The base of a triangle is twice as long as a side of a square and their areas are the same. 
Then the ratio of the altitude of the triangle to the side of the square is: 
$\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 4$
Let the base and altitude of the triangle be $b, h$, the common area $A$, and the side of the square $s$. Then $\frac{2sh}{2}=s^2\implies sh=s^2\implies s=h$, so $\fbox{C}$