Question

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

Answer 1

Area of minor segment $=\pi r^2\times \frac{\theta }{{360}^o}-r^{2}sin\frac{\theta }{2}cos\frac{\theta }{2}$

$=\pi \times {20}^2\times \frac{90}{360}-{20}^{2}sin{45}^{o}cos{45}^o$

$=\frac{22}{7}\times 400\times \frac{1}{4}-400\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}$

$=314.28-200=114.28c{m}^2$

Now, area of major segment = Area of the circle - Area of minor segment

$=\pi r^2-114.28$

$=\frac{22}{7}\times {20}^2-114.28$

$=1257.14-114.28=1142.86c{m}^2$