Question

A track is in the form of a ring whose inner and outer circumferences are $88cm$ and $132cm$ respectively. Find the width of the track.

Answer 1

Hint: In this problem we need to find the radius of the outer and inner circles from the given circumferences of the circles. We have the relation between the radius and circumference as $C=2\pi r$ where $C$ is the circumference of the circle and $r$ is the radius of the circle. The width of the track can be determined by calculating the difference between the radius of outer circle and radius of inner circle.

Complete step by step answer:
Given that, Circumference of inner and outer circles are $88cm$ and $132cm$.
Let us assume the radius of inner circle as $r$ and the radius of outer circle is $R$, then the ring is shown in below figure

Now the width of the ring is $width=R-r$
Finding the value of $R$.
Given that the circumference of the outer circle is $132cm$, now the radius of circle is
$\begin{align}
  & 2\pi R=132cm \\
 & R=\dfrac{132}{2\pi }cm.....\left( \text{i} \right)
\end{align}$
Finding the value of $r$.
We have circumference of the inner circle as $88cm$, then the value of $r$ is
$\begin{align}
  & 2\pi r=88cm \\
 & r=\dfrac{88}{2\pi }
\end{align}$
Now the width of the ring is given by
$\begin{align}
  & width=R-r \\
 & =\dfrac{132}{2\pi }cm-\dfrac{88}{2\pi }cm \\
 & =\dfrac{44}{2\pi }\left( 3-2 \right) \\
 & =\dfrac{44}{2\pi }
\end{align}$
Now substituting the value of $\pi $ as $\dfrac{22}{7}$, then
$\begin{align}
  & width=\dfrac{44}{2\left( \dfrac{22}{7} \right)} \\
 & =\dfrac{44\times 7}{44} \\
 & =7cm
\end{align}$

Hence the width of the ring is $7cm$.

Note: The problem becomes easier when we convert the given data into diagrammatic form. Don’t use the value of $\pi $ earlier. Based on the calculations according to our convenience we will substitute the value of $\pi $ as $\dfrac{22}{7}$ or $3.14$