Question 902633
<pre>
No dimensions are given, only the 3:2 ratio, so I can only 
assume that the hypotenuse is 5 units long and is divided 
into two parts, one which is 3 units long and the other 
2 units long, like this drawing. Let the radius be R units
long: 

{{{drawing(400,1000/3,-1,5,-1,4, triangle(0,0,4,0,4,3), circle(3,1,1), 
green(line(2.4,1.8,3,1),line(3,1,4,1),line(3,0,3,1)),
locate(1.1,1.2,3),locate(3,36/11-.65,2), locate(3.5,1,R), locate(3.5,0,R),
locate(2.8,.65,R),locate(4.05,.65,R),locate(4.05,2,2),locate(1.5,0,3),
locate(2.75,1.55,R) 


)}}}

So by the Pythagorean theorem:

{{{(3+R)^2+(R+2)^2 = (3+2)^2}}}

{{{9+6R+R^2+R^2+4R+4=5^2}}}

{{{2R^2+10R+13=25}}}

{{{2R^2+10R-12=0}}}

Divide through by 2:

{{{R^2+5R-6=0}}}

{{{(R+6)(R-1)=0}}}

{{{R+6=0}}}, {{{R-1=0}}}
  {{{R=-6}}},  {{{R=1}}}

We ignore the negative answer, and
the radius is R = 1 unit long.

Anything you don't understand you can ask 
me in the thank-you note below and I will 
get back to you.

Edwin</pre>