Question

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

Answer 1

Area of square lawn$={56}^2=3136m^2$

As diagonals of square perpendicular bisect each other and are equal,

$\Rightarrow OA=OB$,

$\angle AOB={90}^o$

In $\Delta AOB,2A{O}^2=A{B}^2$

$\Rightarrow AO=\frac{56}{\sqrt{2}}=28\sqrt{2}$m

AO is radius of sector AOB

area$\left(AOB\right)=\pi r^2\times \frac{90}{360}=\frac{\pi (28\sqrt{2}{)}^2}{4}$

$=\frac{2\pi \times 28\times 28}{4}$

$=\frac{22}{7}\times 14\times 28$

$=22\times 56=1232m^2$

Area of $\Delta AOB=\frac{1}{2}\left(AO\right)\left(OB\right)=\frac{1}{2}\left(28\sqrt{2}\right)\left(28\sqrt{2}\right)={28}^2=784$

$\Rightarrow$ Area of shaded region$=2\times (1232-784)=896m^2$.