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