Question

In Fig., $ABC$ is a quadrant of a circle of radius $14cm$ and a semicircle is drawn with $BC$ as diameter. Find the area of the shaded region

Answer 1

Area of shaded region $=$ Area of $△ABC$ $+$ Area of semicircle with diameter $BC$ $-$ Area of quadrant ABC
$\therefore$ Area of $△ABC$ $=\frac{1}{2}\times 14\times 14=98c{m}^2$
In $△ABC,$ by Pythagoras theorem,
$BC=14\sqrt{2}$
$\therefore$Area of semicircle with diameter $BC$$=\frac{\pi \times r^2}{2}$ $=\frac{1}{2}\times \frac{22}{7}\times 7\times 7\times 2=154c{m}^2$
Area of quadrant ABC $=\frac{\pi \times r^2}{4}$ $=\frac{1}{4}\times \frac{22}{7}\times 14\times 14=154c{m}^2$
$\therefore$ Area of shaded region $=98+154-154=98c{m}^2$