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

Solution:

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×">××10 = 250 m
17×">××10 = 170 m
12×">××10 = 120 m

Now,

Let:

$a=250 \mathrm{~m}, b=170 \mathrm{~m}$ and $c=120 \mathrm{~m}$

$\therefore s=\frac{540}{2}=270 \mathrm{~m}$

By Heron's formula, we have:

Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{270(270-250)(270-170)(270-120)}$

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

$=\sqrt{30 \times 3 \times 3 \times 20 \times 20 \times 5 \times 30 \times 5}$

 

$=30 \times 3 \times 20 \times 5$

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

Cost of ploughing 1 m2 field = Rs 5

Cost of ploughing $9000 \mathrm{~m}^{2}$ field $=5 \times 9000=\operatorname{Rs} 45000$.