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
10.
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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