2009年全国中考数学压轴题精选精析(一)

2009年全国中考数学压轴题精选精析（二十）73.（2009年山西）26．（本题14分）如图，已知直线128:33lyx与直线2:216lyx相交于点Cll12，、分别交x轴于AB、两点．矩形DEFG的顶点DE、分别在直线12ll、上，顶点FG、都在x轴上，且点G与点B重合．（1）求ABC△的面积；（2）求矩形DEFG的边DE与EF的长；（3）若矩形DEFG从原点出发，沿x轴的反方向以每秒1个单位长度的速度平移，设移动时间为(012)tt≤≤秒，矩形DEFG与ABC△重叠部分的面积为S，求S关于t的函数关系式，并写出相应的t的取值范围．[来源:学科网]（2009年山西26题解析）（1）解：由28033x，得4xA．点坐标为40，．由2160x，得8xB．点坐标为80，．∴8412AB．········································································（2分）由2833216yxyx，．解得56xy，．∴C点的坐标为56，．·······························（3分）∴111263622ABCCSABy△·．··················································（4分）（2）解：∵点D在1l上且2888833DBDxxy，．∴D点坐标为88，．··········································································（5分）又∵点E在2l上且821684EDEEyyxx，．．∴E点坐标为48，．··········································································（6分）∴8448OEEF，．·································································（7分）（3）解法一：①当03t≤时，如图1，矩形DEFG与ABC△重叠部分为五边形CHFGR（0t时，为四边形CHFG）．过C作CMAB于M，则RtRtRGBCMB△∽△．ADBEOCFxyy1ly2l（G）（第26题）ADBEORFxyy1ly2lMGCADBEOCFxyy1ly2lGRMADBEOCFxyy1ly2lGRM∴BGRGBMCM，即36tRG，∴2RGt．RtRtAFHAMC△∽△，∴11236288223ABCBRGAFHSSSStttt△△△．即241644333Stt．··························································（10分）75.（2009年山西太原）29．（本小题满分12分）问题解决如图（1），将正方形纸片ABCD折叠，使点B落在CD边上一点E（不与点C，D重合），压平后得到折痕MN．当12CECD时，求AMBN的值．类比归纳在图（1）中，若13CECD，则AMBN的值等于；若14CECD，则AMBN的值等于；若1CECDn（n为整数），则AMBN的值等于．（用含n的式子表示）联系拓广如图（2），将矩形纸片ABCD折叠，使点B落在CD边上一点E（不与点CD，重合），压平后得到折痕MN，设111ABCEmBCmCDn，，则AMBN的值等于．（用含mn，的式子表示）（2009年山西太原29题解析）解：方法一：如图（1-1），连接BMEMBE，，．方法指导：为了求得AMBN的值，可先求BN、AM的长，不妨设：AB=2图（2）NABCDEFM图（1）ABCDEFMNN图（1-1）ABCDEFM由题设，得四边形ABNM和四边形FENM关于直线MN对称．∴MN垂直平分BE．∴BMEMBNEN，．·····································1分∵四边形ABCD是正方形，∴902ADCABBCCDDA°,．∵112CECEDECD，．设BNx，则NEx，2NCx．在RtCNE△中，222NECNCE．∴22221xx．解得54x，即54BN．·········································3分在RtABM△和在RtDEM△中，222AMABBM，222DMDEEM，2222AMABDMDE．·····························································5分设AMy，则2DMy，∴2222221yy．解得14y，即14AM．·····································································6分∴15AMBN．·····················································································7分方法二：同方法一，54BN．································································3分如图（1－2），过点N做NGCD∥，交AD于点G，连接BE．∵ADBC∥，∴四边形GDCN是平行四边形．∴NGCDBC．同理，四边形ABNG也是平行四边形．∴54AGBN．∵90MNBEEBCBNM，°．90NGBCMNGBNMEBCMNG，°，．在BCE△与NGM△中N图（1-2）ABCDEFMG90EBCMNGBCNGCNGM，，°．∴BCENGMECMG△≌△，．·························５分∵114AMAGMGAM5，=．4·····················································6分∴15AMBN．····················································································7分类比归纳25（或410）；917；2211nn································································10分联系拓广2222211nmnnm·····················································································12分评分说明：1．如你的正确解法与上述提供的参考答案不同时，可参照评分说明进行估分．2．如解答题由多个问题组成，前一问题解答有误或未答，对后面问题的解答没有影响，可依据参考答案及评分说明进行估分．s