$ABCD$ is a rectangle. If $AB = 12 \, cm$ and $BC = 7 \, cm$,then find the area of $ABCD$ in $cm^2$.

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

In $\Delta ABC$,$AD$ is a median. If $\operatorname{ar}(ADB) = 53 \, cm^2$,then find $\operatorname{ar}(ABC)$ in $cm^2$.

In $\Delta PQR$,$M$ and $N$ are the midpoints of $PQ$ and $PR$ respectively. $X$ is any point on $QR$. Prove that,$ar(MXN) = \frac{1}{4} ar(PQR)$.

In $\Delta ABC$,point $D$ lies on side $BC$. $E$ is the midpoint of $AD$. Prove that,$ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.