In $\Delta PQR$,$\angle Q = 90^{\circ}$,$QR = 21 \text{ cm}$ and $PR = 29 \text{ cm}$,then find the area of $\Delta PQR$ in $\text{cm}^2$.

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 $\triangle ABC,$ if $L$ and $M$ are the points on $AB$ and $AC,$ respectively such that $LM \parallel BC.$ Prove that $\operatorname{ar}(\triangle LOB) = \operatorname{ar}(\triangle MOC).$

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

In the figure,if parallelogram $ABCD$ and rectangle $ABEM$ are of equal area,then:

Write True or False and justify your answer:
$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. Then $ar(\triangle BDE) = \frac{1}{4} ar(\triangle ABC).$