湘潭市2009中考数学试题及答案

（1）猜想：BG?DE ·········································································································· 1分 ?BC?DC ?BCG??DCE?90° CG?CE ····························································································· 3分 ?△BCG≌△DCE（SAS） ·（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?k??0.2解答? ······················································································································· 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?AE?BE?(AO?OE)(OB?OE) ??CEEB······························································································ 6分 ?y?(4?x)(4?x)?16?x2 ·（3）解法一：?tan?D?33即tan?A? 33CE3CE2116?x21则?解得x?2或x??4（舍去） ???即223AE3AE3(4?x)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） ??316?x2故当x?2时，tan?D?26．（本题满分10分） （1）证明：?PO?PQ，??APO??BPQ?90°， 在Rt△AOP中，?APO??AOP?90°， 3 ····························································································· 10分 3??BPQ??AOP APBQ即OA?BQ?AP?BP ················································· 3分 ?OABPm(4?m)（2）?OA?BP?AP?BP，即BQ? 3?△AOP∽△BPQ，则4m?m212515?l?3??(m?4m?4)??(m?2)2? 333335···························································································· 6分 ?当m?2时，l有最小值． ·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 ?m1?1，m2?7，m??9（舍去） 2Q O 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．