Question

In the given figure, ABCD is a square of side 4 cm. A quadrant of a circle of radius 1 cm is drawn at each vertex of the square and a circle of diameter 2 cm is also drawn. Find the area of the shaded region.
[Use $\mathrm{\pi }$ = 3.14.]

Answer 1

Area of the square ABCD = $(\mathrm{Side}{)}^2$
$$\begin{gathered} =4^2 \\ =16{\mathrm{cm}}^2 \end{gathered}$$

Area of the circle = ${\mathrm{\pi r}}^2$
Radius = 1 cm
$$\begin{gathered} \mathrm{Area}=3.14\times (1{)}^2 \\ =3.14{\mathrm{cm}}^2 \end{gathered}$$

Area of the quadrant of one circle = $\frac{1}{4}{\mathrm{\pi r}}^2$
$$\begin{gathered} =\frac{1}{4}\times 3.14\times 1^2 \\ =0.785{\mathrm{cm}}^2 \end{gathered}$$

Area of the quadrants of four circles = $0.785\times 4=3.14{\mathrm{cm}}^2$
Area of the shaded region = Area of the square $-$ Area of the circle $-$ Area of the quadrants of four circles
=$16-3.14-3.14=9.72{\mathrm{cm}}^2$