Question

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

Answer 1

Draw $AD\perp BC$

$\Rightarrow △ABD\cong △ACD$ by R.H.S

$\therefore BD=DC$ by CPCT

$\because BC=a$

$\therefore BD=DC=\frac{a}{2}$

In right angled triangle $△ABD,{AD}^2={AB}^2-{BD}^2$

${AD}^2=a^2-{\left(\frac{a}{2}\right)}^2=a^2-\frac{a^2}{4}=\frac{3a^2}{4}$

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

Area of $△ABC=\frac{1}{2}\times BC\times AD=\frac{1}{2}\times a\times \frac{a\sqrt{3}}{2}=\frac{a^2\sqrt{3}}{4}$