Question

In the given figure, PQRS is a square lawn with side PQ=42 m. Two circular flower beds are there on the sides PS and QR with center at O, the intersection of its diagonals. Find the total area of the two flower beds(shaded parts).

Answer 1

Given,PQRS is a square with side 42 m.

Let its diagonals intersect at O.

Then, OP=OQ=OR=OS

$\text{and}\angle POS=\angle QOR={90}^{\circ }$

$P{R}^2=P{Q}^2+Q{R}^2$

$PR=(\sqrt{2}\times 42)m$

Now, OP=$\frac{1}{2}\times \left(diagonal\right)=21\sqrt{2}m$


$\because$ Area of flower bed PAS=Area of flower bed QBR

$\because$ Total Area of the two flower beds=Area of flower bed PAS+Area of flower bed QBR

=$2\times \left[\text{Area of sector OPAS-Area of}△POS\right]$

=$2\times [\pi r^2\frac{\theta }{{360}^{\circ }}-\frac{1}{2}\times OP\times OQ]$

=$2\times [\pi r^2\frac{\theta }{{360}^{\circ }}-\frac{1}{2}{r}^2]$

$[where,\theta ={90}^{\circ }]$

=$2\times [\frac{22}{7}\times (21\sqrt{2}{)}^2\frac{{90}^{\circ }}{{360}^{\circ }}-\frac{1}{2}\times 21\sqrt{2}\times 21\sqrt{2}]$

$[\because sin{90}^{\circ }=1]$

$=2\times [\frac{22}{7}\times 21\times 21\times 2\times \frac{1}{4}-\frac{1}{2}\times 21\times 21\times 2]$

$=2[33\times 21-441]$

=2[693-441]

=$504m^2$

Hence area of flower beds is $504m^2$.