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

In the given figure,$P$ is a point in the interior of parallelogram $ABCD$. Show that,
$(1) \operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(2) \operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$

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

Which of the following figures lie on the same base and between the same parallels? Write the common base and the two parallels for the figure for which the answer is affirmative.

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

In the figure,$X$ and $Y$ are the mid-points of $AC$ and $AB$ respectively,$QP \parallel BC$ and $CYQ$ and $BXP$ are straight lines. Prove that $\text{ar}(ABP) = \text{ar}(ACQ).$