Question

In the figure, the inside perimeter of a running track with semicircular ends and straight parallel lines is 312 m. The length of the straight portion of the track is 90 m. If the track has a uniform width of 2 m throughout, find its area.

a. 686.57 ${{m}^{2}}$
b. 656.57 ${{m}^{2}}$
c. 636.57 ${{m}^{2}}$
d. 616.57 ${{m}^{2}}$

Answer 1

Hint: In order to solve this question, we should know that the perimeter of a field is the total length of the boundary. By using this concept we will find the radius of the semicircle and then to find the area of the track we will subtract the area covered by the inside boundary from the area covered by the outside boundary and we will get the answer.

Complete step-by-step answer:
In this question, we have been asked to find the area of the track in the figure below for a few given conditions.

Now, we have been given that the perimeter of inner boundary is 312 m and we know that perimeter of boundary in total length of boundary and we know that perimeter of circle is $2\pi r$, so we can say, for radius = r of the inner semicircle,
$\begin{align}
  & \pi r+90+\pi r+90=312 \\
 & \Rightarrow 2\pi r+180=312 \\
 & \Rightarrow 2\pi r=312-180 \\
 & \Rightarrow \pi r=\dfrac{132}{2}=66 \\
 & r=\dfrac{66}{\pi } \\
\end{align}$
Now, we know that, $\pi =\dfrac{22}{7}$. So, we get, $r=\dfrac{66}{22}\times 7=21m$.
Now, we have been given that the outer boundary of the track is 2 m away from the inner boundary throughout. So, we can say that the radius r of the outer semicircle will be 21 + 2 = 23 m.
We know that the area covered by a circle is $\pi {{r}^{2}}$ and the area of a rectangle is given by, length $\times $ breadth. So, we can say that the area covered by outer boundary is,
2 (Area of semicircle of radius r’) + Area of rectangle of length 90 m and width 2r’
Now, we know that the width of the outer rectangle is the same as the diameter of the outer semicircle. So, we can say that, the width of the outer rectangle = $23\times 2=46m$.
Therefore, we get the area covered by the outer boundary as,
$\begin{align}
  & 2\times \dfrac{\pi r{{'}^{2}}}{2}+90\times 46 \\
 &= \pi {{\left( 23 \right)}^{2}}+90\times 46 \\
 & =529\pi +4140.........\left( i \right) \\
\end{align}$
Similarly, we will calculate the area covered by the inner boundary. So, we can say that area covered by inner boundary is,
2 (Area of semicircle of radius r) + Area of rectangle of length 90 m and width 2r
Now, we know that the width of the inner rectangle is the same as the diameter of the inner semicircle. So, we can say that, the width of the inner rectangle = $21\times 2=42m$.
Therefore, we get the area covered by the inner boundary as,
$\begin{align}
  & 2\times \dfrac{\pi {{r}^{2}}}{2}+90\times 42 \\
 & =\pi {{\left( 21 \right)}^{2}}+3780 \\
 &= 441\pi +3780.........\left( ii \right) \\
\end{align}$
Now, we know that the area of the tract can be calculated by subtracting the area covered by the inner boundary from the area covered by the outer boundary. So, on subtracting equation (ii) from equation (i), we get,
$\begin{align}
  & \left( 529\pi +4140 \right)-\left( 441\pi +3780 \right) \\
 &= 529\pi -441\pi +4140-3780 \\
 &= 88\pi +360 \\
\end{align}$
Now, we know that $\pi =\dfrac{22}{7}$. Therefore, we can say that the area of the tract is,
$\begin{align}
  & 88\times \dfrac{22}{7}+360 \\
 & =276.57+360 \\
 &= 636.57{{m}^{2}} \\
\end{align}$
Therefore, we get the area of the tract as $636.57{{m}^{2}}$. Hence, we get option (c) as the correct answer.

Note: While solving this question, the possible mistake we can make is the calculation mistake because this question contains a lot of calculations. Also, we need to remember that the area of the rectangle is given by, length $\times $ breadth and area of circle is given by, $\pi {{r}^{2}}$. Also, sometimes, we make mistakes by writing the outer semicircle radius as 21 + 1 which is wrong. It should be 21 + 2 ,23 m. So, we have to be careful while solving the question.