Question 392725
<pre><font face = "batangche" color = "indigo" size = 4><b>
{{{drawing(400,650/3,-12,12,-2,11,

arc(0,0,20,-20,0,180), line(-10,0,10,0), line(-6,8,6,8),
line(6,8,10,0), line(-6,8,-10,0),
locate(0,0,20), locate(0,8,12),
locate(-6.5,9,A), locate(-10,0,B), locate(10,0,C), locate(6,9,D)


 )}}} 

Oh, yes, there is enough information.

Let O be the center of the semicircle.  BC, which is 20 units long
is a diameter of the semicircle, so OB is a radius and is 10 unit
long.  

Draw radius OA, which is also 10 units long.

Draw OE perpendicular to both bases of the trapezoid, splitting the 
upper 12-unit base AD of the trapezoid into two segments AE and DE, 
each 6 units long.
 

{{{drawing(400,650/3,-12,12,-2,11, locate(0,9,E),
green(line(0,0,-6,8),line(0,0,0,8)),
arc(0,0,20,-20,0,180), line(-10,0,10,0), line(-6,8,6,8),
line(6,8,10,0), line(-6,8,-10,0), locate(0,0,O),
locate(-3,8,6), locate(3,8,6), locate(-4.5,4.5,10),
locate(-6.5,9,A), locate(-10,0,B), locate(10,0,C), locate(6,9,D),
 locate(-6,0,10), locate(5,0,10)

 )}}} 

Triangle AEO is a right triangle and we can use the
Pythagorean theorem to find OE. 

{{{OA^2=OE^2+AE^2}}}
{{{10^2=OE^2+6^2}}}
{{{100=OE^2+36}}}
{{{64=OE^2}}}
{{{8=OE}}}

So we label OE as 8 units long.
Draw AF parallel and equal in length to OE, which
is also 8 units long:

{{{drawing(400,650/3,-12,12,-2,11, locate(0,9,E),
green(line(0,0,-6,8),line(0,0,0,8)),
arc(0,0,20,-20,0,180), line(-10,0,10,0), line(-6,8,6,8),
line(6,8,10,0), line(-6,8,-10,0), locate(0,0,O),
locate(-3,8,6), locate(3,8,6), locate(-4.5,4.5,10),
locate(-6.5,9,A), locate(-10,0,B), locate(10,0,C), locate(6,9,D),
 locate(5,0,10), red(line(-6,8,-6,0)), locate(.2,4,8), locate(-5.6,4,8), 
locate(-6,0,F)

 )}}} 

Since AE is the same length as OF, OF is also 6 units long,
and since OB is 10 units long, BF is 10-6 or 4 units long:

{{{drawing(400,650/3,-12,12,-2,11, locate(0,9,E),
green(line(0,0,-6,8),line(0,0,0,8)),
arc(0,0,20,-20,0,180), line(-10,0,10,0), line(-6,8,6,8),
line(6,8,10,0), line(-6,8,-10,0), locate(0,0,O),
locate(-3,8,6), locate(3,8,6), locate(-4.5,4.5,10),
locate(-6.5,9,A), locate(-10,0,B), locate(10,0,C), locate(6,9,D),
 locate(5,0,10), red(line(-6,8,-6,0)), locate(.2,4,8), locate(-5.6,4,8), 
locate(-6,0,F), locate(-8,0,4), locate(-3,0,6) 

 )}}} 
 
Now since ABF is a right triangle, we can find AB using the
Pythagorean theorm:

{{{AB^2=BF^2+AF^2}}}
{{{AB^2=4^2+8^2)}}}
{{{AB^2=16+64}}}
{{{AB^2=80}}}
{{{AB=sqrt(80)}}}
{{{AB=sqrt(16*5)}}}
{{{AB=sqrt(16)sqrt(5)}}}
{{{AB=4sqrt(5)}}}

And CD is also {{{4sqrt(5)}}}

or about 8.94427191 units long

Edwin</pre>