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

Write True or False and justify your answer:
$ABCD$ is a parallelogram and $X$ is the mid-point of $AB$. If $\text{ar}(AXCD) = 24 \text{ cm}^2$,then $\text{ar}(ABC) = 24 \text{ cm}^2$.

In $\Delta ABC,$ $P$ and $Q$ are the points of trisection of $BC$ (i.e.,points dividing $BC$ into three equal parts). Prove that,$\operatorname{ar}(ABP) = \operatorname{ar}(APQ) = \operatorname{ar}(AQC) = \frac{1}{3} \operatorname{ar}(ABC).$

$XYZW$ is a square. If $XY = 17 \text{ cm}$,then find the area of $XYZW$ in $\text{cm}^2$.

In the figure,the area of parallelogram $ABCD$ is:

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