A path of 4m width runs round a semicircular grassy plot whose c… - Vidyadip

Question

A path of 4m width runs round a semicircular grassy plot whose circumference is 81 $\big(\frac{5}{7}\big)$m. Find:

The area of the path.
The cost of gravelling the path at the rate of ₹ 1.50 per square metre.
The cost of turfing the plot at the rate of 45 paise per $m^2$.

✓

Answer

Width of path around the semicircular grassy plot = 4m
Circumference of the plot $= 81 \big(\frac{5}{7}\big)$m.
$=\big(\frac{572}{7}\big)\text{m}$
Let r be the radius of the plot, then
$\pi\text{r}=\frac{572}{7}\Rightarrow\frac{22}{7}\text{r}=\frac{572}{7}$
$(\text{Semicircumference}=\pi\text{r})$
$\Rightarrow\text{r}=\frac{572}{7}\times\frac{7}{22}=26$
$\therefore$ Radius of plot (r) = 26m
Width of the parh = 4m
Outer radius (R) = 26 + 4 = 30m

$\therefore\text{Area of path}=\frac{1}{2}\pi(\text{R}^2-\text{r}^2)$
$=\frac{1}{2}\times\frac{22}{7}(30^2-26^2)\text{m}$
$=\frac{11}{7}(30+26)(30-26)\text{m}^2$
$=\frac{11}{7}\times56\times4=352\text{m}^2$
Cost of Gravelling at the rate of 1.50 per $m^2$
$=₹352\times\frac{150}{100} =₹\frac{352\times3}{2}$
$=₹176\times3=₹528$
Area of the plot $=\frac{1}{2}\pi\text{r}^2$
$=\frac{1}{2}\times\frac{22}{7}\times(26)^2\text{m}^2$
$=\frac{11}{7}\times676=\frac{7436}{7}\text{m}^2$
Rate of turFing the plot = 45 paisa per $m^2$
$\therefore\text{Total cost}=\text{Rs}.\frac{7436}{7}\times\frac{45}{100}$
= ₹ 478.02 = ₹ 478 (approx)

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