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.