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 $\Delta ABC$,$\angle B = 90^{\circ}$,$AB = 14 \, cm$ and $AC = 50 \, cm$,then find the area of $\Delta ABC$ in $cm^2$.

$PQRS$ is a rectangle. If $PQ = 20 \, cm$ and $\operatorname{ar}(PQRS) = 300 \, cm^2$,then $SP = \dots \, cm$.

In the figure,$ABCDE$ is any pentagon. $BP$ is drawn parallel to $AC$ and meets $DC$ produced at $P$,and $EQ$ is drawn parallel to $AD$ and meets $CD$ produced at $Q$. Prove that $\operatorname{ar}(ABCDE) = \operatorname{ar}(APQ)$.

In trapezium $ABCD$,$AB \parallel DC$ and $L$ is the mid-point of $BC$. Through $L$,a line $PQ \parallel AD$ has been drawn which meets $AB$ in $P$ and $DC$ produced in $Q$. Prove that $\operatorname{ar}(ABCD) = \operatorname{ar}(APQD)$.