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 $DM = \ldots \text{ cm}$.

In parallelogram $ABCD$,$P$ is the midpoint of $CD$. Then,$\operatorname{ar}(ABCD) : \operatorname{ar}(PBC) = \dots$

The diagonals of a parallelogram $ABCD$ intersect at a point $O$. Through $O$,a line is drawn to intersect $AD$ at $P$ and $BC$ at $Q$. Show that $PQ$ divides the parallelogram into two parts of equal area.

In parallelogram $ABCD$,diagonals $AC$ and $BD$ intersect at point $O$. Point $P$ lies on line segment $BO$. Prove that,$ar(ADO) = ar(CDO)$.

Write True or False and justify your answer:
$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC.$ Then $\operatorname{ar}(\triangle BDE) = \frac{1}{4} \operatorname{ar}(\triangle ABC).$

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