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 $\operatorname{ar}(ABC) = \dots cm^2$.

In parallelogram $XYZW$,$XY = 24 \, cm$. Altitudes $WP$ and $WQ$ are corresponding to bases $XY$ and $YZ$ respectively. If $WP = 6 \, cm$ and $WQ = 8 \, cm$,then find $YZ$ and the perimeter of parallelogram $XYZW$.

In parallelogram $ABCD$,$AB = 20 \, cm$. Altitudes $AY$ and $DX$ are corresponding to bases $BC$ and $AB$ respectively. If $DX = 12 \, cm$ and $AY = 15 \, cm$,then find $BC$ and the perimeter of $ABCD$.

In $\Delta ABC$,$\angle B = 90^{\circ}$ and $BM$ is an altitude to the hypotenuse $AC$. If $AB = 12 \, cm$ and $BC = 16 \, cm$,then find the length of $BM$ in $cm$.

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 $\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(EFGD).$

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 $\operatorname{ar}(ABCD) = \operatorname{ar}(APQD)$.