The perimeter of square $ABCD$ is $16 \, cm$,then $ar(ABCD) = \ldots \ldots \ldots \, cm^2$.

In $\Delta ABC$,$\angle B = 90^{\circ}$,$BC = 8 \, \text{cm}$,and $AC = 17 \, \text{cm}$. $BE$ is a median of the triangle and $M$ is the midpoint of $BE$. Find the area of $\Delta BMC$ in $\text{cm}^2$.

In $\Delta ABC$,$AD$ is a median. If $ar(\Delta ABC) = 50 \, cm^2$,then $ar(\Delta ADC) = \dots \dots \dots cm^2$.

In $\Delta ABC$,$\angle B = 90^{\circ}$,$AB = 14 \, cm$ and $AC = 50 \, cm$,then find the area of $\Delta ABC$ in $cm^2$.

$ABCD$ is a parallelogram. If $\operatorname{ar}(ABC) = 42 \, \text{cm}^2$,then find $\operatorname{ar}(ABCD)$ in $\text{cm}^2$.

In $\Delta ABC$,$\angle B = 90^{\circ}$ and $BM$ is an altitude to the hypotenuse $AC$. If $AB = 12 \, cm$ and $BC = 16 \, cm$,then find the length of $BM$ in $cm$.