Question

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

Answer 1

$A\to$ triangle

$A+B\to$Quadrant

$B+C\to$ Semicircle

$C=A+B+C-(A+B)$

$=A+(B+C)-(A+B)$

Radius of quadrant $=14cm$

$A+B=\frac{1}{4}\pi \times 14\times 14=14\sqrt{2}cm$

Area of semicircle$=B+C=\frac{1}{2}\pi .\frac{14\sqrt{2}}{2}.\frac{14\sqrt{2}}{2}=49\pi$

$A=\frac{1}{2}\times 14\times 14=98$

$C=98+49\pi -49\pi =98{cm}^2$