Question

In figure two circular flower beads have been shown on two sides of a square lawn ABCD of side 56 m. If the center of each circular flower bed is the point of intersection of the diagonals of the square lawn, find the sum of the areas of the lawns and the flower beds.

Answer 1

We have,

In $\Delta ABC$ by Pythagoras theorem

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

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

$\therefore$ $OA=OB=$ $\frac{1}{2}$ $AC=28$ $\sqrt{2}$ m

So, the radius of the circle having a center at the point of intersection of diagonals is $28\sqrt{2}$ m.

Now,

Area of one circular end $=$ Area of a segment of angle ${90}^{\circ }$ in a circle of radius $28\sqrt{2}$.

$Area=(\frac{\pi \theta }{360}-sin\frac{\theta }{2}cos\frac{\theta }{2})r^2$

$Substitutingr=28\sqrt{2}and\theta -{90}^{\circ }$

$=\{\frac{22}{7}\times \frac{90}{360}-sin{45}^{\circ }cos{45}^{\circ }\}\times (28\sqrt{2}{)}^2{m}^2$

$=28\times 28\times 2\times \frac{4}{14}c{m}^2=448m^2$

$\therefore$ Area of two flower beds = $2$ $\times$ $448$ $m^2=896m^2$

Area of the square lawn = $56\times 56m^2=4032m^2$

Total area $=896+4032=4928m^2$