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 $\mathrm{\pi }=3.14$)

Answer 1

We know area of minor segment of the circle is $A=\left\{\frac{\mathrm{\pi \theta }}{360}-\sin \frac{\theta }{2}\cos \frac{\theta }{2}\right\}{r}^2$
$$\begin{gathered} \Rightarrow A=\left\{\frac{\mathrm{\pi }\times 90°}{360}-\sin \frac{90}{2}\cos \frac{90}{2}\right\}{\left(20\right)}^2 \\ \Rightarrow A=\left(\frac{\mathrm{\pi }}{4}-\frac{1}{2}\right)\left(400\right) \end{gathered}$$
Area of the major segment = Area of the circle − area of the minor segment
$$\begin{gathered} =\mathrm{\pi }{\left(20\right)}^2-\left(400\right)\left[\frac{\mathrm{\pi }}{2}-\frac{1}{2}\right] \\ =\left(400\right)\left[\mathrm{\pi }-\frac{\mathrm{\pi }}{2}+\frac{1}{2}\right] \\ =1142{\mathrm{cm}}^2 \end{gathered}$$

DISCLAIMER: The answer given in the book is incorrect. If we take the radius 10 instead of 20, then the answer will match.