The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals $12 \, cm$ and $16 \, cm$ is (in $cm^2$)

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 $\operatorname{ar}(\triangle BPQ) = \frac{1}{2} \operatorname{ar}(\triangle ABC)$.

Write True or False and justify your answer:
$PQRS$ is a rectangle inscribed in a quadrant of a circle of radius $13 \, cm$. $A$ is any point on $PQ$. If $PS = 5 \, cm$,then $\text{ar}(PAS) = 30 \, cm^2$.

Write True or False and justify your answer:
$ABCD$ is a parallelogram and $X$ is the mid-point of $AB$. If $\text{ar}(AXCD) = 24 \text{ cm}^2$,then $\text{ar}(ABC) = 24 \text{ cm}^2$.

In rhombus $ABCD$,$AC = 12 \, cm$ and $BD = 15 \, cm$,then $\operatorname{ar}(ABCD) = \dots \, cm^2$.

$A$ point $E$ is taken on the side $BC$ of a parallelogram $ABCD$. $AE$ and $DC$ are produced to meet at $F$. Prove that $\operatorname{ar}(\triangle ADF) = \operatorname{ar}(ABFC)$.