Question

In Figure, two circular flower beds have been shown on two sides of a square lawn $ABCD$ of side $56m$. 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

Considering $△ABC,$

By Pythagoras Theorem,

$AC=\sqrt{A{B}^2+B{C}^2}$

$AC=\sqrt{{56}^2+{56}^2}$

$AC=56\sqrt{2}$

Radius of circle $=\frac{AC}{2}=\frac{56\sqrt{2}}{2}=\frac{56}{\sqrt{2}}$m. $[R=\frac{DiagonalAC}{2}]$

Total area $=(side{)}^2+$ Area of sector ODC and OAB

Total area $=(56{)}^2+$ Area of sector ODC and OAB

$=(56{)}^2+\frac{\pi /2}{2\pi }\times \pi r^2+\frac{\pi /2}{2\pi }\pi r^2$

$=(56{)}^2+\frac{\pi }{2}{r}^2$

$=(56{)}^2+\frac{(56{)}^2\pi }{4}$

$=3136+2463$

Total area $\simeq 5599m^2$.