If ABCD is a square, as shown in the figure, then calculate the area of the shaded region depicted in the figure. [Take $\pi =3.14]$

From the figure, area of square ABCD

$=\mathrm{Side}{}^2$

$=4^2$

$=16{ \mathrm{cm}}^2$

The radius(r₁) for the circle at the centre is $\frac{2}{2}=1 \mathrm{cm}$  

Therefore, the area of the circle at the centre is:

$\begin{array}{c}\begin{array}{c}={\mathrm{\pi r}}^2\\ =\mathrm{\pi }\times 1^2\\ = \mathrm{\pi } {\mathrm{cm}}^2\end{array}\end{array}$

The radius for the four quadrants r₂ $=$ 1 cm

And therefore, the area of the four quadrants is:

$\begin{array}{c}=4\times \frac{1}{4}\times \pi \times r^2\\ =\mathrm{\pi }\times 1^2\\ =\mathrm{\pi } {\mathrm{cm}}^2\end{array}$   

Now,  the area of the shaded portion 

$= \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{square} - (\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{at} \mathrm{the} \mathrm{centre} + \mathrm{Area} \mathrm{of} \mathrm{the} 4 \mathrm{quadrants})$

$\begin{array}{c}=16-(3.14+3.14)\\ =16-6.28\\ =9.72{ \mathrm{cm}}^2\end{array}$