Solution: ∵ Let the center of the great circle be F, the diameter be CD, the center of the great circle be M, the tangent point between AB and semicircle be N, the connection MN be EF⊥AB, the vertical foot be E, and the connection FA be the radius of the great circle.

∴MN⊥AB，

∫AB∨CD，

∴EF=MN，

∵EF⊥AB，

∴ae= 1/2ab= 1/2× 12=6(cm)，

∴FA2-EF2=FA2-MN2=AE2=36(cm2)，

∫ The area of the shaded part is equal to the semicircle area minus the semicircle area.

∴ The area of the shaded part is:1/2π (Fa2-Mn2) =18π (cm2). Comments: This topic examines the properties of tangents, vertical diameter theorems, properties of parallel lines and Pythagorean theorems. This question is moderately difficult. Pay attention to the method of auxiliary lines and the application of the idea of combining numbers with shapes and the whole idea.