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

$(1)$ If a planar region formed by a figure $T$ is made up of two non-overlapping planar regions formed by figures $P$ and $Q$,then $\operatorname{ar}(T) = \dots$
$(2)$ Area of a parallelogram $= \dots$

The median of a triangle divides it into two:

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 parallelogram whose area is $180 \, cm^{2}$ and $A$ is any point on the diagonal $QS$. The area of $\triangle ASR = 90 \, cm^{2}$.

In the given figure,$PQM$ is a line and $SQ || RM$. Prove that $ar(PQR) = ar(PMS)$.