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 $\triangle ABC,$ if $L$ and $M$ are the points on $AB$ and $AC,$ respectively such that $LM \parallel BC.$ Prove that $\operatorname{ar}(\triangle LOB) = \operatorname{ar}(\triangle MOC).$

$PQRS$ is a square. $T$ and $U$ are the mid-points of $PS$ and $QR$ respectively. Find the area of $\Delta OTS$,if $PQ = 8 \, cm$,where $O$ is the point of intersection of $TU$ and $QS$.

The area of the parallelogram $ABCD$ is $90 \, cm^{2}$ (see figure). Find:
$(i) \; ar(ABEF)$
$(ii) \; ar(ABD)$
$(iii) \; ar(BEF)$

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

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