Question

If $BC$ passes through the centre of the circle and $AB=AC=a$, then the area of the shaded region in the given figure is:

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

Answer 1

Answer: (D)

$\frac{a^2}{2}(\frac{\pi }{2}-1)$

Here $O$ is the centre of the circle.
From right angled $\Delta BAC$,
$B{C}^2=A{C}^2+A{B}^2=2a^2$
$\therefore BC=\sqrt{2}a$
$\therefore$ Radius of circle, OB $=\frac{1}{2}\sqrt{2}a=\frac{1}{\sqrt{2}}a$
$\therefore$ Area of semi-circle $=\frac{1}{2}\pi (OB{)}^2$
$=\frac{1}{2}\pi (\frac{1}{\sqrt{2}}a{)}^2=\frac{1}{4}\pi a^2$
Area of $\Delta BAC=\frac{1}{2}\times a\times a=\frac{a^2}{2}$
Hence, area of shaded region $=\frac{1}{4}\pi a^2-\frac{a^2}{2}$
$=\frac{a^2}{2}(\frac{\pi }{2}-1)$