Question

The perpendicular drawn from the one vertex to opposite side of an equilateral triangle is $12cm$. Find the perimeter and area of this triangle.

Answer 1

Here $AD=12$cm

$\therefore$ In $\Delta ABD$

$\angle B={60}^o$

$\angle AOD={90}^o$

$\angle BAD={30}^o$

$\therefore \sin {60}^o=\frac{AD}{AB}$

$\therefore \frac{\sqrt{3}}{2}=\frac{12}{AB}$

$\therefore AB=\frac{24}{\sqrt{3}}$

$\therefore AB=8\sqrt{3}$cm

$\therefore$ Perimeter of triangle$=8\sqrt{3}\times 3$

$=24\sqrt{3}$cm

$\therefore$ Area of triangle $=\frac{\sqrt{3}}{4}\times (side{)}^{2}c{m}^2$

$\therefore$ Area of triangle$=\frac{\sqrt{3}}{4}\times (8\sqrt{3}{)}^2$

$\therefore$ Area of triangle$=\frac{\sqrt{3}}{4}\times 64\times 3$

$\therefore$ Area $=46\sqrt{3}c{m}^2$.