Question

 A chord of a circle of radius 20 cm sub tends an angle of 90⁰ at the centre . Find the area of the corresponding major segment of the circle
( Use \[\pi = 3 . 14\]) 

Answer 1

We know area of minor segment of the circle is \[A = \left\{ \frac{\pi\theta}{360} - \sin\frac{\theta}{2}\cos\frac{\theta}{2} \right\} r^2\]

\[\Rightarrow A = \left\{ \frac{\pi \times 90°}{360} - \sin\frac{90}{2}\cos\frac{90}{2} \right\} \left( 20 \right)^2 \]
\[ \Rightarrow A = \left( \frac{\pi}{4} - \frac{1}{2} \right)\left( 400 \right)\] 

Area of the major segment = Area of the circle − area of the minor segment

\[= \pi \left( 20 \right)^2 - \left( 400 \right)\left[ \frac{\pi}{2} - \frac{1}{2} \right]\]
\[ = \left( 400 \right)\left[ \pi - \frac{\pi}{2} + \frac{1}{2} \right]\]
\[ = 1142 {cm}^2\]