Parallel to CD, draw line EF through B
We have four triangles EBA, FBA, EBC and FBD....and their combined area is 15
And because triangle EAF is similar to triangle CAD.....then EB = 5/12 CE and BF = 5/12 ED ....and the height of EBA and FBA = BA = 5, while the height of EBC and FBD = EB = 7
So we have
Area EBA + Area of FBA + Area of EBC + Area of FBD = 15
(1/2)(5/12)CE*BA + (1/2)(5/12)ED *BA + (1/2)(5/12)CE *EB + (1/2)(5/12)ED *EB = 15
(1/2)(5/12)CE*5 + (1/2)(5/12)ED *5 + (1/2)(5/12)CE *7 + (1/2)(5/12)ED *7 = 15
[ (1/2)(5/12) CE * (5 + 7) ] + [ (1/2)(5/12) ED * (5 + 7) ] = 15
[ (1/2)(5/12) CE * (12) ] + [ (1/2)(5/12) ED * (12) ] = 15
[ (1/2)(5/12)* 12 [ CE + ED] = 15
(60/24) [ CD] = 15
CD = 15 * (24/60) = (1/4) * 24 = 6
So.....area CDA = (1/2)CD * EA = (1/2) * 6 * 12 = 36 units^2