In an equilateral triangle with side a, prove that area

Let ABC be the equilateral triangle with each side equal to a.
Let AD be the altitude from A, meeting BC at D.
Therefore, D is the midpoint of BC.
Let AD be h.
Applying Pythagoras theorem in right-angled triangle ABD, we have: 

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

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

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

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

Therefore,

Area of triangle $\mathrm{ABC}=\frac{1}{2} \times$ base $\times$ height $=\frac{1}{2} \times a \times \frac{\sqrt{3}}{2} a$

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

This completes the proof.