Question

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：

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．

Answer

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