Question

If BC passes through the center of the circle, then area of the shaded region in the given figure is

• A. $\frac{a^2}{2}(3-x)$
• B. $a^2(\frac{\pi }{2}-1)$
• C. $2a^2(\pi -1)$
• D. $\frac{a^2}{2}(\frac{\pi }{2}-1)$

Answer 1

The correct option is D $\frac{a^2}{2}(\frac{\pi }{2}-1)$

$BC$ passes through the center of the circle.

Thus, $BC$ is the diameter of the circle and subtends a right angle at the circumference of the circle.

$\therefore \angle BAC=90$

In $△BAC$,

$B{C}^2=A{C}^2+B{A}^2$

$=a^2+a^2-2a^2$

$=>BC=\sqrt{2}a$

Area of $△BAC=\frac{1}{2}\times BA\times AC$$=\frac{1}{2}\times a\times a$$=\frac{1}{2}\times a^2$

Area of semicircle$=\frac{\pi {\left(\frac{D}{2}\right)}^2}{2}$ $=\frac{\pi {\left(\frac{BC}{2}\right)}^2}{2}$$=\frac{\pi {\left(\frac{\sqrt{2}a}{2}\right)}^2}{2}$ $=\frac{\pi a^2}{4}$

$\therefore$ Area of shaded region $=\frac{\pi a^2}{4}-\frac{a^2}{2}$$=\frac{a^2}{2}(\frac{\pi }{2}-1)$