An equilateral triangle having length of its side as 5 cm. Find its height.

At point D, the equilateral triangle ABC's height divides the triangle's base into two identical halves.

Therefore,

$BD=DC=2.5 \mathrm{cm}$

In triangle $\mathrm{ABD}$, using Pythagoras theorem,

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

$5^2=A{D}^2+2.5^2$

$A{D}^2=25-6.25$

$A{D}^2=18.75$

$AD=4.33 \mathrm{cm}$