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

The perimeter of square $ABCD$ is $16 \, cm$,then $ar(ABCD) = \ldots \ldots \ldots \, cm^2$.

The diagonals of a parallelogram $ABCD$ intersect at a point $O$. Through $O$,a line is drawn to intersect $AD$ at $P$ and $BC$ at $Q$. Show that $PQ$ divides the parallelogram into two parts of equal area.

$ABCD$ is a quadrilateral whose diagonal $AC$ divides it into two parts of equal area. Then $ABCD$:

Write True or False and justify your answer:
$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. Then $ar(\triangle BDE) = \frac{1}{4} ar(\triangle ABC).$

In the given figure,$P$ is a point in the interior of parallelogram $ABCD$. Show that,
$(1) \operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(2) \operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$