Find the altitude of an equilateral triangle of side 5 $\sqrt{3}$ cm.

In △ ABC,

Draw altitude AD perpendicular to BC.

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

∴ BD = $\dfrac{1}{2}$ × BC = $\dfrac{1}{2} \times 5\sqrt{3} = \dfrac{5\sqrt{3}}{2}$

In right angled △ABD,

Using Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

⇒ AB² = AD² + BD²

$\Rightarrow (5\sqrt{3})^2 = \text{AD}^2 + \Big(\dfrac{5\sqrt{3}}{2}\Big)^2 \\[1em] \Rightarrow 25 \times 3 = \text{AD}^2 + \dfrac{25 \times 3}{4} \\[1em] \Rightarrow 75 = \text{AD}^2 + \dfrac{75}{4} \\[1em] \Rightarrow \text{AD}^2 = 75 - \dfrac{75}{4} \\[1em] \Rightarrow \text{AD}^2 = \dfrac{300 - 75}{4} \\[1em] \Rightarrow \text{AD}^2 = \dfrac{225}{4} \\[1em] \Rightarrow \text{AD} = \sqrt{\dfrac{225}{4}} \\[1em] \Rightarrow \text{AD} = \dfrac{15}{2} \\[1em] \Rightarrow \text{AD} = 7.5 \text{ cm}.$

Hence, the altitude is 7.5 cm.