A chord of a circle of radius 14 cm makes a right angle at the centre.

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 of the circle of radius $r=14 \mathrm{~cm}$ makes right angle at the centre.

So, $\theta=90^{\circ}$

Substituting the value of r and angle θ in above formula,

Area of minor segment

$A=\left\{\frac{90^{\circ} \pi}{360^{\circ}}-\sin \frac{90^{\circ}}{2} \cos \frac{90^{\circ}}{2}\right\} \times 14 \times 14$

$=\left\{\frac{\pi}{4}-\sin 45^{\circ} \cos 45^{\circ}\right\} \times 196$

$=\frac{22 \times 196}{7 \times 4}-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times 196$

$=154-98$

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

Hence, area of minor segment is $56 \mathrm{~cm}^{2}$

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

$=\frac{22}{7} \times 14 \times 14$

$=616 \mathrm{~cm}^{2}$

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

$=616-56$

$=560 \mathrm{~cm}^{2}$