中考数学压轴题100题精选以及答案

2010年中考数学压轴题100题精选（1-10题）答案【001】解：（1）抛物线2(1)33(0)yaxa经过点(20)A，，309333aa·······································································································1分二次函数的解析式为：232383333yxx·························································3分（2）D为抛物线的顶点(133)D，过D作DNOB于N，则33DN，2233(33)660ANADDAO，°···························································4分OMAD∥①当ADOP时，四边形DAOP是平行四边形66(s)OPt·······················································5分②当DPOM时，四边形DAOP是直角梯形过O作OHAD于H，2AO，则1AH（如果没求出60DAO°可由RtRtOHADNA△∽△求1AH）55(s)OPDHt··········································································································6分③当PDOA时，四边形DAOP是等腰梯形26244(s)OPADAHt综上所述：当6t、5、4时，对应四边形分别是平行四边形、直角梯形、等腰梯形．··7分（3）由（2）及已知，60COBOCOBOCB°，，△是等边三角形则6262(03)OBOCADOPtBQtOQtt，，，过P作PEOQ于E，则32PEt···················································································8分113633(62)222BCPQStt=233633228t·····································9分当32t时，BCPQS的面积最小值为6338············································································10分此时3339333324444OQOPOEQEPE，=，xyMCDPQOABNEH222233933442PQPEQE······························································11分【002】解：（1）1，85；（2）作QF⊥AC于点F，如图3，AQ=CP=t，∴3APt．由△AQF∽△ABC，22534BC，得45QFt．∴45QFt．∴14(3)25Stt，即22655Stt．（3）能．①当DE∥QB时，如图4．∵DE⊥PQ，∴PQ⊥QB，四边形QBED是直角梯形．此时∠AQP=90°．由△APQ∽△ABC，得AQAPACAB，即335tt．解得98t．②如图5，当PQ∥BC时，DE⊥BC，四边形QBED是直角梯形．此时∠APQ=90°．由△AQP∽△ABC，得AQAPABAC，即353tt．解得158t．（4）52t或4514t．【注：①点P由C向A运动，DE经过点C．方法一、连接QC，作QG⊥BC于点G，如图6．PCt，222QCQGCG2234[(5)][4(5)]55tt．由22PCQC，得22234[(5)][4(5)]55ttt，解得52t．方法二、由CQCPAQ，得QACQCA，进而可得BBCQ，得CQBQ，∴52AQBQ．∴52t．②点P由A向C运动，DE经过点C，如图7．22234(6)[(5)][4(5)]55ttt，4514t】【003】解.(1)点A的坐标为（4，8）…………………1分将A(4,8)、C（8，0）两点坐标分别代入y=ax2+bxACBPQED图4AC)BPQD图3E)FACBPQED图5AC(E))BPQD图6GAC(E))BPQD图7G8=16a+4b得0=64a+8b解得a=-12,b=4∴抛物线的解析式为：y=－12x2+4x…………………3分（2）①在Rt△APE和Rt△ABC中，tan∠PAE=PEAP=BCAB,即PEAP=48∴PE=12AP=12t．PB=8-t．∴点Ｅ的坐标为（4+12t，8-t）.∴点G的纵坐标为：－12（4+12t）2+4(4+12t）=－18t2+8.…………………5分∴EG=－18t2+8-(8-t)=－18t2+t.∵-18＜0，∴当t=4时，线段EG最长为2.…………………7分②共有三个时刻.…………………8分t1=163，t2=4013，t3=8525．…………………11分【004】（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△∽△．ADBEORFxyy1ly2lM（图3）GCADBEOCFxyy1ly2lG（图1）RMADBEOCFxyy1ly2lG（图2）RM∴BGRGBMCM，即36tRG，∴2RGt．RtRtAFHAMC△∽△，∴11236288223ABCBRGAFHSSSStttt△△△．即241644333Stt．（10分）【005】（1）如图1，过点E作EGBC于点G．·····················1分∵E为AB的中点，∴122BEAB．在RtEBG△中，60B∠，∴30BEG∠．···············2分∴22112132BGBEEG，．即点E到BC的距离为3．·················································3分（2）①当点N在线段AD上运动时，PMN△的形状不发生改变．∵PMEFEGEF，，∴PMEG∥．∵EFBC∥，∴EPGM，3PMEG．同理4MNAB．············································································································4分如图2，过点P作PHMN于H，∵MNAB∥，∴6030NMCBPMH∠∠，∠．∴1322PHPM．∴3cos302MHPM．则35422NHMNMH．在RtPNH△中，222253722PNNHPH．∴PMN△的周长=374PMPNMN．···················································6分②当点N在线段DC上运动时，PMN△的形状发生改变，但MNC△恒为等边三角形．当PMPN时，如图3，作PRMN于R，则MRNR．类似①，32MR．∴23MNMR．··············································································································7分图1ADEBFCG图2ADEBFCPNMGH∵MNC△是等边三角形，∴3MCMN．此时，6132xEPGMBCBGMC．···············································8分当MPMN时，如图4，这时3MCMNMP．此时，61353xEPGM．当NPNM时，如图5，30NPMPMN∠∠．则120PMN∠，又60MNC∠，∴180PNMMNC∠∠．因此点P与F重合，PMC△为直角三角形．∴tan301MCPM．此时，6114xEPGM．综上所述，当2x或4或53时，PMN△为等腰三角形．【006】解：（1）OC=1,所以,q=-1,又由面积知0.5OC×AB=45,得AB=52，设A（a,0）,B(b,0)AB=ba=2()4abab=52，解得p=32,但p0,所以p=32。所以解析式为：2312yxx（2）令y=0，解方程得23102xx，得121,22xx,所以A(12,0),B(2,0),在直角三角形AOC中可求得AC=52,同样可求得BC=5，显然AC2+BC2=AB2，得△ABC是直角三角形。AB为斜边，所以外接圆的直径为AB=52,所以5544m。（3）存在，AC⊥BC，①若以AC为底边，则BD//AC,易求AC的解析式为y=-2x-1,可设BD的解析式为y=-2x+b，把B(2,0)代入得BD解析式为y=-2x+4，解方程组231224yxxyx得D（52，9）②若以BC为底边，则BC//AD,易求BC的解析式为y=0.5x-1,可设AD的解析式为y=0.5x+b，把图3ADEBFCPNM图4ADEBFCPMN图5ADEBF（P）CMNGGRGA(12，0)代入得AD解析式为y=0.5x+0.25，解方程组23120.50.25yxxyx得D(53,22)综上，所以存在两点：（52,9）或(53,22)。【007】【008】证明：（1）∵∠ABC=90°，BD⊥EC，∴∠1与∠3互余，∠2与∠3互余，∴∠1=∠2…………………………………………………1分∵∠ABC=∠DAB=90°，AB=AC∴△BAD≌△CBE…………………………………………2分∴AD=BE……………………………………………………3分（2）∵E是AB中点，∴EB=EA由（1）AD=BE得：AE=AD……………………………5分∵AD∥BC∴∠7=∠ACB=45°∵∠6=45°∴∠6=∠7由等腰三角形的性质，得：EM=MD，AM⊥DE。即，AC是线段ED的垂直平分线。……………………7分（3）△DBC是等腰三角（CD=BD）……………………8分理由如下：由（2）得：CD=CE由（1）得：CE=BD∴CD=BD∴△DBC是等腰三角形。……………………………10分【009】解：（1）①ACx⊥轴，AEy⊥轴，四边形AEOC为矩形．BFx⊥轴，BDy⊥轴，四边形BDOF为矩形．ACx⊥轴，BDy⊥轴，四边形AEDKDOCKCFBK，，均为矩形．·············1分1111OCxACyxyk，，，11AEOCS