In square $ABCD$,$AC = 16 \text{ cm}$,then $\operatorname{ar}(ABCD) = \dots \text{ cm}^2$.

In the figure,$ABCD$ and $AEFD$ are two parallelograms. Prove that $\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.

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

$D, E$ and $F$ are the midpoints of the sides $BC, CA$ and $AB$ respectively of $\Delta ABC$. Show that:
$(i)$ $BDEF$ is a parallelogram.
$(ii)$ $ar(DEF) = \frac{1}{4} ar(ABC)$
$(iii)$ $ar(BDEF) = \frac{1}{2} ar(ABC)$

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

In parallelogram $ABCD$,$DM$ is an altitude corresponding to base $AB$. If $AB = 15 \text{ cm}$ and $\text{ar}(ABCD) = 360 \text{ cm}^2$,then $DM = \ldots \text{ cm}$.