The area of a circle $A$ is in direct variation with the square of its radius $r$, which means $A = k r^2$, where $k$ is the constant of proportionality (in this case, $k=\pi$). Given: $A_1 = 616$ sq.cm. when $r_1 = 14$ cm. We can find the constant $k$ (which is $\pi$): $616 = k (14)^2$ $616 = k (196)$ $k = \frac{616}{196} = \frac{22}{7}$. So, $k=\pi$. Now, we need to find the area $A_2$ when $r_2 = 42$ cm. $A_2 = k r_2^2$ $A_2 = \frac{22}{7} (42)^2$ $A_2 = \frac{22}{7} (42 \times 42)$ $A_2 = 22 \times (\frac{42}{7}) \times 42$ $A_2 = 22 \times 6 \times 42$ $A_2 = 132 \times 42$ $A_2 = 5544$ sq.cm. Alternatively, since $A = k r^2$, we have $\frac{A_1}{r_1^2} = \frac{A_2}{r_2^2}$. We notice that $r_2 = 42 = 3 \times 14 = 3 r_1$. So, $A_2 = k r_2^2 = k (3r_1)^2 = k (9r_1^2) = 9 (k r_1^2) = 9 A_1$. $A_2 = 9 \times 616 = 5544$ sq.cm. Therefore, the area of the circle whose radius is $42$ cm is $5544$ sq.cm.