?BCG??DCE?90° CG?CE ?△BCG≌△DCE(SAS) ·············································································· 3分 (2)在△BCG与△DHG中 由(1)得?CBG??CDE ················································································· 4分 ························································································· 5分 ?CGB??DGH ·??DHB??BCG?90° ································································································ 6分 ?BH?DE ·22．(本题满分6分) 解:方程两边同乘以x(x?2),得 ·········································································· 1分 ·············································································· 2分 40(x?2)?40x?x(x?2) ·整理得x?2x?80?0 ··················································································· 3分 ·························································································· 5分 ?x??10或x?8 经检验x??10,x?8都是原方程的根．······························································ 6分 23．(本题满分8分) (1)设y?kx?b依题意得 ·················································································· 1分 2?15.6?k?b ······························································································ 2分 ??15.4?2k?b解答??k??0.2 ······························································································ 3分 ?b?15.8························································································ 4分 ?y??0.2x?15.8 ·(2)当x?6时,y??0.2?6?15.8 ······································································ 5分 ········································································································ 6分 ?14.6 ·(3)不能 ········································································································· 7分 略(理由合理) ·································································································· 8分 24．(本题满分8分) (1)解:掷两枚硬币出现的情况是(正,正)、(正,反)、(反,正)、(反,反),故出现两枚硬币都朝上的概率是1； ··································································································· 3分 4(2)25,125,75 ··································································································· 6分 (3)略(只要有理就行) ························································································ 8分 25．(本题满分10分) (1)证明:AB为O直径,??ACB?90°即?ACE??BCE?90° 又CD?AB,??A??ACE?90° ························································· 3分 ??A??ECB,?Rt△ACE∽Rt△CBE ·(2)△ACE∽△CBE ?AECE2即CE?AEBE?(AO?OE)(OB?OE) ?CEEB?y?(4?x)(4?x)?16?x2 ··········································································· 6分 (3)解法一:tan?D?33即tan?A? 33CE3CE2116?x21?解得x?2或x??4(舍去) ???即则AE3AE23(4?x)23故当x?2时,tan?D?3 ············································································ 10分 3解法二:tan?D?3BEBE4?x ???23DECE16?x即34?x ?2316?x1(4?x)2??解得x?2或x?4(舍去) 2316?x故当x?2时,tan?D?26．(本题满分10分) (1)证明:3 ············································································ 10分 3PO?PQ,??APO??BPQ?90°, 在Rt△AOP中,?APO??AOP?90°, ??BPQ??AOP APBQ即OABQ?APBP ········································· 3分 ?OABPm(4?m)(2)OABP?APBP,即BQ? 3?△AOP∽△BPQ,则4m?m212515?l?3??(m?4m?4)??(m?2)2? 333335?当m?2时,l有最小值． ············································································ 6分 3(3)解法一 △POQ是等腰三角形 ①若P在线段AB上,?OPQ?90° ?PO?PQ,又△AOP∽△BPQ,?△AOP≌△BPQ ············································· 8分 3) ·?PB?AO,即3?4?m,?m?1,即P点坐标(1，②若P在线段AB的延长线上,PQ交CB的延长线于Q,PO?PQ, 又△AOP∽△BPQ,?△AOP≌△BPQ, 3), ?AO?PB,即3?m?4,即P点的坐标(7，，，3)P2(7，3)使△POQ为等腰三角形． ·故存在P············································· 10分 1(1解法二△POQ是等腰三角形 ?PO?PQ, 即PA?AO?PB?BQ ············································································· 7分 2222?4m?m2?222则m?3?(4?m)??··································································· 8分 ? 3??整理得m?8m?16m?72m?63?0 4322y A P B m4?8m3?7m2?9m2?72m?63?0 m2(m2?8m?7)?9(m2?8m?7)?0 (m?1)(m?7)(m2?9)?0 Q ?m1?1,m2?7,m??9(舍去) 2O C x 3)使△POQ为等腰三角形． ·，3),P2(7，故存在P·············································· 10分 1(1注:以上各题的其它解法,请参照此标准记分．
2009年湘潭市初中毕业学业考试题及答案
