Answer
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Hint: To solve the question, we will first find out the area of the outer circle which is formed from the grass walk around the pond and then we will find the area of the pond which is the area of the smaller circle. When we subtract the area of the smaller circle (pond) from the area of outer circle, we will get the required area of the pond.
Complete step-by-step answer:
We are given that the diameter of a pond is 100m. so the radius of the pond is given by the formula:
\[\begin{align}
& radius(r)=\dfrac{diameter}{2} \\
& \Rightarrow \pi =\dfrac{100}{2} \\
& \Rightarrow \pi =50m \\
\end{align}\]
We are also given that outside the pon, there is a grass walk which has a width of 1m. so, we can think of this as an outer circle and an inner circle. This can be shown as:
Therefore, as shown in the figure above, the area of the inner circle can be considered as the area of the pond and the area of shaded portion can be considered as the area of the walk which we have to find. Now the diameter of the outer circle= 100+2+2=104m. thus, its radius will be equal to $\dfrac{104}{2}m=52m$
Now, to find the area of the shaded portion, we will find the area of the outer circle and subtract the area of inner circle from it. The area of a circle with radius ‘r’ is given by:
$area=\pi {{\left( r \right)}^{2}}$
The area of the outer circle
$\begin{align}
& =\pi {{\left( 52 \right)}^{2}} \\
& =27004r{{m}^{2}} \\
\end{align}$
The area of inner circle=
$\begin{align}
& =\pi {{\left( 50 \right)}^{2}} \\
& =2500r{{m}^{2}} \\
\end{align}$
We are not calculating the values of the area by putting the value of \[\pi \] because all the options in the question is given in terms of \[\pi \].
So, now we have to find the area of the shaded portion which is equal to the area of grass walk. Thus, area of grass walk is given by:
Area of grass walk= area of outer circle- area of inner circle
Area of grass walk= 2704 \[\pi \]-2500 \[\pi \]
Area of grass walk= $204r{{m}^{2}}$
So, the correct answer is “Option C”.
Note: We can also calculate the area of the shaded portion without finding the radius. The formula in this case would be:
Area of shaded region:
$r\left[ {{\left( \dfrac{d2}{2} \right)}^{2}}-{{\left( \dfrac{d1}{2} \right)}^{2}} \right]$
Complete step-by-step answer:
We are given that the diameter of a pond is 100m. so the radius of the pond is given by the formula:
\[\begin{align}
& radius(r)=\dfrac{diameter}{2} \\
& \Rightarrow \pi =\dfrac{100}{2} \\
& \Rightarrow \pi =50m \\
\end{align}\]
We are also given that outside the pon, there is a grass walk which has a width of 1m. so, we can think of this as an outer circle and an inner circle. This can be shown as:
Therefore, as shown in the figure above, the area of the inner circle can be considered as the area of the pond and the area of shaded portion can be considered as the area of the walk which we have to find. Now the diameter of the outer circle= 100+2+2=104m. thus, its radius will be equal to $\dfrac{104}{2}m=52m$
Now, to find the area of the shaded portion, we will find the area of the outer circle and subtract the area of inner circle from it. The area of a circle with radius ‘r’ is given by:
$area=\pi {{\left( r \right)}^{2}}$
The area of the outer circle
$\begin{align}
& =\pi {{\left( 52 \right)}^{2}} \\
& =27004r{{m}^{2}} \\
\end{align}$
The area of inner circle=
$\begin{align}
& =\pi {{\left( 50 \right)}^{2}} \\
& =2500r{{m}^{2}} \\
\end{align}$
We are not calculating the values of the area by putting the value of \[\pi \] because all the options in the question is given in terms of \[\pi \].
So, now we have to find the area of the shaded portion which is equal to the area of grass walk. Thus, area of grass walk is given by:
Area of grass walk= area of outer circle- area of inner circle
Area of grass walk= 2704 \[\pi \]-2500 \[\pi \]
Area of grass walk= $204r{{m}^{2}}$
So, the correct answer is “Option C”.
Note: We can also calculate the area of the shaded portion without finding the radius. The formula in this case would be:
Area of shaded region:
$r\left[ {{\left( \dfrac{d2}{2} \right)}^{2}}-{{\left( \dfrac{d1}{2} \right)}^{2}} \right]$
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