Is the area of the largest circle that can be drawn inside a rectangle of length $a \, cm$ and breadth $b \, cm$ $(a > b)$ equal to $\pi b^{2} \, cm^{2}$? Why?

(B) The statement is False.
The largest circle that can be drawn inside a rectangle of length $a$ and breadth $b$ (where $a > b$) is limited by the smaller dimension of the rectangle,which is the breadth $b$.
The diameter of this largest circle is equal to the breadth of the rectangle,$d = b$.
Therefore,the radius of the circle is $r = \frac{b}{2}$.
The area of the circle is given by the formula $A = \pi r^{2}$.
Substituting the value of $r$,we get $A = \pi \left(\frac{b}{2}\right)^{2} = \frac{\pi b^{2}}{4} \, cm^{2}$.
Thus,the area is $\frac{\pi b^{2}}{4} \, cm^{2}$,not $\pi b^{2} \, cm^{2}$.