$A$ point $P$ lies on the side $CD$ of parallelogram $ABCD$. If $ar(ABCD) = 56 \, cm^2$,then $ar(PAB) = \dots \dots \dots cm^2$.

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

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

The median of a triangle divides it into two:

In $\Delta ABC$,$\angle B = 90^{\circ}$,$AB = 8\, \text{cm}$ and $BC = 15\, \text{cm}$,then $\text{ar}(\Delta ABC) = \dots \text{cm}^2$.