In the below figure, $ A B=36 \mathrm{~cm} $ and $ M $ is mid-point of $ A B . $ Semi-circles are drawn on $ A B, A M $ and $ M B $ as diameters. A circle with centre $ C $ touches all the three circles. Find the area of the shaded region.
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Given:

\( A B=36 \mathrm{~cm} \) and \( M \) is mid-point of \( A B . \)

Semi-circles are drawn on \( A B, A M \) and \( M B \) as diameters.

A circle with centre \( C \) touches all the three circles.

To do: 

We have to find the area of the shaded region.

Solution:

Diameter of the largest semicircle $= 36\ cm$

This implies,

Radius $R =\frac{36}{2}$

$= 18\ cm$

Diameter of each of the smaller semicircles $= 18\ cm$

This implies,

Radius $r_{1}=\frac{18}{2}$

$=9 \mathrm{~cm}$

Diameter of the smallest circle $=\frac{1}{3} \times 36$

$=12 \mathrm{~cm}$

This implies,

Radius $r_{2}=\frac{12}{2}$

$=6 \mathrm{~cm}$

Therefore,

Area of the shaded region $=$ Area of the largest semicircle $-$ (Area of two smaller semicircles $+$ Area of the smallest circle)

$=\frac{1}{2} \pi \mathrm{R}^{2}-(2 \times \frac{1}{2} \pi r_{1}^{2}+\pi r_{2}^{2})$

$=\frac{1}{2} \pi \mathrm{R}^{2}-(\pi r_{1}^{2}+\pi r_{2}^{2})$

$=\frac{1}{2} \pi \mathrm{R}^{2}-\pi r_{1}^{2}-\pi r_{2}{ }^{2}$

$=\pi[\frac{1}{2} \mathrm{R}^{2}-r_{1}^{2}-r_{2}^{2}]$

$=\pi[\frac{1}{2}(18)^{2}-(9)^{2}-(6)^{2}]$

$=\pi[\frac{324}{2}-81-36]$

$=\pi(\frac{324}{2}-117)$

$=\pi(162-117)$

$=45 \pi \mathrm{cm}^{2}$

The area of the shaded region is $45 \pi\ cm^2$.

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Updated on: 10-Oct-2022

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