题目内容
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.
![]()
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