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

If the mid-points of the sides of a quadrilateral are joined in order,prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral.

$(1)$ If two figures are congruent,they must have $\ldots \ldots$ areas.
$(2)$ Area of figure $A$ is denoted as $\ldots \ldots$ symbolically.

In the figure,$PSDA$ is a parallelogram. Points $Q$ and $R$ are taken on $PS$ such that $PQ = QR = RS$ and $PA \parallel QB \parallel RC$. Prove that $\operatorname{ar}(PQE) = \operatorname{ar}(CFD)$.

In the figure,$ABCD$ and $AEFD$ are two parallelograms. Prove that $\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.

State whether each of the following statements is true or false:
$(1)$ If $ar(ABC) = 96 \, cm^2$ for the parallelogram $ABCD$,then $ar(ABCD) = 192 \, cm^2$.
$(2)$ Area of a right triangle = Product of the sides forming the right angle.