发布时间 : 星期日 文章江苏省南京市秦淮区2018届中考一模数学试题(含答案)更新完毕开始阅读00dc1f78e0bd960590c69ec3d5bbfd0a7956d5d4
24．（本题8分）
解：设铁棒的长为x m．
OB
， ········································································· 1分 AB
1∴OB?AB·cos∠ABO?x·cos60°···································································· 3分 ?x． ·
2OD在Rt△COD中，cos∠CDO?， ········································································· 4分
CD在Rt△AOB中，cos∠ABO?
∴OD?CD·cos∠CDO?x·cos51°18′?0.625x． ··························································· 6分 ∵BD?OD?OB，
1∴0.625x?x?1． ······························································································ 7分
2解这个方程，得x?8．
答：该铁棒的长为8m． ······························································································ 8分 25．（本题8分）
解：（1）AB与⊙D相切． ·········································· 1分
证明：过点D作DF⊥AB，垂足为F． ·············· 2分 ∵AD是Rt△ABC的角平分线，∠C?90°，
∴DF?DC，················································ 3分 即d?r，
∴AB与⊙D相切． ······································· 4分
C D E A B F
（2）∵∠C?90°，AC?BC?1，∴∠BAC?∠B?45°，AB?2．
∵DF⊥AB，∴∠BDF?∠B?45°，∴BF?DF． ∵AB、AC分别与⊙D相切，∴AF?AC?1． 设⊙D的半径为r．易得BF?2?1，BD?1?r，
∴2(2?1)?1?r，∴r?2?1． ·································································· 6分 ∵AD是Rt△ABC的角平分线，∠BAC?45°，
1
∴∠DAC?∠BAC?22.5°．
2
又∵∠C?90°，∴∠CDE?67.5°． ··································································· 7分
π(2?1)?67.5(32?3)π∴lCE． ······························································· 8分 ???180826．（本题9分）
解：（1）∵四边形ABCD是矩形，∴∠ABC?∠C?90°．
∵第一次折叠使点C落在AB上的F处，并使折痕经过点B， ∴∠CBE?∠FBE?45°，∴∠CBE?∠CEB?45°，
∴BC?CE?a，BE?2a． ·············································································· 2分 ∵第二次折叠纸片，使点A落在E处，得到折痕BG，
AB∴AB?BE?2a，∴······································································ 3分 ?2． ·
BC21（2）根据题意和（1）中的结论，有AH?BH?a，AM?a．
22AMAH2∴． ······················································································ 4分 ??BHBC2
∵四边形ABCD是矩形，∴∠A?∠B?90°，∴△MAH∽△HBC， ··························· 5分 ∴∠AHM?∠BCH． ······················································································· 6分 ∵∠BCH?∠BHC?90°，∴∠AHM?∠BHC?90°，∴∠MHC?90°，
∴HC⊥HM． ································································································ 7分 2722（3）··································································································· 9分 a． ·
3227．（本题9分）
解：（1）③④． ········································································································· 2分
（说明：只答对1个得1分，答错一个不给分） （2）证明：连接CO并延长，交⊙O于点E，连接BE．
∵PT是⊙O的切线，切点为C， ∴∠PCE?90°． ∴∠PCB?∠ECB?90°． ∵CE是⊙O的直径， ∴∠CBE?90°， ∴∠BEC?∠ECB?90°， ∴∠BEC?∠PCB．
又∵∠BEC?∠BDC，∴∠PCB?∠BDC．
又∵∠BPC?∠CPD，∴△PBC∽△PCD，
CBPC∴．······················································································ 3分 ?CDPDABPA同理，． ··············································································· 4分 ?ADPD∵PA、PC为⊙O的切线，
∴PA?PC， ························································································· 5分 ∴
P B C T A O S E M D
CBAB． ?CDAD∴AB?CD?AD?BC．
∴四边形ABCD是和谐四边形． ······························································ 6分
（3）AB∥CD ，CD?3AB． ··················································································· 9分
（说明：结论“AB∥CD”1分，“CD?3AB”2分）