In rhombus $ABCD$,$AC = 12 \, cm$ and $BD = 15 \, cm$,then $\operatorname{ar}(ABCD) = \dots \, cm^2$.

Observe the given figure. Does the parallelogram $ABCD$ and the triangle $QBC$ lie on the same base and between the same parallels? If yes,write the common base and the two parallels.

In the figure,$BD \parallel CA$,$E$ is the mid-point of $CA$,and $BD = \frac{1}{2} CA$. Prove that $\operatorname{ar}(ABC) = 2 \operatorname{ar}(DBC)$.

Write True or False and justify your answer:
$PQRS$ is a rectangle inscribed in a quadrant of a circle of radius $13 \, cm$. $A$ is any point on $PQ$. If $PS = 5 \, cm$,then $\text{ar}(PAS) = 30 \, cm^2$.

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

State whether each of the following statements is true or false:
$(1)$ If $ar(ABC) = 96 \, cm^2$ for the parallelogram $ABCD$,then $ar(ABCD) = 192 \, cm^2$.
$(2)$ Area of a right triangle = Product of the sides forming the right angle.