The perimeter of a triangular field is 540 m, and its sides are in the ratio 25:17:12.

Solution:

Let the sides of the triangular field be $25 x, 17 x$ and $12 x$.

As, perimeter $=540 \mathrm{~m}$

$\Rightarrow 25 x+17 x+12 x=540$

$\Rightarrow 54 x=540$

$\Rightarrow x=\frac{540}{54}$

$\Rightarrow x=10$

So, the sides are $250 \mathrm{~m}, 170 \mathrm{~m}$ and $120 \mathrm{~m}$.

Now, semi $-$ perimeter, $s=\frac{250+170+120}{2}=\frac{540}{2}=270 \mathrm{~m}$

So, area of the field $=\sqrt{270(270-250)(270-170)(270-120)}$

$=\sqrt{270 \times 20 \times 100 \times 150}$

$=\sqrt{3^{3} \times 10 \times 2 \times 10 \times 10^{2} \times 3 \times 5 \times 10}$

$=3^{2} \times 10^{3}$

$=9000 \mathrm{~m}^{2}$

Also, the cost of ploughing the field $=\frac{9000 \times 40}{100}=₹ 3,600$