Question

A racetrack is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m. Find the width and the area of the track.

Answer 1

Let r m and R m be the inner and outer boundaries, respectively.
Thus, we have:

$$\begin{gathered} 2\mathrm{\pi }r=352 \\ \Rightarrow r=\frac{352}{2\mathrm{\pi }} \\ \mathrm{Also}, \\ 2\mathrm{\pi R}=396 \\ \Rightarrow R=\frac{396}{2\mathrm{\pi }} \end{gathered}$$

Width of the track$=\left(R-r\right)$
$$\begin{gathered} =\left(\frac{396}{2\mathrm{\pi }}-\frac{352}{2\mathrm{\pi }}\right)\text{m} \\ =\frac{1}{2\mathrm{\pi }}\left(396-352\right)\text{m} \\ =\left(\frac{1}{2}\times \frac{7}{22}\times 44\right)\text{m} \\ =7\text{m} \end{gathered}$$

Area of the track$=\mathrm{\pi }\left({\mathrm{R}}^2-{\mathrm{r}}^2\right)$
$$\begin{gathered} =\mathrm{\pi }\left(\mathrm{R}+\mathrm{r}\right)\left(\mathrm{R}-\mathrm{r}\right) \\ =\left[\mathrm{\pi }\left(\frac{396}{2\mathrm{\pi }}+\frac{352}{2\mathrm{\pi }}\right)\times \left(\frac{396}{2\mathrm{\pi }}-\frac{352}{2\mathrm{\pi }}\right)\right]{\text{m}}^2 \\ =\left(\mathrm{\pi }\times \frac{748}{2\mathrm{\pi }}\times 7\right){\text{m}}^2 \\ =\frac{748}{2}\times 7{\text{m}}^2 \\ =2618{\text{m}}^2 \end{gathered}$$