Write True or False and justify your answer:
$PQRS$ is a rectangle inscribed in a quadrant of a circle of radius $13 \, cm$. $A$ is any point on $PQ$. If $PS = 5 \, cm$,then $\text{ar}(PAS) = 30 \, cm^2$.

(B) In the rectangle $PQRS$,the diagonal $PR$ is the radius of the circle,so $PR = 13 \, cm$.
Given $PS = 5 \, cm$,in right-angled $\triangle PSR$,by Pythagoras theorem:
$PR^2 = PS^2 + SR^2$
$13^2 = 5^2 + SR^2$
$169 = 25 + SR^2$
$SR^2 = 144 \implies SR = 12 \, cm$.
Since $PQRS$ is a rectangle,$PQ = SR = 12 \, cm$.
The area of $\triangle PQS = \frac{1}{2} \times PQ \times PS = \frac{1}{2} \times 12 \times 5 = 30 \, cm^2$.
Since $A$ is any point on $PQ$,the triangle $PAS$ has base $PA$ and height $PS = 5 \, cm$.
Area of $\triangle PAS = \frac{1}{2} \times PA \times PS$.
Since $PA < PQ$ (as $A$ is on $PQ$),$\text{ar}(PAS) < \text{ar}(PQS) = 30 \, cm^2$.
Therefore,the statement $\text{ar}(PAS) = 30 \, cm^2$ is False.