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

Solution:

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

$2 \pi r=352$

$\Rightarrow r=\frac{352}{2 \pi}$

Also,

$2 \pi \mathrm{R}=396$

$\Rightarrow R=\frac{396}{2 \pi}$

Width of the track $=(R-r)$

$=\left(\frac{396}{2 \pi}-\frac{352}{2 \pi}\right) \mathrm{m}$

$=\frac{1}{2 \pi}(396-352) \mathrm{m}$

$=\left(\frac{1}{2} \times \frac{7}{22} \times 44\right) \mathrm{m}$

$=7 \mathrm{~m}$

Area of the track $=\pi\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)$

$=\pi(\mathrm{R}+\mathrm{r})(\mathrm{R}-\mathrm{r})$

$=\left[\pi\left(\frac{396}{2 \pi}+\frac{352}{2 \pi}\right) \times\left(\frac{396}{2 \pi}-\frac{352}{2 \pi}\right)\right] \mathrm{m}^{2}$

$=\left(\pi \times \frac{748}{2 \pi} \times 7\right) \mathrm{m}^{2}$

$=\frac{748}{2} \times 7 \mathrm{~m}^{2}$

$=2618 \mathrm{~m}^{2}$