Question

（9分）
操作：小明準備制作棱長為1cm的正方體紙盒，現(xiàn)選用一些廢棄的圓形紙片進行如下設計：
 

紙片利用率=×100%
發(fā)現(xiàn)：（1）方案一中的點A、B恰好為該圓一直徑的兩個端點．你認為小明的這個發(fā)現(xiàn)是否正確，請說明理由．
（2）小明通過計算，發(fā)現(xiàn)方案一中紙片的利用率僅約為38.2%．請幫忙計算方案二的利用率，并寫出求解過程．
探究：（3）小明感覺上面兩個方案的利用率均偏低，又進行了新的設計（方案三），請直
接寫出方案三的利用率．

Answer 1

發(fā)現(xiàn)：（1）小明的這個發(fā)現(xiàn)正確．····················································· 1分
理由：解法一：如圖一：

連接AC、BC、AB，∵AC＝BC＝，AB＝
∴AC²+BC²＝AB²   ∴∠BAC＝90°，····························································· 2分
∴AB為該圓的直徑．················································································ 3分
解法二：如圖二：

連接AC、BC、AB．易證△AMC≌△BNC，∴∠ACM＝∠CBN．
又∵∠BCN＋∠CBN＝90°，∴∠BCN＋∠ACM＝90°，即∠BAC＝90°，·················· 2分
∴AB為該圓的直徑．················································································ 3分
（2）如圖三：

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

略