Question

The perimeter of a triangular field is $135cm$ and its sides are in the ratio $25:17:12$. Find its area.

Answer 1

Perimeter = 135 cm

Let ratio = x

$\begin{array}{c}a=25x\\ b=17x\\ c=12x\\ a+b+c=135\\ 25x+17x+12x=135\\ 54x=135\\ x=\frac{135}{54}\\ x=\frac{45}{18}\\ x=\frac{5}{2}\\ a=25\times \frac{5}{2}=\frac{125}{2}\\ b=17\times \frac{5}{2}=\frac{85}{2}\\ c=12\times \frac{5}{2}=\frac{60}{2}\\ Semiperimeter:\frac{1}{2}\left[\frac{125}{2}+\frac{85}{2}+\frac{60}{2}\right]\\ =\frac{1}{2}\times \frac{1}{2}\left[125+85+60\right]\\ =\frac{1}{4}\left(270\right)\\ =\frac{135}{2}\\ s-a=\frac{135}{2}-\frac{125}{2}\\ =\frac{10}{2}=5\\ s-b=\frac{135}{2}-\frac{85}{2}\\ =\frac{50}{2}=25\\ s-c=\frac{135}{2}-\frac{60}{2}\\ =\frac{75}{2}\\ Area=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ =\sqrt{\frac{135}{2}\left(\frac{10}{2}\right)\left(\frac{50}{2}\right)\left(\frac{75}{2}\right)}\\ =\frac{1}{4}\sqrt{135\times 10\times 50\times 75}\\ =\frac{1}{4}\times 2250\\ =562.50c{m}^2.\end{array}$