A frustum of a right circular cone has a diameter of base $ 20 \mathrm{~cm} $, of top $ 12 \mathrm{~cm} $, and height $ 3 \mathrm{~cm} $. Find the area of its whole surface and volume.

Given:

A frustum of a right circular cone has a diameter of base \( 20 \mathrm{~cm} \), of top \( 12 \mathrm{~cm} \), and height \( 3 \mathrm{~cm} \).

To do:

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

Solution:

Base diameter of the frustum $= 20\ cm$

This implies,

Radius of the frustum $r_1 =\frac{20}{2}$

$= 10\ cm$

Diameter of the top $= 12\ cm$

Radius of the top $r_{2}=\frac{12}{2}$

$=6 \mathrm{~cm}$

Height of the cone $h=3 \mathrm{~cm}$

Therefore,

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

$=\sqrt{(3)^{2}+(10-6)^{2}}$

$=\sqrt{(3)^{2}+(4)^{2}}$

$=\sqrt{9+16}$

$=\sqrt{25}$

$=5 \mathrm{~cm}$

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

$=\pi(10+6) \times 5+\pi(10)^{2}+\pi(6)^{2}$

$=80 \pi+100 \pi+36 \pi$

$=216 \pi$

$=216 \times \frac{22}{7}$

$=678.85 \mathrm{~cm}^{2}$

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

$=\frac{22}{3 \times 7}[(10)^{2}+10 \times 6+(6)^{2}] 3$

$=\frac{22}{21}[100+60+36] \times 3$

$=\frac{22}{21} \times 196 \times 3$

$=616 \mathrm{~cm}^{3}$

The area of its whole surface and volume are $678.85\ cm^2$ and $616\ cm^3$ respectively.

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Updated on: 10-Oct-2022

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