Question

Find the area of the athletic ground given below. (All dimensions are in Meters)

A. $1165.5c{{m}^{2}}$
B. $1165.5{{m}^{2}}$
C. $116.5c{{m}^{2}}$
D. None

Answer 1

Hint: We can see two geometrical shapes in the given diagram. They are one rectangle and two semi circles. We have the width of the rectangle and total length of the ground which includes the radius of the two semi circles. So, to calculate the length of the rectangle we will subtract the radiuses of two semi circles from the total length of ground. Now we will calculate the area of each shape and sum up them to get the required result.

Complete step-by-step answer:
Given that, The athletic ground is given below

The diameter of the two semi circles is $CA=21m$.
Now the area of the one semi circle is given by
$\begin{align}
  & A=\dfrac{1}{8}\pi {{d}^{2}} \\
 & \Rightarrow A=\dfrac{1}{8}\times 3.14\times {{21}^{2}} \\
 & \Rightarrow A=173.0925{{m}^{2}} \\
\end{align}$
Now the area of the two semi circles is given by
$\begin{align}
  & 2A=2\times 173.0925 \\
 & \Rightarrow 2A=346.185{{m}^{2}} \\
\end{align}$
In the problem we have the width of the rectangle as $21m$. Now the length of the rectangle is given by
$\begin{align}
  & l=60-\dfrac{21}{2}-\dfrac{21}{2} \\
 & \Rightarrow l=39m \\
\end{align}$
So the area of the rectangle is calculated as
$\begin{align}
  & a=l\times b \\
 & \Rightarrow a=39\times 21 \\
 & \Rightarrow a=819{{m}^{2}} \\
\end{align}$
Now the total area of the ground is equal to the sum of the area of the rectangle and the total areas of the two semi circles.
$\begin{align}
  & \therefore {{A}_{r}}=a+2A \\
 & \Rightarrow {{A}_{r}}=819+346.185 \\
 & \Rightarrow {{A}_{r}}=1165.185{{m}^{2}} \\
\end{align}$
$\therefore $Area of the athletic ground is $1165.15{{m}^{2}}$.

So, the correct answer is “Option B”.

Note: We can directly use the below formula to get the area of the athletic ground.
$\begin{align}
  & A=b\times \left( EF-CA \right)+\pi {{\left( CA \right)}^{2}} \\
 & \Rightarrow A=21\times \left( 60-21 \right)+\pi \times {{21}^{2}} \\
 & \Rightarrow A=819+346.185 \\
 & \Rightarrow A=1165.15{{m}^{2}} \\
\end{align}$
From both the methods we got the same result.