Question

From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are $14\mathrm{cm}$, $10\mathrm{cm}$ and $6\mathrm{cm}$. Find the area of the triangle.

Answer 1

Step 1: Solve for the side of the equilateral triangle

Let the point in the interior of the triangle be $O$.

Given,

1. $△ABC$ is equilateral i.e. $AB=BC=CA=a\mathrm{cm}$
2. The lengths of perpendiculars are:

$$\begin{gathered} OQ=14\mathrm{cm} \\ OR=10\mathrm{cm} \\ OP=6\mathrm{cm} \end{gathered}$$

We know that, Area of a triangle $=\frac{1}{2}\times$ base $\times$height

Aan nd, area of equilateral triangle$=\frac{\sqrt{3}}{4}{a}^2$

From the figure,

Area of $△ABC=$Area of$△AOB+$ Area of $△BOC+$Area of $△COA$

$$\begin{gathered} \Rightarrow \frac{\sqrt{3}}{4}{a}^2=\frac{1}{2}\times a\times 6+\frac{1}{2}\times a\times 14+\frac{1}{2}\times a\times 10 \\ \Rightarrow \sqrt{3}{a}^2=60a \\ \Rightarrow a=\frac{60}{\sqrt{3}} \end{gathered}$$

Step 2: Solve for area of the triangle

Area of $△ABC=\frac{\sqrt{3}}{4}{a}^2$

$$\begin{gathered} =\frac{\sqrt{3}}{4}\times \frac{60}{\sqrt{3}}\times \frac{60}{\sqrt{3}} \\ =300\sqrt{3}{\mathrm{cm}}^2 \end{gathered}$$

Hence, the area of the given equilateral triangle is $300\sqrt{3}{\mathrm{cm}}^2$