$\text{ In }\mathrm{\Delta ABC}\text{, if }\mathrm{AB}=\mathrm{BC}=\mathrm{CA}=2$ units, then the length of the median of $∆$ABC is

Since it is an equilateral ${\mathrm{\Delta }}^{\mathrm{le}}$

AD is median as well as altitude
$\therefore \mathrm{AD}\perp \mathrm{BC},\mathrm{BD}=\frac{1}{2}\mathrm{BC}$
BD = 1 unit
By Pythagoras theorem
$\begin{array}{l}{\mathrm{AD}}^2+{\mathrm{BD}}^2={\mathrm{AB}}^2\Rightarrow {\mathrm{AD}}^2+1^2=2^2\\ {\mathrm{AD}}^2=4-1\Rightarrow \mathrm{AD}=\sqrt{3}\text{ units }\end{array}$