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

(a) 10 cm

Explanation:

Let ABCD be the rhombus.
∴ AB = BC = CD = DA
Here, AC and BD are the diagonals of ABCD, where AC = 16 cm and BD = 12 cm.
Let the diagonals intersect each other at O.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ ∆​AOB is a right angle triangle, in which OA = AC /2 = 16/2 = 8 cm and OB = BD/2 = 12/2 = 6 cm.

Now, $A B^{2}=O A^{2}+O B^{2}$      [Pythagoras theorem]

$\Rightarrow A B^{2}=(8)^{2}+(6)^{2}$

$\Rightarrow A B^{2}=64+36=100$

$\Rightarrow A B=10 \mathrm{~cm}$

Hence, the side of the rhombus is 10 cm.