ADEC is the trapezoid in this specific case. ADEB is a parallelogram where DE=AB=14=b1 and AD=BE=11=s1. The height h of the trapezoid is EF. AC=20=b2 and CE=14=s2. Let x=BF, then h2+x2=112 (Pythagoras). CF=x+6, and h2+(x+6)2=142, that is, h2+x2+12x+62=142, or algebraically h2+x2+2(b2-b1)x+(b2-b1)2=s22. We can substitute for h2+x2=s12, so:
2(b2-b1)x+(b2-b1)2=s22-s12, and x=(s22-s12-(b2-b1)2)/(2(b2-b1)). Putting in figures, x=(196-121-36)/12=39/12=13/4. Since h2+x2=s12, h=√(s12-x2).
We can substitute for x giving us a rather complicated algebraic formula. Again, using figures h=√(121-169/16)=√1767/4=10.51 approx.
You're welcome to put together the formula but it will not be one you can easily remember. Best to split into two formulae, one for x and the other for h. The formula should work for every trapezoid but in some cases x will be negative, meaning that point B will be to the left of F. Its square, of course, will be positive so h can still be calculated.