A racetrack is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m.
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.
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 $=\pi\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)$
$=\pi(\mathrm{R}+\mathrm{r})(\mathrm{R}-\mathrm{r})$
$=\left[\pi\left(\frac{396}{2 \pi}+\frac{352}{2 \pi}\right) \times\left(\frac{396}{2 \pi}-\frac{352}{2 \pi}\right)\right] \mathrm{m}^{2}$
$=\left(\pi \times \frac{748}{2 \pi} \times 7\right) \mathrm{m}^{2}$
$=\frac{748}{2} \times 7 \mathrm{~m}^{2}$
$=2618 \mathrm{~m}^{2}$
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