解:(1)法一:過O作OE⊥AB于E,則AE=

AB=2

.····················· 1分
在Rt

AEO中,∠BAC=30°,cos30°=

.
∴OA=

=

=4. …………………………2分
又∵OA=OB,∴∠ABO=30°.∴∠BOC=60°.∵AC⊥BD,∴

.
∴∠COD =∠BOC=60°.∴∠BOD=120°.······················································· 3分
∴S
陰影=

=

.································································· 4分
法二:連結AD.∵AC⊥BD,AC是直徑,

∴AC垂直平分BD. ……………………1分∴AB=AD,BF=FD,

. ∴∠BAD=2∠BAC=60°,
∴∠BOD=120°. ……………………2分
∵BF=

AB=2

,sin60°=

,AF=AB·sin60°=4

×

=6.
∴OB
2=BF
2+OF
2.即

.∴OB=4. ···························· 3分
∴S
陰影=

S
圓=

. ········································································ 4分
法三:連結BC.∵AC為⊙O的直徑,∴∠ABC=90°.……………………1分

∵AB=4

,∴

. ……………………2分
∵∠A=30°, AC⊥BD,∴∠BOC=60°,∴∠BOD=120°.
∴S
陰影=

π·OA
2=

×4
2·π=

.……………………4分
以下同法一.
(2)設圓錐的底面圓的半徑為r,則周長為2πr,
∴

. ∴

. ···················································· 6分
(3)

<8

-12,故能得到兩個這樣的底面!8分