Question

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

Answer 1

Proof- Since all the $3$ sides of an equilateral triangle are same $AB=BC=CA=a$

For altitude of $△ABC$

Draw a $\perp$ from point $C$ on $AB$.

So that $CO\perp AB$. Let $OC=h$

In $△AOC$

By using Pythagoras Theorem

$A{C}^2=A{O}^2+O{C}^2$

$O{C}^2=A{C}^2-A{O}^2$

$h^2=a^2-{\left(\frac{a}{2}\right)}^2$

$h^2=a^2-\frac{a^2}{4}$

$h^2=\frac{3a^2}{4}\Rightarrow h=\frac{\sqrt{3}}{2}a$

Now Area of $△ABC=\frac{1}{2}\times Base\times Height$

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

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

So Area of equilateral $△$ comes is $\frac{\sqrt{3}}{4}{a}^2$