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,

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

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

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

$\Rightarrow A{D}^2=18.75$

$\Rightarrow \sqrt{A{D}^2} = \sqrt{18.75}$

We know that ,

 $\sqrt{16} < \sqrt{18.75} <\sqrt{25}$

$\Rightarrow 4 < \sqrt{18.75} < 5$

Only one of the given options matches our condition,

$\Rightarrow AD=4.33 \mathrm{cm}$