Question

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

Answer 1

Given :

An equlateral triangle ABC such that AB=BC=CA=a.

To prove :

$ar\left(△ABC\right)=\frac{\sqrt{3}}{4}{a}^2$

Construction :

Draw $AD\perp BC$.

Proof :

In $△sABD$ and $ACD$, we have

$AB=AC$

$\angle ADB=\angle ADC={90}^o$

$AD=AD$

So, by RHS criterion of congruence,

$△ABD\cong △ACD$

$⟹BD=DC$

But, $BD+DC=a$

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

Now, in right triangle ABD, we have

$A{B}^2=A{D}^2+B{D}^2$

$a^2=A{D}^2+(\frac{a}{2}{)}^2$

$A{D}^2=a^2-\frac{a^2}{4}=\frac{3a^2}{4}$

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

Therefore,

$ar\left(△ABC\right)=\frac{1}{2}(BC\times AD)=\frac{1}{2}(a\times \frac{\sqrt{3}}{2}a)=\frac{\sqrt{3}}{4}{a}^2$