Question

A point O is taken inside an equilateral $\Delta ABC$. If $OM\perp AC$, $OL\perp BC$ and $ON\perp AB$ such that OL = 14 cm, OM = 10 cm and ON = 6 cm, find the area of $\Delta ABC$.

Answer 1

Let each side of $\Delta ABC$ be 'a' cm.

Then $\text{Area}\left(\Delta ABC\right)=\text{Area}\left(\Delta OBC\right)+\text{Area}\left(\Delta AOC\right)+\text{Area}\left(\Delta AOB\right)$

We know area of a triangle with base 'b' and height 'h' id given by '$\frac{1}{2}\times \text{base}\times \text{height}$'
So, $\text{Area}\left(\Delta ABC\right)=\frac{1}{2}a(OL+OM+ON)$

$=\frac{1}{2}a(14+10+6)$

$=15a$
Given, $\Delta ABC$ is equailateral.
Area of an equilateral triangle with side 'a' is given by $\frac{\sqrt{3}}{4}{a}^2$.

$\Rightarrow \frac{\sqrt{3}}{4}{a}^2=15a$

$\Rightarrow a=15\times \frac{4}{\sqrt{3}}$

$\Rightarrow a=60\times \frac{\sqrt{3}}{3}=20\sqrt{3}$

$\therefore Area\left(\Delta ABC\right)=15\times 20\sqrt{3}=300\sqrt{3}c{m}^2$