∫AD∨BC, AC bisection ∠BCD, ∠ ADC = 120,

∴∠BCD=60，

∫ AC split ∠BCD,

∴∠ACD=30，

∴∠AOD=2∠ACD=60，∠OAC=∠ACO=30。

∴∠BAC=90，

∴BC is diameter,

OA = OD = OB = OC，

Then △AOD, △AOB and △COD are equilateral triangles.

∴AB=AD=CD.

The perimeter of the quadrilateral ABCD is 10cm,

The radius of a circle is 10 ÷ 5 = 2 (cm).

∴ The area of the shaded part = 60 π× 4360-3× 22 = 2 π 3-3 (cm2).

So the answer is (2π 3-3) cm2.