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

$ABCD$ is a quadrilateral whose diagonal $AC$ divides it into two parts of equal area. Then $ABCD$:

If in the figure,$PQRS$ and $EFRS$ are two parallelograms,then $\operatorname{ar}(MFR) = \frac{1}{2} \operatorname{ar}(PQRS)$. State whether the statement is True or False.

In $\Delta PQR$,$PM$ is a median and $N$ is the midpoint of $PM$. If $\text{ar}(PQN) = 36 \text{ cm}^2$,then $\text{ar}(PQR) = \dots \text{ cm}^2$.

In $\Delta ABC$,$AD$ is a median. If $ar(\Delta ABC) = 50 \, cm^2$,then $ar(\Delta ADC) = \dots \dots \dots cm^2$.