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A race track is in the form of a ring whose inner and outer circumferences are 437m and 503m respectively. The area of the track isA. 66 sq.mB. 4935 sq.mC. 9870 sq.mD. None of these

Last updated date: 10th Aug 2024
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Hint: To do this problem, we should know the formula for circumference and area of a circle. The circumference and area of a circle whose radius is r units are $2\pi r$ and $\pi {{r}^{2}}$ respectively. In the question, we are given a race track which is between two concentric circles as shown in below figure. Using the given circumferences, we should find the radius of the inner circle and outer circle. The area of the race track will be equal to the difference of areas of outer circle and inner circle.

In the question, we are given a race track with concentric circles and their circumferences are given. We are asked to find the area of the race track.
Let us consider a circle of radius r units. Its circumference is given by $2\pi r$ units. Its area is given by $\pi {{r}^{2}}$ sq units. Using these relations, we can find the area of the race track.
Let us consider the inner circle with circumference 437 units. Let the radius be ${{r}_{1}}$ units.
From the above formulae
\begin{align} & 2\pi {{r}_{1}}=437 \\ & {{r}_{1}}=\dfrac{437}{2\pi }\text{ units} \\ \end{align}
Let us consider the outer circle with circumference 503 units. Let the radius be ${{r}_{2}}$ units.
From the above formulae
\begin{align} & 2\pi {{r}_{2}}=503 \\ & {{r}_{2}}=\dfrac{503}{2\pi }\text{ units} \\ \end{align}
The area of the race track is given by the difference between the areas of two circles.
Required area = Area of outer circle – Area of inner circle.
Required area = $\pi {{r}_{2}}^{2}-\pi {{r}_{1}}^{2}$
We can use the formula of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ and the required area becomes
$\pi \left( {{r}_{2}}^{2}-{{r}_{1}}^{2} \right)=\pi \left( {{r}_{2}}-{{r}_{1}} \right)\left( {{r}_{2}}+{{r}_{1}} \right)$
Substituting the values of ${{r}_{2}}\text{ and }{{r}_{1}}$ in the above equation, we get
Required area = $\pi \left( \dfrac{503}{2\pi }-\dfrac{437}{2\pi } \right)\left( \dfrac{503}{2\pi }+\dfrac{437}{2\pi } \right)=\pi \left( \dfrac{503-437}{2\pi } \right)\left( \dfrac{503+437}{2\pi } \right)$
Area of the track = $\pi \left( \dfrac{66}{2\pi } \right)\left( \dfrac{940}{2\pi } \right)=\dfrac{66\times 940}{4\pi }$
Using $\pi =\dfrac{22}{7}$ in the above equation, we get
Area of track = $\dfrac{66\times 940}{4\times \dfrac{22}{7}}=\dfrac{66\times 940\times 7}{4\times 22}=3\times 235\times 7=235\times 21=4935$
So, the correct answer is “Option B”.

Note: Students can make a mistake by calculating diameters from given circumferences and using them in the area formula as radii. To avoid this confusion, we should write the formula clearly in either radius or diameter but not both and calculate the required values. To avoid confusion, students can calculate the areas separately with ${{r}_{1}}\text{ and }{{r}_{2}}$ and subtract them.