Solution:

(b) 10 cm

Explanation:

We all know that,

A rhombus is a simple quadrilateral with four equal-length sides and diagonals that are perpendicular bisector of each other.

Now according to the question, we get,

$BD\text{ }=\text{ }12\text{ }cm$ and $AC\text{ }=\text{ }16\text{ }cm$

$\angle AOB\text{ }=\text{ }90{}^\circ $

Since, AC and BD bisects each other

BO = ½ BD and AO = ½ AC

Then, we obtain,

BO = 6 cm and AO = 8 cm

In right angled ∆AOB,

Now, using the Pythagoras theorem,

We get,

AB² = AO² + OB²

$A{{B}^{2}}~=\text{ }{{8}^{2}}~+\text{ }{{6}^{2}}~=\text{ }64\text{ }+\text{ }36\text{ }=\text{ }100$

Therefore, $AB\text{ }=~\surd 100\text{ }=\text{ }10\text{ }cm$

We know that the four sides of a rhombus are equal.

As a result, we get, one side of rhombus = 10 cm.