A chord of a circle of radius $ 20 \mathrm{~cm} $ subtends an angle of $ 90^{\circ} $ at the centre. Find the area of the corresponding major segment of the circle. (Use $ \pi=3.14 $ )

Given:

A chord of a circle of radius \( 20 \mathrm{~cm} \) subtends an angle of \( 90^{\circ} \) at the centre.

To do:

We have to find the area of the corresponding major segment of the circle.

Solution:

Let $AB$ be the chord of a circle of radius $10\ cm$ and $O$ be the centre of the circle.

$\angle \mathrm{AOB}=90^{\circ}$

This implies,

Angle of the major sector $=360^{\circ}-90^{\circ}$

$=270^{\circ}$

Area of the major sector $=\frac{270}{360} \times \pi \times(10)^{2}$

$=\frac{3}{4} \times 3.14 \times 100$

$=75 \times 3.14$

$=235.5 \mathrm{~cm}^{2}$

Draw $\mathrm{OM} \perp \mathrm{AB}$

$\mathrm{AM}=\frac{1}{2} \mathrm{AB}$

$\angle \mathrm{AOM}=\frac{1}{2} \times 90^{\circ}$

$=45^{\circ}$

$\frac{\mathrm{AM}}{\mathrm{OA}}=\sin 45^{\circ}$

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

$\mathrm{AM}=10 \times \frac{1}{\sqrt{2}} \mathrm{~cm}$

Therefore,

$\mathrm{AB}=10 \sqrt{2} \mathrm{~cm}$ and $\mathrm{OM}=\mathrm{OA}$

$\cos 45^{\circ}=10 \times \frac{1}{\sqrt{2}}$

$=5 \sqrt{2} \mathrm{~cm}$

Area of $\Delta \mathrm{OAB}=\frac{1}{2}\times10 \sqrt{2} \times 5 \sqrt{2}$

$=50 \mathrm{~cm}^{2}$

The area of the major segment $=235.5+50$

$=285.5 \mathrm{~cm}^{2}$

The area of the major segment is $285.5 \mathrm{cm}^{2}$.