The lengths of the diagonals of a rhombus are 16 cm and 12 cm respectively.

Let $A B C D$ be a rhombus.

$L$ et $A C$ and $B D$ be the diagonals of the rhombus intersecting at a point $O$.

Let $A C=16 \mathrm{~cm}$

$\mathrm{BD}=12 \mathrm{~cm}$

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

$\therefore A O=\frac{1}{2} A C$

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

$\quad=8 \mathrm{~cm}$

$B O=\frac{1}{2} B D$

$\quad=\left(\frac{1}{2} \times 12\right) \mathrm{cm}$

$\quad=6 \mathrm{~cm}$ From the right $\Delta A O B:$

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

$=\left\{(8)^{2}+(6)^{2}\right\} \mathrm{cm}^{2}$

$=(64+36) \mathrm{cm}^{2}$

$=100 \mathrm{~cm}^{2}$

$\Rightarrow A B=\sqrt{100} \mathrm{~cm}$

$=10 \mathrm{~cm}$

Hence, the length of the side $A B$ is $10 \mathrm{~cm}$.

$A B=B C=C D=D A=10 \mathrm{~cm} \quad$ (all sides of a rhombus are equal)