Question

It is proposed to add to a square lawn with each side $58m$ two circular ends the center of each circle being the point of intersection of the diagonals of the square. The area of the whole lawn is ?

Answer 1

Step 1: Construct the figure of the lawn according to given conditions

Consider $ABCD$ is a square lawn whose side is $58m$.

Also, $AED$ and $BFC$ are the two circular ends

Also, the diagonal of the lawn $=\sqrt{{\left(58\right)}^2+{\left(58\right)}^2}=58\sqrt{2}m$

Step 2: Find the radius of the circular segments

The diagonals of a square are perpendicular bisectors of each other

$\Rightarrow$ $OA=OB=OD=OC=\frac{AC}{2}=29\sqrt{2}m$.

Thus, the radius of a circle that has a center at the point of intersection of diagonal is given by

$r=\frac{58\sqrt{2}}{2}=29\sqrt{2}m$

Step 3:Find The area of circular segments

Segments of the circle , $AED$ and $BFC$ are equal. Hence their areas are equal.

Area of segment $AED$$=$Area of sector$OAED$ $-$Area of triangle $OAD$ …[ $\mathbf{}\mathbf{\because }\frac{\mathbf{\theta }}{{\mathbf{360}}^{\mathbf{\circ }}}\mathbf{\times }{\mathbf{\pi r}}^{\mathbf{2}}\mathbf{=}$Area of sector]

$\Rightarrow$ Area of segment $AED$$=$$\left(\frac{90°}{360°}\times \mathrm{\pi }\times {\mathrm{r}}^2\right)-\left(\frac{1}{2}\times \mathrm{AO}\times \mathrm{OD}\right)$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \angle \mathrm{AOD}={90}^{\circ }\right]$

$\Rightarrow$Area of segment $AED$$=$$\left(\frac{1}{2}\times \mathrm{\pi }\times {29}^2\right)-\left({29}^2\right)$

$\Rightarrow$Area of segment $AED$$=480.039m^2$

Step 4: Find the area of the lawn

Area of lawn $=$Area of square $ABCD$ $+$Area of segment $AED$$+$Area of segment $BFC$

$\Rightarrow$Area of lawn $=$Area of square $ABCD$ $+2\times$Area of segment $AED$

$\Rightarrow$ Area of lawn $={58}^2+2\times 480.039$

$\Rightarrow$ Area of lawn $=4324.078m^2$

Hence, the area of the whole lawn is $4324.078m^2$.