)^{2}}$

$=\sqrt{(12)^{2}+(9)^{2}}$

$=\sqrt{144+81}$

$=\sqrt{225}$

$=15 \mathrm{~cm}$

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

$=\pi[(r_{1}+r_{2}) l+r_{1}^{2}+r_{2}^{2}]$

$=\pi[(12+3) \times 15+(12)^{2}+(3)^{2}]$

$=\pi[15 \times 15+144+9]$

$=\pi[225+144+9]$

$=\pi \times 378$

$=378 \pi \mathrm{cm}^{2}$

Volume of the frustum $=\frac{\pi}{3}(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}) h$

$=\frac{\pi}{3}[(12)^{2}+12 \times 3+(3)^{2}] \times 12$

$=\frac{\pi}{3} \times 12[144+36+9]$

$=4 \pi(189) \mathrm{cm}^{3}$

$=756 \pi \mathrm{cm}^{3}$

The total surface area and the volume of the frustum are $378 \pi\ cm^2$ and $756 \pi\ cm^3$ respectively.

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Updated on: 10-Oct-2022

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