Question 1178916
<pre><b>
Let r = the radius of the small semicircle.

{{{drawing(400,240,-2.5,2.5,-.5,2.5, 
line(-2,0,2,0), red(line(-4/3,0,0,1)),green(line(0,0,0,2)),
arc(0,0,4,-4,0,180), circle(0,1,1), arc(-1.32,0,1.34,-1.34,0,180),
locate(-5/3,0,r),locate(-1.3,-.12,sqrt((r+1)^2-1^2)),locate(-1.1,.4,r),
locate(-.5,.9,1),locate(.05,.57,1),locate(1,0,2)
 )}}}

For the right triangle:

The red hypotenuse = r+1.
The green right verticle leg = 1.
So, by the Pythagorean theorem, 
the bottom horizontal leg = {{{sqrt((r+1)^2-1^2)}}}

We add the two parts of the left horizontal radius of the
large semicircle, and get {{{r+sqrt((r+1)^2-1^2)}}}  This 
sum must be equal to the right radius of the large 
semicircle, which is 2.  So we have:

{{{r+sqrt((r+1)^2-1^2)}}}{{{""=""}}}{{{2}}}

{{{sqrt((r+1)^2-1)}}}{{{""=""}}}{{{2-r}}}

{{{sqrt(r^2+2r+1-1)}}}{{{""=""}}}{{{2-r}}}

{{{sqrt(r^2+2r)}}}{{{""=""}}}{{{2-r}}}

Square both sides:

{{{r^2+2r}}}{{{""=""}}}{{{(2-r)^2}}}

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

Subtract r<sup>2</sup> from both sides:

{{{2r}}}{{{""=""}}}{{{4-4r}}}

{{{6r}}}{{{""=""}}}{{{4}}}

{{{r}}}{{{""=""}}}{{{4/6}}}

{{{r}}}{{{""=""}}}{{{2/3}}}cm.    <---answer

Edwin</pre></b>