Two circular flower beds have been …

Given side of the square = 56 m
From the fig in {tex}\Delta{/tex}BCD,
BD² = BC² + DC²
{tex}\Rightarrow{/tex} BD² = 2(56)²
{tex}\Rightarrow \mathrm{BD}=56 \sqrt{2} \;\mathrm{m}{/tex}
We know that diaganols of a square are equal and perpendicular bisector of each other,
{tex}\Rightarrow{/tex} AC = BD = {tex}56 \sqrt{2} m{/tex}
and AO = BO = CO = DO = {tex}\frac{56 \sqrt{2}}{2}=28 \sqrt{2} \;m{/tex}
Also area({tex}\Delta{/tex}AOD) = area({tex}\Delta{/tex}DOC) = area ({tex}\Delta{/tex}BOC) = area({tex}\Delta{/tex}AOB)
{tex}=\frac{1}{4} \operatorname{area}(\mathrm{ABCD})=\frac{1}{4} \times(56)^{2}=\frac{3136}{4}=784 \;\mathrm{m}^{2}{/tex}
and Area a sector OAB = Area sector OCD {tex}=\frac{1}{4} \pi \times(28 \sqrt{2})^{2}{/tex}
{tex}=\frac{1}{4} \times \frac{22}{7} \times 1568{/tex}
= 1232 m²
Thus required area = Area of {tex}\Delta{/tex}AOD + Area of sector ODC + Area of {tex}\Delta{/tex}BOC + Area of sector OAB
= 784 + 1232 + 784 + 1232
= 4032 m²