$ABCD$ is a parallelogram in which $BC$ is produced to $E$ such that $CE = BC$ $(Fig.)$. $AE$ intersects $CD$ at $F.$ If $\text{ar}(\Delta DFB) = 3 \, \text{cm}^2$,find the area of the parallelogram $ABCD$ (in $\text{cm}^2$).

In which of the following figures,do you find two polygons on the same base and between the same parallels?

In the figure,$X$ and $Y$ are the mid-points of $AC$ and $AB$ respectively,$QP \parallel BC$ and $CYQ$ and $BXP$ are straight lines. Prove that $\text{ar}(ABP) = \text{ar}(ACQ).$

In the figure,$l, m,$ and $n$ are straight lines such that $l \parallel m$ and $n$ intersects $l$ at $P$ and $m$ at $Q$. $ABCD$ is a quadrilateral such that its vertex $A$ is on $l$. The vertices $C$ and $D$ are on $m$ and $AD \parallel n$. Show that $\operatorname{ar}(ABCQ) = \operatorname{ar}(ABCDP).$

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 $\triangle ASR = 90 \, cm^{2}$.

If $E, F, G$ and $H$ are respectively the midpoints of the sides of a parallelogram $ABCD$,show that $ar(EFGH) = \frac{1}{2} ar(ABCD)$.