In square $ABCD$,$AC = 16 \text{ cm}$,then $\operatorname{ar}(ABCD) = \dots \text{ cm}^2$.

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

If a triangle and a parallelogram are on the same base and between the same parallels,then the ratio of the area of the triangle to the area of the parallelogram is:

In $\Delta ABC$,$AD$ is a median and $AM$ is an altitude. The side $BA$ of $\Delta ABC$ is produced to any point $E$,so that $AB = AE$. If $BC = 16\, cm$ and $AM = 8\, cm$,then find the area of $\Delta EBD$ in $cm^2$.

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,$ar(ADO) = ar(CDO)$.