∴OC=6cm，

∴S sector =12π (oc) 2 =12π× 62 =18π cm2.

(2)∵CD= 12cm，∠C=90，

∴S sector DCE = 90 π×122360 = 36 π cm2;

(3))∫BC = 18cm，

∴OC=9cm，

∴S sector =12π (oc) 2 =12π× 92 = 812π cm2.

So the answer is: 812π;

As shown in fig. 4, connecting OE,

∫AD and CF are tangent to point e,

∴OE⊥AD，

A quadrilateral is a square,

∴OE=OC=CD= 12cm，

Eoc of sector s =14π (oc) 2 =14π×122 = 36π;

∫OB = BC-OC = 18- 12 = 6cm，OF=CD= 12cm，∠B=90，

∴∠OFB=30，

∴∠BOF=90 -30 =60，

∴∠EOF=30，

∴S sector eof = 30 π×122360 =12π,

∴S fan COF=S fan EOC+S fan EOF = 36π+ 12π = 48π cm2.

So the answer is:18 π; 36π.