A circle with diameter PQ of length 10 is internally tangent at P to a circle of radius 20. Square ABCD is constructed with A and B on the large circle,$\overline{\mathrm{CD}}$tangent at Q to the smaller circle, and the  smaller circle outside ABCD. The length of $\overline{\mathrm{AB}}$can be written in the form $\mathrm{m}+\sqrt{\mathrm{n}}$where m and n are integers. Find $\frac{\mathrm{m}}{9\mathrm{m}+4}$

Call the center of the larger circle O. extend the diameter $\overline{\mathrm{PQ}}$to the other side of the square (at point E), and draw $\overline{\mathrm{AO}}$We now have right tirangle, with
hypotenuse of length 20.
Since OQ = OP - PQ = 20 -10 = 10 ,
we know that OE = ABH -OQ = AB -10 .
The other leg. AE, is just $\frac{1}{2}$AB
Apply the Pythagorean theorem:
$\begin{array}{l}(\mathrm{AB}-10{)}^2+\left(\frac{1}{2}\mathrm{AB}\right)={20}^2\\ {\mathrm{AB}}^2-20\mathrm{AB}+100+\frac{1}{4}{\mathrm{AB}}^2-400=0\\ {\mathrm{AB}}^2-16\mathrm{AB}-240=0\end{array}$
The quadratic formula shows that the answer is
$\frac{16\pm \sqrt{{16}^2+4.240}}{2}=8\pm \sqrt{304}$
Discard the negative root,
so out answer is 8 + 304 = 312