Question

A path of 4 m width runs round a semi-circular grassy plot whose circumference is 163 $\frac{3}{7}$ m Find:

(i) the area of the path

(ii) the cost of gravelling the path at the rate of Rs 1.50 per square metre

(iii) the cost of turfing the plot at the rate of 45 paise per m².

Answer 1

We have given AB = 4m and circumference of semicircle with radius OA as.

We are asked to find the area between the two semi-circles.

For that we will first find OA.

$\mathrm{\pi }r=163\frac{3}{7}$

Now we will substitute

$$\begin{gathered} \frac{22}{7}\times r=163\frac{3}{7} \\ \Rightarrow r=\frac{1144}{7}\times \frac{7}{22} \\ \Rightarrow r=52 \\ \Rightarrow OA=52m \end{gathered}$$

Now we will find OB.

$$\begin{gathered} \therefore OB=52+4 \\ \therefore OB=56\mathrm{m} \end{gathered}$$

Now we will find the area between two semi-circles as given below,
$$\begin{gathered} \therefore \mathrm{Area}=\frac{\mathrm{\pi }\times 56\times 56}{2}-\frac{\mathrm{\pi }\times 52\times 52}{2} \\ =\frac{\mathrm{\pi }}{2}\left(1568-1352\right){\mathrm{m}}^2 \\ =\frac{\mathrm{\pi }}{2}\times 216 \\ =\frac{22}{7\times 2}\times 216 \\ =339.43{\mathrm{m}}^2 \end{gathered}$$

Therefore, area of the path is 339.43 m².

Now we will find the cost of gravelling the path.

$$\begin{gathered} \mathrm{Cost}=339.43\times 1.50 \\ =\mathrm{Rs}509.14 \end{gathered}$$

Therefore, cost of gravelling the path is Rs 509.14.

Now we will find the cost of turfing the plot. For that we will find the area of the plot.

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{plot}=\frac{1}{2}{\mathrm{\pi r}}^2 \\ =\frac{1}{2}\times \frac{22}{7}{\left(52\right)}^2 \\ =\frac{1}{2}\times \frac{22}{7}\times 52\times 52 \\ =4249.14 \\ \mathrm{Cost}\mathrm{of}\mathrm{turfing}\mathrm{the}\mathrm{plot}=4249.14\times 0.45 \\ =\mathrm{Rs}1912.11 \end{gathered}$$

Therefore, cost of the turfing the plot is Rs 1912.11.
Disclaimer: Due to some error in the question, we get the different answers.