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 $cm^2$)

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 parallelogram $ABCD$,$AB = 12 \, cm$. Altitudes $DM$ and $DN$ correspond to bases $AB$ and $BC$ respectively. If $DM = 5 \, cm$ and $DN = 6 \, cm$,then find the length of $BC$ in $cm$.

$ABCD$ is a square. $E$ and $F$ are respectively the midpoints of $BC$ and $CD$. If $R$ is the midpoint of $EF$,prove that $\operatorname{ar}(\triangle AER) = \operatorname{ar}(\triangle AFR)$.

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 $\operatorname{ar}(\Delta APQ) = \dots \text{cm}^2$.

In $\Delta ABC$,$AD$ is a median,$M$ and $N$ are the midpoints of $BD$ and $MD$ respectively. If $\operatorname{ar}(AND) = 20\, cm^2$,then $\operatorname{ar}(ABC) = \dots cm^2$.