摘要：如图14.从一个直径是2的圆形铁皮中剪下一个圆心角为的扇形． (1)求这个扇形的面积(结果保留)． (2)在剩下的三块余料中.能否从第③块余料中剪出一个圆作为底面与此扇形围成一个圆锥?请说明理由． (3)当的半径为任意值时.(2)中的结论是否仍然成立?请说明理由． 解:(1)连接.由勾股定理求得: ·································································· 1分 ································································· 2分 (2)连接并延长.与弧和交于. ····························································································· 1分 弧的长:······················································································ 2分 圆锥的底面直径为:··················································································· 3分 .不能在余料③中剪出一个圆作为底面与此扇形围成圆锥．·············· 4分 (3)由勾股定理求得: 弧的长:···················································································· 1分 圆锥的底面直径为:················································································ 2分 且 ···································································································· 3分 即无论半径为何值.··················································································· 4分 不能在余料③中剪出一个圆作为底面与此扇形围成圆锥．

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