ABCD is a rhombus whose diagonals intersect at O.

We know that the diagonals of a rhombus bisect each other at right angles.

$\therefore \mathrm{BO}=\frac{1}{2} \mathrm{BD}=\left(\frac{1}{2} \times 16\right) \mathrm{cm}$

$=8 \mathrm{~cm}$

$\mathrm{AB}=10 \mathrm{~cm}$ and $\angle \mathrm{AOB}=90^{\circ}$

From right $\Delta \mathrm{OAB}:$

$\mathrm{AB}^{2}=\mathrm{AO}^{2}+\mathrm{BO}^{2}$

$\Rightarrow \mathrm{AO}^{2}=\left(\mathrm{AB}^{2}-\mathrm{BO}^{2}\right)$

$\Rightarrow \mathrm{AO}^{2}=(10)^{2}-(8)^{2} \mathrm{~cm}^{2}$

$\Rightarrow \mathrm{AO}^{2}=(100-64) \mathrm{cm}^{2}=36 \mathrm{~cm}^{2}$

$\Rightarrow \mathrm{AO}=\sqrt{36} \mathrm{~cm}=6 \mathrm{~cm}$

$\therefore \mathrm{AC}=2 \times \mathrm{AO}=(2 \times 6) \mathrm{cm}=12 \mathrm{~cm}$