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.

Answer 1

Perimeter of a triangle = 540 m
Ratio in sides = 25:17:12
Sum of ratios = 25+17+12=54
$\therefore$ First side = $\frac{540\times 25}{54}=250m$
Second side = $\frac{540\times 17}{54}=170m$
Third side = $\frac{540\times 12}{54}=120m$
$\therefore s=\frac{Perimeter}{2}=\frac{540}{2}=270\therefore Area=\sqrt{s(s-a)(s-b)(s-c)}$
=$\sqrt{270(270-250)(270-170)(270-120)m^2}=\sqrt{270\times 20\times 100\times 150}$
=$\sqrt{\begin{array}{c}3\times 3\times 3\times 2\times 5\times 5\times 2\times 2\times 10\times 10\times 2\times 3\times 5\times 5\end{array}}{m}^2$
=$3\times 3\times 2\times 5\times 2\times 10\times 5=9000m^2$