Question

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

Answer 1

Given: Perimeter of triangular field $=240dm$ Two sides are $78dm$ and $50dm$

$\therefore$ Third side $=240-(78+50)=240-128$

$=112dm$

$s=\frac{\text{Perimeter}}{2}=\frac{240}{2}=120dm$

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

$=\sqrt{120(120-78)(120-50)(120-112)}$

$=\sqrt{120\times 42\times 70\times 8}$

$=\sqrt{2\times 2\times 2\times 3\times 5\times 7\times 2\times 3\times 7\times 2\times 5\times 2\times 2\times 2}$

$=2\times 2\times 2\times 2\times 3\times 5\times 7=1680d{m}^2$

Let ${}^{\prime }{x}^{\prime }dm$ be the length of perpendicular on $50dm$

We know that,

Area $=\frac{1}{2}\times \text{base}\times \text{height}$

$\Rightarrow \text{height}\left(x\right)=\frac{area\times 2}{base}=\frac{1680\times 2}{50}=\frac{336}{5}=67.2dm$

Hence, the required length is $67.2dm$.