Question

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 flower beds.

Answer 1

Side of square ABCD$=56$ m

AC$=$BD(diagonals of a square are equal in lengths)

Diagonal of square (AC) $=\sqrt{2}\times$ side square

$=\sqrt{2}\times 56=56\sqrt{2}m$

OA$=$OB$=\frac{1}{2}$AC$=\frac{1}{2}\left(56\sqrt{2}\right)=28\sqrt{2}$ m

Let OA$=$OB$=r$ m[radius of the sector]

Area of sector OAB$=\left[\frac{{90}^o}{{360}^o}\right]\pi r^2$

$=\left(\frac{1}{4}\right)\pi r^2=\frac{1}{4}\times \frac{22}{7}\times {\left(\frac{28}{\sqrt{2}}\right)}^2{m}^2$

$=[\frac{1}{4}\times \frac{22}{7}\times 28\times 28\times 2]m^2$

$=1232m^2$

Area of flower bed AB$=$area of sector OAB$-$ area of $\Delta$OAB

$\Rightarrow 1232-\frac{1}{2}\times OB\times OA\Rightarrow 1232-\frac{1}{2}\times 28\sqrt{2}\times 28\sqrt{2}$

$\Rightarrow 1232-784=448m^2$

Similarly, Area of other flower bed CD$=448m^2$

Total area $=$ Area of square ABCD$+$area of flower bed AB$+$Area of flower bed CD

$=(56\times 56)+448+448=4032m^2$

Hence, sum of the area of lawns and the flower beds are $4032m^2$.