Question 894882
<pre>
I changed your BO to OC because if it were BO then AC would not be divided by
the center O as is stated. However if it was really supposed to be BO, then tell 
me in the thank-you note form below and I'll redo it using BO = 20 instead of 
OC = 20.  


{{{drawing(400,6200/19,-5+3,33+3,-5,26,
arc(16,12,24,-24,217,397),locate(22,0,r),locate(22,12,r),
triangle(0,0,16,12,16,0),triangle(16,12,28,12,28,21),rectangle(16,0,28,12),
locate(0,0,C), locate(28,0,B), locate(28,22.5,A),locate(20.5,18,15),
locate(7,8,20),locate(16.25,6.2,r),locate(15,13,O),locate(3,-.2,sqrt(20^2-r^2)),
locate(16,0,D),locate(28.3,12.4,E),locate(28.3,18,sqrt(15^2-r^2))




)}}}

By similar triangles AEO and ODC:

{{{sqrt(15^2-r^2)/r}}}{{{""=""}}}{{{r/sqrt(20^2-r^2)}}}

Cross multiply. 

(Upper left times lower right equal upper right times lower left)

{{{sqrt(15^2-r^2)*sqrt(20^2-r^2)}}}{{{""=""}}}{{{r*r}}}

{{{sqrt(225-r^2)*sqrt(400-r^2)}}}{{{""=""}}}{{{r^2}}}

Square both sides.  Squaring a square root takes away the
square root and just leaves what's under it.  Squaring {{{r^2}}}
gives {{{r^4}}}

{{{(225-r^2)(400-r^2)}}}{{{""=""}}}{{{r^4}}}

FOIL the left side:

{{{90000-225r^2-400r^2+r^4}}}{{{""=""}}}{{{r^4}}}

Combine the middle terms on the left:

{{{90000-625r^2+r^4}}}{{{""=""}}}{{{r^4}}}

Subtract {{{r^4}}} from both sides:

{{{90000-625r^2}}}{{{""=""}}}{{{"0"}}}

Add -90000 to both sides:

{{{-625r^2}}}{{{""=""}}}{{{-90000}}}

Divide both sides by -625

{{{r^2}}}{{{""=""}}}{{{144}}}

Take positive square roots of both sides:

{{{r}}}{{{""=""}}}{{{12}}}

Edwin</pre>