Question

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

Answer 1

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:
$$\begin{gathered} A{B}^2=A{D}^2+B{D}^2 \\ \Rightarrow a^2=h^2+(\frac{a}{2}{)}^2 \\ \Rightarrow h^2=a^2-\frac{a^2}{4}=\frac{3}{4}{a}^2 \\ \Rightarrow h=\frac{\sqrt{3}}{2}a \end{gathered}$$

Therefore,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}=\frac{1}{2}\times a\times \frac{\sqrt{3}}{2}a \\ =\frac{\sqrt{3}}{4}{a}^2 \end{gathered}$$

This completes the proof.