The perimeter of square $ABCD$ is $16 \, cm$,then $ar(ABCD) = \ldots \ldots \ldots \, cm^2$.

Prove that the area of a rhombus is half the product of its diagonals.

Write True or False and justify your answer:
In the figure,$ABCD$ and $EFGD$ are two parallelograms and $G$ is the mid-point of $CD.$ Then $\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(EFGD).$

In $\triangle ABC$,$D$ is the mid-point of $AB$ and $P$ is any point on $BC$. If $CQ \parallel PD$ meets $AB$ in $Q$,then prove that $\operatorname{ar}(\triangle BPQ) = \frac{1}{2} \operatorname{ar}(\triangle ABC)$.

$ABCD$ is a rhombus. If $AC = 16 \, cm$ and $BD = 30 \, cm$,then find the area of $ABCD$ in $cm^2$.

In which of the following figures,do you find two polygons on the same base and between the same parallels?