Square $ABCD$ has sides of length 3. Segments $CM$ and $CN$ divide the square's area into three equal parts. How long is segment $CM$?

$\text{(A)}\ \sqrt{10} \qquad \text{(B)}\ \sqrt{12} \qquad \text{(C)}\ \sqrt{13} \qquad \text{(D)}\ \sqrt{14} \qquad \text{(E)}\ \sqrt{15}$
Since the square has side length $3$, the area of the entire square is $9$.
The segments divide the square into 3 equal parts, so the area of each part is $9 \div 3 = 3$.
Since $\triangle CBM$ has area $3$ and base $CB = 3$, using the area formula for a triangle:
$A_{tri} = \frac{1}{2}bh$
$3 = \frac{1}{2}3h$
$h = 2$
Thus, height $BM = 2$.
Since $\triangle CBM$ is a right triangle, $CM = \sqrt{BM^2 + BC^2} = \sqrt{2^2 + 3^2} = \boxed{\text{(C)}\ \sqrt{13}}$.