Question
A racetrack is in the form of a ring whose inner circumference is 352m and outer circumference is 396m. Find the Width and the area of the track.
✓
Answer
Let r m and R m be the inner and outer boundaries, respectively.
Thus, we have:
$2\pi\text{r}=352$
$\Rightarrow\text{r}=\frac{352}{2\pi}$
Also,
$2\pi\text{R}=396$
$\Rightarrow\text{R}=\frac{396}{2\pi}$
Width of the track $=(\text{R}-\text{r})$
$=\Big(\frac{396}{2\pi}-\frac{352}{2\pi}\Big)\text{m}$
$=\frac{1}{2\pi}(396-352)\text{m}$
$=\Big(\frac{1}{2}\times\frac{7}{22}\times44\Big)\text{m}$
$=7\text{m}$
Area of the track $=\pi\big(\text{R}^2-\text{r}^2\big)$
$=\pi\big(\text{R}+\text{r}\big)\big(\text{R}-\text{r}\big)$
$=\Big[\pi\Big(\frac{396}{2\pi}+\frac{352}{2\pi}\Big)\times\Big(\frac{396}{2\pi}-\frac{352}{2\pi}\Big)\Big]\text{m}^2$
$=\Big(\pi\times\frac{748}{2\pi}\times7\Big)\text{m}^2$
$=\frac{748}{2}\times7\text{m}^2$
$=2618\text{m}^2$
Thus, we have:
$2\pi\text{r}=352$
$\Rightarrow\text{r}=\frac{352}{2\pi}$
Also,
$2\pi\text{R}=396$
$\Rightarrow\text{R}=\frac{396}{2\pi}$
Width of the track $=(\text{R}-\text{r})$
$=\Big(\frac{396}{2\pi}-\frac{352}{2\pi}\Big)\text{m}$
$=\frac{1}{2\pi}(396-352)\text{m}$
$=\Big(\frac{1}{2}\times\frac{7}{22}\times44\Big)\text{m}$
$=7\text{m}$
Area of the track $=\pi\big(\text{R}^2-\text{r}^2\big)$
$=\pi\big(\text{R}+\text{r}\big)\big(\text{R}-\text{r}\big)$
$=\Big[\pi\Big(\frac{396}{2\pi}+\frac{352}{2\pi}\Big)\times\Big(\frac{396}{2\pi}-\frac{352}{2\pi}\Big)\Big]\text{m}^2$
$=\Big(\pi\times\frac{748}{2\pi}\times7\Big)\text{m}^2$
$=\frac{748}{2}\times7\text{m}^2$
$=2618\text{m}^2$
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