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(c) $10 \mathrm{~cm}$

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 $O$. $A C=16 \mathrm{~cm}$

$B D=12 \mathrm{~cm}$

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

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

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

$=8 \mathrm{~cm}$

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

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

$=6 \mathrm{~cm}$

From the right $\Delta A O B$ :

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

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

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

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

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

= 10cm

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

Therefore, the length of each side of the rhombus is $10 \mathrm{~cm}$ because all the sides of $a$ rhombus are equal.