In $\Delta ABC$,$\angle B = 90^{\circ}$,$AB = 8\, \text{cm}$ and $BC = 15\, \text{cm}$,then $\text{ar}(\Delta ABC) = \dots \text{cm}^2$.

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

The medians $BE$ and $CF$ of a triangle $ABC$ intersect at $G$. Prove that the area of $\triangle GBC = \text{area of the quadrilateral } AFGE.$

If a triangle and a parallelogram are on the same base and between the same parallels,then the ratio of the area of the triangle to the area of the parallelogram is:

In square $ABCD$,$AC = 16 \text{ cm}$,then $\operatorname{ar}(ABCD) = \dots \text{ cm}^2$.

$ABCD$ is a parallelogram and $BC$ is produced to a point $Q$ such that $AD = CQ$. If $AQ$ intersects $DC$ at $P$,show that $\operatorname{ar}(BPC) = \operatorname{ar}(DPQ)$.