Question
A path of width 3.5m runs around a semicircular grassy plot whose perimeter is 72m. Find the area of the path. $\Big(\text{Use }\pi=\frac{22}{7}\Big)$
✓
Answer
Perimeter of semicircle grassy plot = 72m
Let r be the radius of the plot
$\therefore\pi\text{r}+2\text{r}=72\Rightarrow\frac{22}{7}\text{r}+2\text{r}=72$
$\Rightarrow\frac{36}{7}\text{r}=72\Rightarrow\text{r}=\frac{72\times7}{36}=14\text{m}$
$\therefore$ Inner radius = 14m
and outer radius = 14+3.5 = 17.5 $=\frac{35}{2}\text{m}$
$\text{Now area of path}=\frac{1}{2}\pi\text{R}^2-\frac{1}{2}\pi\text{r}^2=\frac{1}{2}\pi$
$(\text{R}^2-\text{r}^2)$
$=\frac{1}{2}\times\frac{22}{7}\bigg[\Big(\frac{35}{2}\Big)^2-(14)^2\bigg]$
$=\frac{11}{7}\Big[\frac{1225}{4}-196\Big]\text{m}^2$
$=\frac{11}{7}\Big[\frac{1225-784}{4}\Big]$
$=\frac{11}{7}\times\frac{441}{4}\text{m}^2=\frac{693}{4}\text{m}^2$
$=173.25\text{m}^2$
Let r be the radius of the plot
$\therefore\pi\text{r}+2\text{r}=72\Rightarrow\frac{22}{7}\text{r}+2\text{r}=72$
$\Rightarrow\frac{36}{7}\text{r}=72\Rightarrow\text{r}=\frac{72\times7}{36}=14\text{m}$
$\therefore$ Inner radius = 14m
and outer radius = 14+3.5 = 17.5 $=\frac{35}{2}\text{m}$
$\text{Now area of path}=\frac{1}{2}\pi\text{R}^2-\frac{1}{2}\pi\text{r}^2=\frac{1}{2}\pi$
$(\text{R}^2-\text{r}^2)$
$=\frac{1}{2}\times\frac{22}{7}\bigg[\Big(\frac{35}{2}\Big)^2-(14)^2\bigg]$
$=\frac{11}{7}\Big[\frac{1225}{4}-196\Big]\text{m}^2$
$=\frac{11}{7}\Big[\frac{1225-784}{4}\Big]$
$=\frac{11}{7}\times\frac{441}{4}\text{m}^2=\frac{693}{4}\text{m}^2$
$=173.25\text{m}^2$
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