Question

In given figure $ABPC$ 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

In right angle $△BAC$, using Pythagoras theorem,

${AB}^2+{AC}^2={BC}^2\Rightarrow BC=\sqrt{{14}^2+{14}^2}=14\sqrt{2}$

Diameter of semi-circle$=14\sqrt{2}cm$

Radius of semi-circle$=\frac{14\sqrt{2}}{2}=7\sqrt{2}$

Area of shaded region = area of semi circle BC - [area of quadrant$-ar△ABC$]

$=\frac{1}{2}\times \pi r^2-\frac{1}{4}\times \pi \times {AC}^2+\frac{1}{2}\times AC\times AB$

$=\frac{1}{2}\times \frac{22}{7}\times 7\sqrt{2}\times 7\sqrt{2}-\frac{1}{4}\times \frac{22}{7}\times 14\times 14+\frac{1}{2}\times 14\times 14$

$=154-154+98=98c{m}^2$