阳光家教网 www.ygjj.com 中考（初三复习）数学资料 8综上可得P点的生标共5个解，分别为P（－12，4）、P（－4，4）、P（－，4）、
3P（8，4）、P（4，4）．
下面提供参考解法二：
以直角进行分类进行讨论（分三类）： 第一类如上解法⑴中所示图
此时（D-b,o)，E(O,2b) ?P为直角：设直线DE：y?2x?2b，b1b的中点坐标为(-,b)，直线DE的中垂线方程：y?b??(x?)，令y?4得
222P(3b3?8,4)．由已知可得2PE?DE即2?(b?8)2?(4?2b)2?b2?4b2化简222得3b?32b?64?0解得 b1?8，b2?83b将之代入（P-8，4）?P（4，4)、 1?32P2(?4,4)；
第二类如上解法②中所示图
此时（D-b,o)，E(O,2b) ?E为直角：设直线DE：y?2x?2b，，直线PE的方程：y??1x?2b，令y?4得P(4b?8,4)．由已知可得PE?DE即2(4b?8)2?(4?2b)2?b2?4b2化简得b2?(2b?8)2解之得 ，
b1?4，b2?84P4(?,4) 将之代入（P4b-8，4）?P3?（8，4)、33第三类如上解法③中所示图
此时（D-b,o)，E(O,2b) ?D为直角：设直线DE：y?2x?2b，，直线PD的方程：y??(x?b)，令y?4得P(?b?8,4)．由已知可得PD?DE即 P-b-8，4）?P（-12，4)、82?42?b2?4b2解得b1?4，b2??4将之代入（5?12P6(?4,4)（P6(?4,4)与P2重合舍去）．
综上可得P点的生标共5个解，分别为P（－12，4）、P（－4，4）、P（－P（8，4）、P（4，4）．
8，4）、 3
阳光家教网 www.ygjj.com 中考（初三复习）数学资料 事实上，我们可以得到更一般的结论：
OA?h、如果得出AB?a、OC?b、设k?
b?a，则P点的情形如下 h直角分类情形 k?1 k?1 P1(h,h) ?P为直角 P1(?h,h) P2(?h,h) P3(?hk,h) 1?khkP4(,h) k?1hP2(?,h) 2?E为直角 P5(?h(k?1),h) ?D为直角 P3(0,h) P4(?2h,h) P6(?h(k?1),h)
9.
阳光家教网 www.ygjj.com 中考（初三复习）数学资料
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阳光家教网 www.ygjj.com 中考（初三复习）数学资料 11. 解：（1）设A地经杭州湾跨海大桥到宁波港的路程为x千米，
x?120x······································································································· 2分 ?， ·
1023解得x?180．
由题意得
?A地经杭州湾跨海大桥到宁波港的路程为180千米． ······················································ 4分 （2）1.8?180?28?2?380（元），
····························· 6分 ?该车货物从A地经杭州湾跨海大桥到宁波港的运输费用为380元． ·
（3）设这批货物有y车，
由题意得y[800?20?(y?1)]?380y?8320， ································································· 8分 整理得y2?60y?416?0，
解得y1?8，y2?52（不合题意，舍去）， ······································································ 9分 ··········································································································· 10分 ?这批货物有8车． ·
12. 解：（1）2，21···························································································· 3分 a，a． ·
44（2）相等，比值为2． ·················· 5分（无“相等”不扣分有“相等”，比值错给1分） （3）设DG?x，
在矩形ABCD中，?B??C??D?90，
???HGF?90?，
??DHG??CGF?90???DGH，
?△HDG∽△GCF，
DGHG1???， CFGF2?CF?2DG?2x． ············································································································ 6分 同理?BEF??CFG． ?EF?FG，
?△FBE≌△GCF，
1?BF?CG?a?x． ········································································································ 7分
4?CF?BF?BC，
12?2x?a?x?a， ······································································································· 8分
44
阳光家教网 www.ygjj.com 中考（初三复习）数学资料 解得x?2?1a． 42?1·············································································································· 9分 a． ·
4即DG?（4）
32a， ······················································································································· 10分 1627?1822a． 12分 813. 解：（1）分别过D，C两点作DG⊥AB于点G，CH⊥AB于点H． ……………1分 ∵ AB∥CD， ∴ DG＝CH，DG∥CH．
∴ 四边形DGHC为矩形，GH＝CD＝1．
∵ DG＝CH，AD＝BC，∠AGD＝∠BHC＝90°，
C D ∴ △AGD≌△BHC（HL）．
M N AB?GH7?1∴ AG＝BH＝＝3． ………2分 ?22∵ 在Rt△AGD中，AG＝3，AD＝5， ∴ DG＝4．
A B E G H F
1?7??4?∴ S梯形ABCD??16． ………………………………………………3分 2（2）∵ MN∥AB，ME⊥AB，NF⊥AB，
C D ∴ ME＝NF，ME∥NF．
M N ∴ 四边形MEFN为矩形．
∵ AB∥CD，AD＝BC， ∴ ∠A＝∠B．
∵ ME＝NF，∠MEA＝∠NFB＝90°， A B E G H F ∴ △MEA≌△NFB（AAS）．
∴ AE＝BF． ……………………4分 设AE＝x，则EF＝7－2x． ……………5分 ∵ ∠A＝∠A，∠MEA＝∠DGA＝90°， ∴ △MEA∽△DGA．
AEME∴ ． ?AGDG4∴ ME＝x． …………………………………………………………6分
348?7?49∴ S矩形MEFN?ME?EF?x(7?2x)???x???． ……………………8分
33?4?677当x＝时，ME＝＜4，∴四边形MEFN面积的最大值为49．……………9分
436（3）能． ……………………………………………………………………10分
4由（2）可知，设AE＝x，则EF＝7－2x，ME＝x．
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2010中考数学专题复习 - 压轴题（含答案）
