In $\Delta PQR$,$PM$ is a median and $N$ is the midpoint of $PM$. If $\text{ar}(PQN) = 36 \text{ cm}^2$,then $\text{ar}(PQR) = \dots \text{ cm}^2$.

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 $\operatorname{ar}(ABCD) = \operatorname{ar}(APQD)$.

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 parallelogram $ABCD$,diagonals $AC$ and $BD$ intersect at point $O$. Point $P$ lies on line segment $BO$. Prove that,$ar(ABP) = ar(CBP)$.

In the figure,$CD \parallel AE$ and $CY \parallel BA$. Prove that $\operatorname{ar}(\triangle CBX) = \operatorname{ar}(\triangle AXY)$.

Which of the following figures lie on the same base and between the same parallels? Write the common base and the two parallels for the figure for which the answer is affirmative.