#### Find the area of the shaded region given in Figure.

Solution

Area of square PQRS =(side)2=(14)2

=196 cm2

Area of ABCD (let side a) =(side)2=(a)2

Area of 4 semi circle $\left ( r=\frac{a}{2} \right )=4 \times \frac{1}{2}\pi\left ( \frac{a}{2} \right )^{2}$

Area of semi-circle=$\frac{1}{2}\times \pi \times r^{2}$

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

Total inner area = Area of ABCD + Area of 4 semi circles

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

$\\EF=8cm\\ EF=\frac{a}{2}+a+\frac{a}{2}\\ 8=\frac{a+2a+a}{2}\\ 4a=16\\ a=4cm$

Area of inner region =$4^{2}+\frac{\pi 4^{2}}{2}\\$
$\\=16+\frac{ 16 \pi}{2}\\ =16+8 \pi$

Area of shaded area = Area of PQRS – inner region area

$\\=196-16-8 \pi\\ =180-8 \times 3.14\\ =180-25.12\\ =154.88 cm^{2}$