A right angled triangle whose sides are $ 3 \mathrm{~cm}, 4 \mathrm{~cm} $ and $ 5 \mathrm{~cm} $ is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two cones so formed. Also, find their curved surfaces.

Given:

A right angled triangle whose sides are \( 3 \mathrm{~cm}, 4 \mathrm{~cm} \) and \( 5 \mathrm{~cm} \) is revolved about the sides containing the right angle in two ways.

To do:

We have to find the difference in volumes of the two cones so formed and their curved surfaces.

Solution:

In the first case, when the triangle is revolved along $4\ cm$ side, then

Radius of the base of the cone $r_1 = 3\ cm$

Height of the cone $h_1= 4\ cm$

Slant height $l = 5\ cm$

Volume of the cone formed $=\frac{1}{3} \pi r^2h$

$= \frac{1}{3} \pi 3^2 \times 4$

$=12 \pi\ cm^3$

Curved surface area of the cone formed $= \pi rl$

$= \pi \times 3 \times 5$

$= 15 \pi cm^2$

In the second case, when the triangle is revolved along $3\ cm$ side, then

Radius of the base of the cone $r_2 = 4\ cm$

Height of the cone $h_2= 3\ cm$

Slant height $l = 5\ cm$

Volume of the cone formed $=\frac{1}{3} \pi r^2h$

$= \frac{1}{3} \pi 4^2 \times 3$

$=16 \pi\ cm^3$

Curved surface area of the cone formed $= \pi rl$

$= \pi \times 4 \times 5$

$= 20 \pi cm^2$

Difference of the volumes of the two cones $= I6 \pi- 12 \pi$

$= 4 \pi\ cm^3$

The difference in volumes of the two cones so formed is $4 \pi\ cm^3$.

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Updated on: 10-Oct-2022

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