Question

In an equilateral triangle with side a, prove that area $=\frac{\sqrt{3}}{4}{a}^2$.

Answer 1

An equilateral triangle is symmetrical so if you draw a perpendicular from point A to BC it bisects the side BC.

so CD=a/2

AD is the altitude(height) of the triangle ABC

In triangle ADC by pythagoras theorem

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

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

$=\frac{1}{2}\times a\times \frac{\sqrt{3}a}{2}=\frac{\sqrt{3}{a}^2}{4}$