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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...

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:

6ec8aac122bd4f6e

 

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.

6ec8aac122bd4f6e

 

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
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