Question

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

Answer 1

Side of square = $$\displaystyle 4 cm$$
Radius of circle = $$\displaystyle 4 cm$$
Area of sqare = $$\displaystyle (side)^2 = 4\times 4 = 16 cm^2$$
Area of four quadrants of circle = $$\displaystyle 4(\frac{1}{4} \times 3.14 \times 1 \times 1) = 3.14 cm^2$$
Area of circle with diameter $$\displaystyle 2 cm = \pi r^2 = 3.14 \times 1 \times 1 = 3.14 cm^2$$
Area of circle with diameter $$\displaystyle 2 cm = \pi r^2 = 3.14 \times 1 \times 1 = 3.14 cm^2$$
(diameter = $$\frac{radius}{2}$$)
Now,Area of the shaded region = Area of square - (Area of four quadrants of circle + Area of circle with diameter $$\displaystyle 2 cm$$)
$$\displaystyle 16 -(3.14 + 3.14)$$
$$\displaystyle 9.72$$
Therefore, the area of the shaded region is $$\displaystyle 9.72 cm^2$$