If $P$ is any point on the median $AD$ of a $\triangle ABC$,then $\operatorname{ar}(ABP) = \operatorname{ar}(ACP)$. State whether this statement is True or False.

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

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 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 $cm$.

In $\Delta PQR$,$PM$ is a median and $N$ is the midpoint of $PM$. If $\text{ar}(PQN) = 36 \text{ cm}^2$,then $\text{ar}(PQR) = \dots \text{ cm}^2$.

In parallelogram $XYZW$,$XY = 24 \, cm$. Altitudes $WP$ and $WQ$ are corresponding to bases $XY$ and $YZ$ respectively. If $WP = 6 \, cm$ and $WQ = 8 \, cm$,then find $YZ$ and the perimeter of parallelogram $XYZW$.