In $\Delta ABC$,points $P$ and $Q$ are the points of trisection of $BC$. Then,$\operatorname{ar}(\Delta APQ) : \operatorname{ar}(\Delta ABC) = \dots$

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

$ABCD$ is a square. If $AC = 16 \, cm$,then find the area of $ABCD$ in $cm^2$.

$ABCD$ is a parallelogram in which $BC$ is produced to $E$ such that $CE = BC$ $(Fig.)$. $AE$ intersects $CD$ at $F.$ If $\text{ar}(\Delta DFB) = 3 \, \text{cm}^2$,find the area of the parallelogram $ABCD$ (in $\text{cm}^2$).

In quadrilateral $ABCD$,$AM$ and $CN$ are altitudes on diagonal $BD$ drawn from $A$ and $C$ respectively. Prove that,$\operatorname{ar}(ABCD) = \frac{1}{2} \times BD \times (AM + CN)$.

In the figure,$l, m,$ and $n$ are straight lines such that $l \parallel m$ and $n$ intersects $l$ at $P$ and $m$ at $Q$. $ABCD$ is a quadrilateral such that its vertex $A$ is on $l$. The vertices $C$ and $D$ are on $m$ and $AD \parallel n$. Show that $\operatorname{ar}(ABCQ) = \operatorname{ar}(ABCDP).$