Question

Prove that the area of an equilateral triangle of side $a$ is $\frac{\sqrt{3}}{4}{a}^2.$

Answer 1

$ABC$ is equilateral triangle:-

In an equilateral triangle median =altitude

$\Rightarrow AD\perp BC$ and $BD=DC=\frac{BC}{2}=\frac{a}{2}$

Consider triangle $ABD$

$\sin \angle ABD=\frac{AD}{AB}$

$\frac{\sqrt{3}}{2}=\frac{AD}{a}$

$\Rightarrow AD=\frac{\sqrt{3}a}{2}$

area of triangle $ABC\Rightarrow 1/2\times box\times height$

$\Rightarrow 1/2\times \left(BC\right)\times \left(AD\right)$

$\Rightarrow \frac{1}{2}\left(a\right)\left(\frac{\sqrt{3}}{2}a\right)$

area of triangle $ABC\Rightarrow \frac{\sqrt{3}}{2}{a}^2$

$\therefore$ are of equilateral triangle is $\frac{\sqrt{3}{a}^2}{4}$