题目

题型：不详难度：来源：

（9分）
操作：小明准备制作棱长为1cm的正方体纸盒，现选用一些废弃的圆形纸片进行如下设计：
 

纸片利用率=×100%
发现：（1）方案一中的点A、B恰好为该圆一直径的两个端点．你认为小明的这个发现是否正确，请说明理由．
（2）小明通过计算，发现方案一中纸片的利用率仅约为38.2%．请帮忙计算方案二的利用率，并写出求解过程．
探究：（3）小明感觉上面两个方案的利用率均偏低，又进行了新的设计（方案三），请直
接写出方案三的利用率．

答案

发现：（1）小明的这个发现正确．····················································· 1分
理由：解法一：如图一：

连接AC、BC、AB，∵AC＝BC＝，AB＝
∴AC²+BC²＝AB²   ∴∠BAC＝90°，····························································· 2分
∴AB为该圆的直径．················································································ 3分
解法二：如图二：

连接AC、BC、AB．易证△AMC≌△BNC，∴∠ACM＝∠CBN．
又∵∠BCN＋∠CBN＝90°，∴∠BCN＋∠ACM＝90°，即∠BAC＝90°，·················· 2分
∴AB为该圆的直径．················································································ 3分
（2）如图三：

易证△ADE≌△EHF，∴AD＝EH＝1．·························································· 4分
∵DE∥BC，∴△ADE∽△ACB，∴＝∴＝，∴BC＝8．·················· 5分
∴S_△ACB＝16．························································································ 6分
∴该方案纸片利用率＝×100%＝×100%＝37.5%···················· 7分
探究：（3）······················································································· 9分

解析

略