In $\triangle ABC$ the median from $A$ is given perpendicular to the median from $B$. If $BC=7$ and $AC=6$, find the length of $AB$. 
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ \sqrt{17} \qquad \textbf{(C)}\ 4.25\qquad \textbf{(D)}\ 2\sqrt{5} \qquad \textbf{(E)}\ 4.5$
By SAS Similarity, $\triangle ABC \sim \triangle MNC$, so $AB \parallel MN$.  Thus, by AA Similarity, $\triangle AGB \sim \triangle NGM$.
Let $a = GN$ and $b = GM$, so $AG = 2a$ and $BG = 2b$.  By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem),
\[4a^2 + b^2 = 9\]
\[a^2 + 4b^2 = \frac{49}{4}\]
Adding the two equations yields $5a^2 + 5b^2 = \frac{85}{4}$, so $a^2 + b^2 = \frac{17}{4}$.  Thus, $MN = \frac{\sqrt{17}}{2}$, so $AB = \sqrt{17}$, which is answer choice $\boxed{\textbf{(B)}}$.