A chord $ 10 \mathrm{~cm} $ long is drawn in a circle whose radius is $ 5 \sqrt{2} \mathrm{~cm} $. Find area of both the segments. $ \quad $ (Take $ \pi=3.14) $

Given:

A chord \( 10 \mathrm{~cm} \) long is drawn in a circle whose radius is \( 5 \sqrt{2} \mathrm{~cm} \).

To do:

We have to find the area of both the segments. 

Solution:

Radius of the circle $r = 5\sqrt2\ cm$

Length of the chord $AB = 10\ cm$

Let $O L\ \perp\ A B$ which bisects $A B$ at $L$ and divides $\angle \mathrm{AOB}$ and $\angle \mathrm{AOB}=\theta$

$\mathrm{AL}=\mathrm{LB}$

$=\frac{10}{2}$

$=5 \mathrm{~cm}$

$\angle A O B = \frac{\theta}{2}$

In $\Delta OAL$,

$\sin \frac{\theta}{2}=\frac{A L}{O A}$

$=\frac{5}{5 \sqrt{2}}$

$=\frac{1}{\sqrt{2}}$

$=\sin 45^{\circ}$

This implies,

$\frac{\theta}{2}=45^{\circ}$

$\Rightarrow \theta=45^{\circ} \times 2$

$=90^{\circ}$

Area of the minor segment $\mathrm{ACB}=(\frac{\theta \pi}{360^{\circ}}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}) r^{2}$

$=(\frac{90^{\circ} \pi}{360^{\circ}}-\sin \frac{90^{\circ}}{2} \cos \frac{90^{\circ}}{2})(5 \sqrt{2})^{2}$

$=(\frac{1}{4} \times 3.14-\sin 45^{\circ} \cos 45^{\circ}) \times 50$

$=(\frac{1.57}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) \times 50$

$=(\frac{1.57}{2}-\frac{1}{2})\times 50$

$=50 \times \frac{1.57 - 1.00}{2}$

$=25 \times 0.57$

$=14.25 \mathrm{~cm}^{2}$

Area of the circle $=\pi r^{2}$

$=3.14 \times(5 \sqrt{2})^{2}$

$=3.14 \times 50$

$=157 \mathrm{~cm}^{2}$

Therefore,

Area of the major segment $\mathrm{ADB}=$ Area of the circle

$-$ Area of the minor segment

$=157.00-14.25$

$=142.75 \mathrm{~cm}^{2}$

Tutorialspoint

Updated on: 10-Oct-2022

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