
A 7m wide road runs outside around a circular park, whose circumference is 176m. The area of the road is:
Answer
502.5k+ views
Hint: Start by drawing a rough diagram of the situation given in order to have a better understanding of the situation. Then use the data that the circumference of the circle is equal to 176m and we know that the circumference of the circle is given by $2\pi r$ to get the value of r. Once you get r, i.e., the inner radius adds 7 to it to get the external radius. Now use the formula $\pi {{\left( radius \right)}^{2}}$ for finding the area of the inner and the outer circle and find the difference to get the answer.
Complete step-by-step answer:
To start with the question let us draw the diagram of the park with the road along its boundary for better visualisation.
Now as it is given that the circumference of the inner circle is equal to 176m and we know that the circumference of the circle is given by $2\pi r$.
$\begin{align}
& \therefore 2\pi r=176 \\
& \Rightarrow r=\dfrac{88}{\pi }m \\
\end{align}$
Also, we know that the outer radius R is 7m more than r, as the road is 7m wide. We can say that R=r+7.
We can also see from the figure that the area of the road is equal to the area of the larger circle minus the area of the smaller circle. We also know that the area of a circle is given by $\pi {{\left( radius \right)}^{2}}$ .
$\text{Area of the road =}\left( \pi {{R}^{2}}-\pi {{r}^{2}} \right)=\pi \left( {{\left( r+7 \right)}^{2}}-{{r}^{2}} \right)$
Now we will use the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . On doing so, we get
$\text{Area of the road =}\pi \left( r+7-r \right)\left( r+7+r \right)=7\pi \left( 2r+7 \right)$
Now we will substitute the value of r, on doing so, we get
$\text{Area of the road =}7\pi \left( 2\times \dfrac{88}{\pi }+7 \right)=7\times 176+49\pi $
Now if we put the value of $\pi =\dfrac{22}{7}$ , we get
$\text{Area of the road =}7\times 176+49\times \dfrac{22}{7}=1232+7\times 22=1232+154=1386{{m}^{2}}$
Therefore, the answer to the above question is 1386 sq. m.
Note: The key to this question is to draw the correct diagram of the situation given in the question, as once you have drawn the diagram, we just have to put the formula of area of circle and find the difference to get the answer. Also, be careful about the calculation part as there is a high chance of making a mistake in the calculation part.
Complete step-by-step answer:
To start with the question let us draw the diagram of the park with the road along its boundary for better visualisation.

Now as it is given that the circumference of the inner circle is equal to 176m and we know that the circumference of the circle is given by $2\pi r$.
$\begin{align}
& \therefore 2\pi r=176 \\
& \Rightarrow r=\dfrac{88}{\pi }m \\
\end{align}$
Also, we know that the outer radius R is 7m more than r, as the road is 7m wide. We can say that R=r+7.
We can also see from the figure that the area of the road is equal to the area of the larger circle minus the area of the smaller circle. We also know that the area of a circle is given by $\pi {{\left( radius \right)}^{2}}$ .
$\text{Area of the road =}\left( \pi {{R}^{2}}-\pi {{r}^{2}} \right)=\pi \left( {{\left( r+7 \right)}^{2}}-{{r}^{2}} \right)$
Now we will use the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . On doing so, we get
$\text{Area of the road =}\pi \left( r+7-r \right)\left( r+7+r \right)=7\pi \left( 2r+7 \right)$
Now we will substitute the value of r, on doing so, we get
$\text{Area of the road =}7\pi \left( 2\times \dfrac{88}{\pi }+7 \right)=7\times 176+49\pi $
Now if we put the value of $\pi =\dfrac{22}{7}$ , we get
$\text{Area of the road =}7\times 176+49\times \dfrac{22}{7}=1232+7\times 22=1232+154=1386{{m}^{2}}$
Therefore, the answer to the above question is 1386 sq. m.
Note: The key to this question is to draw the correct diagram of the situation given in the question, as once you have drawn the diagram, we just have to put the formula of area of circle and find the difference to get the answer. Also, be careful about the calculation part as there is a high chance of making a mistake in the calculation part.
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