A truncated cone has horizontal bases with radii $18$ and $2$. A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere? 
$\mathrm{(A)}\ 6 \qquad\mathrm{(B)}\ 4\sqrt{5} \qquad\mathrm{(C)}\ 9 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 6\sqrt{3}$
Solution 1
Consider a [trapezoid](https://artofproblemsolving.com/wiki/index.php/Trapezoid) (label it $ABCD$ as follows) cross-section of the truncate cone along a diameter of the bases:


Above, $E,F,$ and $G$ are points of [tangency](https://artofproblemsolving.com/wiki/index.php/Tangent_(geometry)). By the Two Tangent Theorem, $BF = BE = 18$ and $CF = CG = 2$, so $BC = 20$. We draw $H$ such that it is the foot of the altitude $\overline{HD}$ to $\overline{AB}$: 


By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), 
\[r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = 12\]
Therefore, the answer is $\boxed{{A (6)}}.$

Solution 2
Create a trapezoid with inscribed circle $O$ exactly like in Solution #1, and extend lines $\overline{AD}$ and $\overline{BC}$ from the solution above and label the point at where they meet $H$. Because $\frac{\overline{GC}}{\overline{BE}}$ = $\frac{1}{9}$, $\frac{\overline{HG}}{\overline{HE}}$ = $\frac{1}{9}$. Let $\overline{HG} = x$ and $\overline{GE} = 8x$. 
Because these are radii, $\overline{GO} = \overline{OE} = \overline{OF} = 4x$. $\overline{OF} \perp \overline{BH}$ so $\overline{OF}^2 + \overline{FH}^2 = \overline{OH}^2$. Plugging in, we get $4x^2 + \overline{FH}^2 = 5x^2$ so $\overline{FH} = 3x$.Triangles $OFH$ and $BEH$ are similar so $\frac{\overline{OF}}{\overline{BE}}  = \frac{\overline{FH}}{\overline{EH}}$ which gives us $\frac{4x}{18} = \frac{3x}{9x}$. Solving for x, we get \[x = 1.5\] and \[4x =6\]. Thus, the answer is $\boxed{{A (6)}}$.