The perimeter of one square is $3$ times the perimeter of another square.  The area of the larger square is how many times the area of the smaller square?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 9$
Let $a$ be the sidelength of one square, and $b$ be the sidelength of the other, where $a>b$. If the perimeter of one is $3$ times the other's, then $a=3b$. The area of the larger square over the area of the smaller square is
\[\frac{a^2}{b^2} = \frac{(3b)^2}{b^2} = \frac{9b^2}{b^2} = \boxed{\text{(E)}\ 9}\]