In an equilateral triangle with side a, prove that area


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


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: 

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

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

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

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


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.


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