∴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.