A chord 10 cm long is drawn in a circle whose radius is

Solution:

We know that the area of minor segment of angle θ in a circle of radius r is,

$A=\left\{\frac{\pi \theta}{360^{\circ}}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right\} r^{2}$

It is given that the chord AB divides the circle in two segments.

We have $O A=5 \sqrt{2} \mathrm{~cm}$ and $A B=10 \mathrm{~cm}$. So,

$A L=\frac{A B}{2} \mathrm{~cm}$

$=\frac{10}{2} \mathrm{~cm}$

$=5 \mathrm{~cm}$

Let $\angle A O B=2 \theta$. Then,

$\angle A O L=\angle B O L$

$=\theta$

$\ln \Delta O L A$, we have

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

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

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

$\theta=\sin ^{-1} \frac{1}{\sqrt{2}}$

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

$\theta=\sin ^{-1} \frac{1}{\sqrt{2}}$

$=45^{\circ}$

Hence, $\angle A O B=90^{\circ}$

 

Now using the value of $r$ and $\theta$, we will find the area of minor segment

$A=\left\{\frac{90^{\circ} \pi}{360^{\circ}}-\sin \frac{90^{\circ}}{2} \cos \frac{90^{\circ}}{2}\right\} \times 5 \sqrt{2} \times 5 \sqrt{2}$

$=\left\{\frac{\pi}{4}-\sin 45^{\circ} \cos 45^{\circ}\right\} \times 50$

$=\frac{3.14 \times 50}{4}-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times 50$

$=39.25-25$

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

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

$=3.14 \times 5 \sqrt{2} \times 5 \sqrt{2}$

 

$=157.15 \mathrm{~cm}^{2}$

Area of major segment = Area of circle-Area of minor segment

$=157-14.25$

 

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