Ans: D
Let $c_0$ and $c_1$ be the original and new circumference of the circle respectively. Let $A_0$ and $A_1$ be the original and new area of the circle respectively.
Let $c_0$ and $c_1$ be the original and new circumference of the circle respectively. Let $A_0$ and $A_1$ be the original and new area of the circle respectively.
$\begin{array}{rcl}
c_1 & = & c_0\times(1+40\%) \\
& = & 1.4c_0
\end{array}$
Since the original circle and the new circle is similar, we have
$\begin{array}{rcl}
\dfrac{A_1}{A_0} & = & \left(\dfrac{c_1}{c_0}\right)^2 \\
& = & \left( \dfrac{1.4c_0}{c_0}\right)^2 \\
A_1 & = & 1.96 A_0
\end{array}$
Therefore, the percentage increase of the area of the circle
$\begin{array}{cl}
= & \dfrac{A_1-A_0}{A_0} \times 100\% \\
= & \dfrac{1.96A_0-A_0}{A_0} \times 100\% \\
= & 96\%
\end{array}$
