In $\Delta ABC$,$AD$ is a median,$M$ and $N$ are the midpoints of $BD$ and $MD$ respectively. If $\operatorname{ar}(AND) = 20\, cm^2$,then $\operatorname{ar}(ABC) = \dots cm^2$.

$ABCD$ is a trapezium in which $AB \parallel DC$,$DC = 30 \, cm$ and $AB = 50 \, cm$. If $X$ and $Y$ are,respectively,the mid-points of $AD$ and $BC$,prove that $\operatorname{ar}(DCYX) = \frac{7}{9} \operatorname{ar}(XYBA)$.

$(1)$ If a planar region formed by a figure $T$ is made up of two non-overlapping planar regions formed by figures $P$ and $Q$,then $\operatorname{ar}(T) = \dots$
$(2)$ Area of a parallelogram $= \dots$

In trapezium $ABCD$,$AB \parallel DC$ and $L$ is the mid-point of $BC$. Through $L$,a line $PQ \parallel AD$ has been drawn which meets $AB$ in $P$ and $DC$ produced in $Q$. Prove that $\operatorname{ar}(ABCD) = \operatorname{ar}(APQD)$.

The area of the parallelogram $ABCD$ is $90 \, cm^{2}$ (see figure). Find:
$(i) \; ar(ABEF)$
$(ii) \; ar(ABD)$
$(iii) \; ar(BEF)$

Write True or False and justify your answer:
In the figure,$ABCD$ and $EFGD$ are two parallelograms and $G$ is the mid-point of $CD.$ Then $\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(EFGD).$