The centers of two circles are $41$ inches apart. The smaller circle has a radius of $4$ inches and the larger one has a radius of $5$ inches. 
The length of the common internal tangent is: 
$\textbf{(A)}\ 41\text{ inches} \qquad \textbf{(B)}\ 39\text{ inches} \qquad \textbf{(C)}\ 39.8\text{ inches} \qquad \textbf{(D)}\ 40.1\text{ inches}\\ \textbf{(E)}\ 40\text{ inches}$
Let $A$ be the center of the circle with radius $5$, and $B$ be the center of the circle with radius $4$. Let $\overline{CD}$ be the common internal tangent of circle $A$ and circle $B$. Extend $\overline{BD}$ past $D$ to point $E$ such that $\overline{BE}\perp\overline{AE}$. Since $\overline{AC}\perp\overline{CD}$ and $\overline{BD}\perp\overline{CD}$, $ACDE$ is a rectangle. Therefore, $AC=DE$ and $CD=AE$.
Since the centers of the two circles are $41$ inches apart, $AB=41$. Also, $BE=4+5=9$. Using the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) on right triangle $ABE$, $CD=AE=\sqrt{41^2-9^2}=\sqrt{1600}=40$. The length of the common internal tangent is $\boxed{\textbf{(E) } 40\text{ inches}}$