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.
 

 

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







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

 

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:

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










And CD is also 

or about 8.94427191 units long

Edwin