If $ar(PQRS) = 80 \, cm^2$ for a parallelogram $PQRS$,then $ar(PSR) = \dots \dots \dots cm^2$.

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

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 $\text{ar}(PAS) = 30 \, cm^2$.

In trapezium $ABCD$,$AB || CD$ and diagonals $AC$ and $BD$ intersect at point $O$. Prove that $ar(AOD) = ar(BOC)$.

In $\Delta PQR$,$M$ and $N$ are the midpoints of $PQ$ and $PR$ respectively. $X$ is any point on $QR$. Prove that,$ar(MXN) = \frac{1}{4} ar(PQR)$.

In $\Delta ABC$,$AD$ is a median. $P$ and $Q$ are the midpoints of $AB$ and $AD$ respectively. If $\operatorname{ar}(\Delta ABC) = 72 \, \text{cm}^2$,then $\operatorname{ar}(\Delta APQ) = \dots \text{cm}^2$.