Question

In fig. 12.15, two circular flower beds have been shown on two sides of a square lawn $ABCD$ of side $56$ m. If the centre of each circular flower bed is the point of intersection $O$ of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.

Answer 1

Radius of hemispheres = $\frac{56}{2}=28cm$

Area of flower beads $=2\times \frac{\left(\pi r^2\right)}{2}$

$\pi r^2=\pi (28{)}^2-2464$

Area of lawn $=a^2=\left({56}^2\right)=3136m^2$

The sum of the areas of the lawn and the flower beds $=2464+3136$

$=5600m^2$