Question
A racetrack is in the form of a ring whose inner circumference is $528m$ and the outer circumference is $616m$. Find the width of the track.
✓
Answer
Let the inner and outer radii of the track be $r$ metres and $R$ metres, respectively.
Then, $2\pi\text{r}=528$
$2\pi\text{R}=616$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=528$
$2\times\frac{22}{7}\times\text{R}=616$
$\Rightarrow\text{r}=\Big(528\times\frac{7}{44}\Big)=84$
$\text{R}=\Big(616\times\frac{7}{44}\Big)=98$
$\Rightarrow (R - r) = (98 - 84)m = 14m$
Hence, the width of the track is $14m.$
Then, $2\pi\text{r}=528$
$2\pi\text{R}=616$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=528$
$2\times\frac{22}{7}\times\text{R}=616$
$\Rightarrow\text{r}=\Big(528\times\frac{7}{44}\Big)=84$
$\text{R}=\Big(616\times\frac{7}{44}\Big)=98$
$\Rightarrow (R - r) = (98 - 84)m = 14m$
Hence, the width of the track is $14m.$
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