Question

The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. Also, find the cost of ploughing the field at Rs 18.80 per 10 m².

Answer 1

Let the sides of the triangle be 25x m, 17x m and 12x m.
We know:
Perimeter = Sum of all sides
or, 540 = 25x + 17x + 12x
or, 54x = 540
or, x = 10
Thus, we obtain the sides of the triangle.
25$\times$10 = 250 m
17$\times$10 = 170 m
12$\times$10 = 120 m

Now,
$$\begin{gathered} \mathrm{Let}: \\ a=250\mathrm{m},b=170\mathrm{m}\mathrm{and}c=120\mathrm{m} \\ \therefore s=\frac{540}{2}=270\mathrm{m} \\ \mathrm{By}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{have}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{270(270-250)(270-170)(270-120)} \\ =\sqrt{270\times 20\times 100\times 150} \\ =\sqrt{30\times 3\times 3\times 20\times 20\times 5\times 30\times 5} \\ =30\times 3\times 20\times 5 \\ =9000{\mathrm{m}}^2 \end{gathered}$$

Cost of ploughing 10 m² field = Rs 18.80
Cost of ploughing 1 m² field = Rs $\frac{18.80}{10}$

Cost of ploughing 9000 m² field = $\frac{18.80}{10}\times 9000=\mathrm{Rs}16920$