In parallelogram $ABCD$,$P$ is the midpoint of $CD$. Then,$\operatorname{ar}(ABCD) : \operatorname{ar}(PBC) = \dots$

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 $\operatorname{ar}(\triangle BDE) = \frac{1}{4} \operatorname{ar}(\triangle ABC).$

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 $XYZW$.

$(1)$ The part of the plane enclosed by a simple closed figure is called a $\ldots \ldots \ldots$
$(2)$ $\ldots \ldots \ldots$ of the planar region corresponding to a closed figure is called its area.

$D, E$ and $F$ are the midpoints of the sides $BC, CA$ and $AB$ respectively of $\Delta ABC$. Show that:
$(i)$ $BDEF$ is a parallelogram.
$(ii)$ $ar(DEF) = \frac{1}{4} ar(ABC)$
$(iii)$ $ar(BDEF) = \frac{1}{2} ar(ABC)$

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