In the below figure, there are three semicircles, $ A, B $ and $ C $ having diameter $ 3 \mathrm{~cm} $ each, and another semicircle $ E $ having a circle $ D $ with diameter $ 4.5 \mathrm{~cm} $ are shown. Calculate the area of the shaded region."

Given:

Diameter of each of the semicircle \( \mathrm{A}, \mathrm{B} \) and \( \mathrm{C}=3 \mathrm{~cm} \)

Diameter of circle \( D=4.5 \mathrm{~cm} \)

To do: 

We have to find the area of the shaded region.

Solution:

Diameter of each of the semicircles $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}=3 \mathrm{~cm}$

This implies,

Radius $r_{1}=\frac{3}{2}\ cm$

Diameter of semicircle $\mathrm{E}=3+3+3=9 \mathrm{~cm}$

This implies,

Radius $r_{2}=\frac{9}{2} \mathrm{~cm}$

Diameter of circle $D=4.5 \mathrm{~cm}$

$=\frac{9}{2} \mathrm{~cm}$

This implies,

Radius $r_{3}=\frac{9}{4} \mathrm{~cm}$

Area of each semicircle $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}=\frac{1}{2} \pi r_{1}^{2}$

$=\frac{1}{2} \times \frac{22}{7} \times (\frac{3}{2})^2$

$=\frac{99}{28} \mathrm{~cm}^{2}$

Area of semicircle $\mathrm{E}=\pi r_{2}^{2}$

$=\frac{1}{2} \times \frac{22}{7} \times (\frac{9}{2})^{2}$

$=\frac{891}{28} \mathrm{~cm}^{2}$

Area of circle $D=\pi r_{3}^{2}$

$=\frac{22}{7} \times (\frac{9}{4})^{2}$

$=\frac{891}{56} \mathrm{~cm}^{2}$

Therefore,

Area of the shaded region $=$ Area of the semicircle $\mathrm{E} +$ Area of semicircle $\mathrm{B}-$ Area of semicircles $\mathrm{A}$ and $\mathrm{C}-$ Area of circle $\mathrm{D}$

$=\frac{891}{28}+\frac{99}{28}-(\frac{99}{28}+\frac{99}{28}+\frac{891}{56})$

$=\frac{990}{28}-\frac{198+198+891}{56}$

$=\frac{990}{28}-\frac{1287}{56}$

$=\frac{1980-1287}{56}$

$=\frac{693}{56}$

$=\frac{99}{8}$

$=12.375 \mathrm{~cm}^{2}$

The area of the shaded region is $12.375\ cm^2$.

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Updated on: 10-Oct-2022

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