解:(1)法一:过O作OE⊥AB于E,则AE=AB=2.····················· 1分 在RtAEO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阴影==.································································· 4分 法二:连结AD.∵AC⊥BD,AC是直径,
∴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. ∴OB2=BF2+OF2.即.∴OB=4. ···························· 3分 ∴S阴影=S圆=. ········································································ 4分 法三:连结BC.∵AC为⊙O的直径,∴∠ABC=90°.……………………1分
∵AB=4,∴. ……………………2分 ∵∠A=30°, AC⊥BD,∴∠BOC=60°,∴∠BOD=120°. ∴S阴影=π·OA2=×42·π=.……………………4分 以下同法一. (2)设圆锥的底面圆的半径为r,则周长为2πr, ∴. ∴. ···················································· 6分 (3)<8-12,故能得到两个这样的底面。……………………8分 |