Question

The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.

Answer 1

Perimeter of a triangular field = 240 dm
Two sides are 78 dm and 50 dm
$\therefore$ Third side = 240 - (78 + 50)
= 240 - 128 = 112 dm
$\therefore s=\frac{Perimeter}{2}=\frac{240}{2}=120\therefore Area=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{120(120-78)(120-50)(120-112)}=\sqrt{120\times 42\times 70\times 8}$
= $\sqrt{\begin{array}{c}2\times 2\times 2\times 3\times 5\times 7\times 2\times 3\times 7\times 2\\ \times 5\times 2\times 2\times 2\end{array}}d{m}^2$
=$2\times 2\times 2\times 2\times 3\times 5\times 7d{m}^2=1680d{m}^2$
Now length of perpendicular on 50 m
= $\frac{Area\times 2}{Base}=\frac{1680\times 2}{50}=\frac{336}{5}=67.2dm$