Question

If AC passes through the centre of the circle, then the area of the shaded region in the given figure is

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

Answer 1

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

Since AC is the diameter

Thus,$\angle ABC={90}^0$ (angle suntended by the diameter at the circumference is ${90}^0$)

$\therefore △ABC$ is a right angled triangle.

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

$=a^2+a^2$

$=>A{C}^2=2a^2$

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

Now,

OA=OC=radius

and AC=OA+OC

$\therefore OA=\frac{AC}{2}$$

$=\frac{\sqrt{2}a}{2}$

$=\frac{a}{\sqrt{2}}$

Now, Area of $△ABC=\frac{1}{2}\times AB\times AC$

$=\frac{1}{2}\times a\times a$

$=\frac{a^2}{2}$

Area of semicircle$=\frac{\pi (OA{)}^2}{2}$

$=\frac{\pi }{2}\times (\frac{a}{\sqrt{2}}{)}^2$

$=\frac{\pi a^2}{4}$

area of shaded part=$=\frac{a^2}{2}[\frac{\pi }{2}-1]$