Write True or False and justify your answer.
If the side of a rhombus is $10 \, cm$ and one diagonal is $16 \, cm$,the area of the rhombus is $96 \, cm^2$.

(A) Let $ABCD$ be the rhombus whose one diagonal $AC$ is $16 \, cm$. Each side of the rhombus is $10 \, cm$.
We know that the diagonals of a rhombus bisect each other at right angles,so
$OA = OC = 8 \, cm$ and $OB = OD$.
In $\Delta AOB$,we have $\angle AOB = 90^{\circ}$.
Therefore,$AB^2 = OA^2 + OB^2$
$\Rightarrow OB^2 = AB^2 - OA^2$
$= (10)^2 - 8^2 = 100 - 64 = 36$
$OB = \sqrt{36} = 6 \, cm$.
$DB = 2(OB) = 2 \times 6 = 12 \, cm$.
Hence,the area of the rhombus $= \frac{1}{2} \times \text{Product of diagonals}$
$= \frac{1}{2} \times 16 \times 12 = 96 \, cm^2$.
Hence,the given statement is true.