Question

From a point in the interior of an equilateral triangle, perpendiculars are drawn to its sides. The lengths of perpendiculars are 14cm, 10cm and 6cm. Find the area of the triangle.

Answer 1

Let DEF be equilateral triangle
Let eah side be 'a' cm.
$Areaof\Delta DEF=Areaof\Delta DOA+Areaof\Delta DOF+Areaof\Delta EOF\frac{\sqrt{3}}{4}{a}^2=\frac{1}{2}\times a\times 14+\frac{1}{2}\times a\times 10+\frac{1}{2}\times a\times 6\Rightarrow \frac{\sqrt{3}}{4}{a}^2=7a+5a+3a=15a\Rightarrow a(\frac{\sqrt{3}}{4}a-15)=0\Rightarrow a=\frac{60}{\sqrt{3}}=20\sqrt{3}cm\therefore AreaoftriangleDEF=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}\times (20\sqrt{3}{)}^2=\frac{\sqrt{3}}{4}\times 20\times \sqrt{3}\times \sqrt{3}=300\sqrt{3}c{m}^2=300\sqrt{3}c{m}^2$.