If the side of one square is the diagonal of a second square, what is the ratio of the area of the first square to the area of the second? 
$\textbf{(A)}\ 2 \qquad  \textbf{(B)}\ \sqrt2 \qquad  \textbf{(C)}\ 1/2 \qquad  \textbf{(D)}\ 2\sqrt2 \qquad \textbf{(E)}\ 4$
Solution by e_power_pi_times_i

Denote the side of one square as $s$. Then the diagonal of the second square is $s$, so the side of the second square is $\dfrac{s\sqrt{2}}{2}$. The area of the second square is $\dfrac{1}{2}s^2$, so the ratio of the areas is $\dfrac{s^2}{\dfrac{1}{2}s^2} = \boxed{\textbf{(A) } 2}$.