Question

The height of an equilateral triangle is $\sqrt{3}$ a units. Then its area is _________.

Answer 1

Given: The height of an equilateral triangle is $\sqrt{3}$a units.

Let ABC be an equilateral triangle of side 'x' units and AD be the altitude of the triangle of length $\sqrt{3}$a units.

In ∆ABD,
AB = x units
BD = $\frac{x}{2}$ units

Using pythagoras theorem,
$$\begin{gathered} A{B}^2=B{D}^2+A{D}^2 \\ \Rightarrow x^2={\left(\frac{x}{2}\right)}^2+{\left(\sqrt{3}a\right)}^2 \\ \Rightarrow 3a^2=x^2-\frac{x^2}{4} \\ \Rightarrow 3a^2=\frac{4x^2-x^2}{4} \\ \Rightarrow 3a^2=\frac{3x^2}{4} \\ \Rightarrow a^2=\frac{x^2}{4} \\ \Rightarrow a=\frac{x}{2} \\ \Rightarrow x=2a...\left(1\right) \end{gathered}$$

Now,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^2 \\ =\frac{\sqrt{3}}{4}\times {\left(2a\right)}^2 \\ =\frac{\sqrt{3}}{4}\times 4a^2 \\ =\sqrt{3}{a}^2 \end{gathered}$$


Hence, its area is $\underline{\sqrt{3}{a}^2\mathrm{sq}.\mathrm{units}}$.