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.

In parallelogram $ABCD$,$P$ is the midpoint of $CD$. Then,$\operatorname{ar}(ABCD) : \operatorname{ar}(PBC) = \dots$

Write True or False and justify your answer:
$ABCD$ is a parallelogram and $X$ is the mid-point of $AB$. If $\text{ar}(AXCD) = 24 \text{ cm}^2$,then $\text{ar}(ABC) = 24 \text{ cm}^2$.

$(1)$ The part of the plane enclosed by a simple closed figure is called a $\ldots \ldots \ldots$
$(2)$ $\ldots \ldots \ldots$ of the planar region corresponding to a closed figure is called its area.

In the figure,$ABCD$ is a parallelogram. Points $P$ and $Q$ on $BC$ trisect $BC$ into three equal parts. Prove that $\operatorname{ar}(APQ) = \operatorname{ar}(DPQ) = \frac{1}{6} \operatorname{ar}(ABCD)$.

$ABCD$ is a parallelogram and $BC$ is produced to a point $Q$ such that $AD = CQ$. If $AQ$ intersects $DC$ at $P$,show that $\operatorname{ar}(BPC) = \operatorname{ar}(DPQ)$.