The altitude of the equilateral triangle of side a units is :

1. $\dfrac{\text{a} \sqrt{3}}{2}$ units

2. $\dfrac{\sqrt{3 \text{a}}}{2}$ units

3. $\dfrac{\sqrt{2 \text{a}}}{2}$ units

4. $\dfrac{3 \text{a}}{2}$ units

In △ ABC,

AB = BC = AC = a units

Draw altitude AD perpendicular to BC.

In an equilateral triangle, the altitude also acts as the median (bisecting the base).

∴ BD = $\dfrac{1}{2} \times BC = \dfrac{1}{2} \times a = \dfrac{\text{a}}{2}$

In right angled △ABD,

Using Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

⇒ AB² = AD² + BD²

$\Rightarrow \text{a}^2 = \text{AD}^2 + \Big(\dfrac{\text{a}}{2}\Big)^2 \\[1em] \Rightarrow \text{AD}^2 = \text{a}^2 - \dfrac{\text{a}^2}{4} \\[1em] \Rightarrow \text{AD}^2 = \dfrac{4\text{a}^2 - \text{a}^2}{4} \\[1em] \Rightarrow \text{AD}^2 = \dfrac{3\text{a}^2}{4} \\[1em] \Rightarrow \text{AD} = \sqrt{\dfrac{3\text{a}^2}{4}} \\[1em] \Rightarrow \text{AD} = \dfrac{\text{a} \sqrt{3}}{2} \text{ units.}$

Hence, option 1 is the correct option.