$AC$ is one of the diagonals of quadrilateral $ABCD$. $BM$ and $DN$ are altitudes on $AC$ from $B$ and $D$ respectively. If $AC = 18 \, cm$,$BM = 10 \, cm$,and $DN = 6 \, cm$,then $ar(ABCD) = \dots \dots \, cm^2$.

$O$ is any point on the diagonal $PR$ of a parallelogram $PQRS$. Prove that $\operatorname{ar}(\triangle PSO) = \operatorname{ar}(\triangle PQO)$.

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 $\operatorname{ar}(ABCQ) = \operatorname{ar}(ABCDP).$

Prove that the line segment joining the midpoints of two opposite sides of a parallelogram divides the parallelogram into two parallelograms with equal area.

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 parallelogram $PQRS$,$PQ = 15 \, cm$. Altitudes $SM$ and $SN$ are corresponding to bases $PQ$ and $QR$ respectively. If $SM = 6 \, cm$ and $SN = 10 \, cm$,find $QR$ and the perimeter of $PQRS$.