In parallelogram $ABCD$,$AB = 12 \, cm$. Altitudes $DM$ and $DN$ correspond to bases $AB$ and $BC$ respectively. If $DM = 5 \, cm$ and $DN = 6 \, cm$,then find the length of $BC$ in $cm$.

The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals $12 \, cm$ and $16 \, cm$ is (in $cm^2$)

The median of a triangle divides it into two:

In $\Delta ABC$,$AD$ is a median and $AM$ is an altitude. The side $BA$ of $\Delta ABC$ is produced to any point $E$,so that $AB = AE$. If $BC = 16\, cm$ and $AM = 8\, cm$,then find the area of $\Delta EBD$ in $cm^2$.

If the medians of a $\Delta ABC$ intersect at $G$,show that $\operatorname{ar}(AGB) = \operatorname{ar}(AGC) = \operatorname{ar}(BGC) = \frac{1}{3} \operatorname{ar}(ABC)$.

$D, E$ and $F$ are the midpoints of the sides $BC, CA$ and $AB$ respectively of $\Delta ABC$. Show that:
$(i)$ $BDEF$ is a parallelogram.
$(ii)$ $ar(DEF) = \frac{1}{4} ar(ABC)$
$(iii)$ $ar(BDEF) = \frac{1}{2} ar(ABC)$