Question

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

Answer 1

Hint: The area between the sector of circle and the triangle is equal to the area of flower bed.
The area of the flowerbed can be determined by subtracting the area of the triangle from the area of the sector of the circle.

Complete step by step solution:
In the figure, the circular flowerbeds are drawn on the sides of the circle.
The side of square PQ is equal to $42$ m and as each side of square is equal to each other, it means that PQ = QR = $42$ m.
The sector of the circle has centre O. This means that the radius ($r$) of the sector is equal to the length of side OP.
Length of side OP can be determined by determining the diagonal of square as OP is equal to half of the diagonal of square, PR. OP is half of diagonal of square because diagonals of square bisect each other because of which OP = OR and thereby OP is half of PR.
Diagonal of square, PR = $\sqrt{2}$ × side of square = $42\sqrt{2}$ m
This indicates OP = r = $\dfrac{1}{2}\times 42\sqrt{2}\text{ m}=21\sqrt{2}\text{ m}$
$\begin{align}
  & \text{Area of sector of circle =}\dfrac{\theta }{360{}^\circ }\pi {{r}^{2}} \\
 & =\dfrac{90{}^\circ }{360{}^\circ }\times \dfrac{22}{7}\times \left( 21\sqrt{2} \right)\text{ }{{\text{m}}^{2}} \\
 & =693\text{ }{{\text{m}}^{\text{2}}}
\end{align}$
 $\begin{align}
  & \text{Area of triangle =}\dfrac{1}{4}\times \text{Area of square} \\
 & =\dfrac{1}{4}\times {{\left( side \right)}^{2}}\text{ }{{\text{m}}^{2}} \\
 & =\dfrac{42\times 42}{4}\text{ }{{\text{m}}^{\text{2}}} \\
 & =441\text{ }{{\text{m}}^{\text{2}}}
\end{align}$
$\begin{align}
  & \text{Area of one circular flower bed = 693 }-\text{ 441 }{{\text{m}}^{2}} \\
 & \text{= 252 }{{\text{m}}^{2}}
\end{align}$
By multiplying the area of one flower bed by two, the area of two flower beds can be calculated.
$\begin{align}
  & \text{Area of two circular flower beds = 2 }\times \text{ 252 }{{\text{m}}^{2}} \\
 & \text{= 504 }{{\text{m}}^{2}}
\end{align}$

Note:
The circular flower beds in the given figure cannot be called semi-circles as they were not drawn on the radius of the circle but flower beds were drawn on the segment of circle.