2009年全国各地中考数学压轴题参考答案及评分标准(二)

2009年全国各地中考数学压轴题参考答案及评分标准（二）101．解：（1）当t＝4－3时，四边形FBCG为正方形．······································1分当0＜t≤4时，四边形AEGD为平行四边形．··································2分（2）由题意知点D、C的坐标分别为（1，3），（5，3）························4分∵抛物线经过原点O（0，0），∴设抛物线的解析式为y＝ax2＋bx（a≠0）．将D、C两点坐标代入得35253＝＋＝＋baba，解得35653＝＝－ba····················································6分∴抛物线的解析式为y＝－53x2＋536x········································7分（3）存在．∵点Q在抛物线上，∴点Q(x，－53x2＋536x)过点Q作QM⊥x轴于M，如图．∵∠DAB＝60°，CD＝4，∴AB＝5．∴S△ABQ＝21AB·QM＝21×5×|－53x2＋536x|＝21|－3x2＋36x|······················································8分∵EG的延长线与抛物线交于x轴的上方∴－53x2＋536x＞0，∴－3x2＋36x＞0．∴S△ABQ＝21(－3x2＋36x)∵S梯形ABCD＝21(CD＋AB)·BC＝21(4＋5)×2×23＝329······································································9分若S△ABQ＝S梯形ABCD，则21(－3x2＋36x)＝329．整理得x2－6x＋9＝0，解得x＝3．················································10分DCB(A)FEGOxyQM把x＝3代入抛物线的解析式，得y＝－53×32＋536×3＝359，即MQ＝359．∵∠QEM＝60°，∴EM＝o60tanMQ＝3359＝59．······························11分∴t＝3－59＝56（秒）．∴存在这样的时刻t，使得△ABQ的面积与梯形ABCD的面积相等，此时t＝56秒．················································································12分102．解：（1）A（0，2），B（4，0）···································································2分设直线AB的解析式为y＝kx＋b，则有042＝＋＝bkb，解得221＝＝－bk∴直线AB的解析式为y＝－21x＋2．······························3分（2）ⅰ）①当点E在原点和x轴正半轴上时，重叠部分为△CDE，如图①∴S＝S△CDE＝21CE·CD＝21BC·CD＝21(4－x)(－21x＋2)＝41x2－2x＋4∵当点E与原点O重合时，CE＝21BO＝2，∴2≤x＜4．···············4分②当点E在x轴的负半轴上时，设DE与y轴交于点F，则重叠部分为梯形CDFO，如图②∵△FEO∽△ABO，∴OEOF＝OBOA＝21，∴OF＝21OE．又∵OE＝4－2x，∴OF＝21(4－2x)＝2－x∴S＝S梯形CDFO＝21(OF＋CD)·OC＝21[2－x＋(－21x＋2)]·x＝－43x2＋2x．············5分∵当点C与原点O重合时，点C的坐标为（0，0）∴0＜x＜2．··········································································6分综合①②得S＝424124322＋＋－－xxxx········································7分ⅱ）①∵当2≤x＜4时，S＝41x2－2x＋4＝41(x－4)2，∴抛物线的对称轴为x＝4∵抛物线开口向上，∴当2≤x＜4时，S随x的增大而减小图①AOBxyDCE（2≤x＜4）（0＜x＜2）图②AOBxyDCEF∴当x＝2时，S的最大值＝41(2－4)2＝1．······························8分②当0＜x＜2时，S＝－43x2＋2x＝－43(x－34)2＋34，∴抛物线的对称轴为x＝34∵抛物线开口向下，∴当x＝34时，S有最大值为34．················9分综合①②，当x＝34时，S有最大值为34．······························10分ⅲ）存在，点C的坐标为（23，0）和（25，0）．·····························14分附：详解：①当Rt△ADE以点A为直角顶点时，作AE⊥AB交x轴负半轴于点E，如图③∵Rt△AOE∽Rt△BOA，∴OAOE＝OBOA＝21∵OA＝2，∴OE＝1∵OE＋2OC＝4，∴OC＝21×(4－1)＝23∴点C的坐标为（23，0）②当Rt△ADE以点E为直角顶点时，如图④∵∠AED＝90°，∴∠AEO＋∠DEC＝90°∵∠DEC＝∠DBC，∴∠AEO＋∠DBC＝90°∵∠BAO＋∠DBC＝90°，∴∠AEO＝∠BAO∴Rt△AOE∽Rt△BOA，∴OAOE＝OBOA＝21，∴OE＝1∴OC＝1＋21×(4－1)＝25∴点C的坐标为（25，0）综上所述，存在这样的点C，使得△ADE为直角三角形，点C的坐标为：C1（23，0）和C2（25，0）103．解：（1）把x＝0代入y＝－2x＋4，得y＝4．∴C（0，4）．··············································································1分∵矩形OABC，∴BC＝OA＝3，AB＝OC＝4．∴B（－3，4）．···········································································2分（2）∵二次函数y＝－21x2＋bx＋c的图象经过B、C两点∴4433212＝＝－－－)(ccb＋解得423＝＝cb······································4分∴二次函数的解析式为y＝－21x2－23x＋4．······································6分图④AOBxyDCE图③AOBxyDCE（3）证明：连结AC，在Rt△AOC中，AC＝22OCOA＋＝2243＋＝5．∵y＝－2x＋4，当y＝0时，x＝2，∴D（2，0）．∵AD＝OA＋OD＝3＋2＝5，∴AC＝AD．∵P是CD的中点，∴AP⊥CD．·····················································9分（4）存在．······················································································10分方法一：假设四边形APCM为矩形，过点M作MN⊥x轴于N．在Rt△COD中，∵CD＝22ODOC＋＝2224＋＝52．∴MA＝PC＝21CD＝5∵MA∥CD，∴∠MAN＝∠CDO．又∵∠MNA＝∠COD＝90°，∴△MNA∽△COD．·····························12分∴COMN＝ODNA＝CDMA．∴MN＝5254＝2，NA＝5252＝1．∵NO＝NA＋OA＝1＋3＝4∴M（－4，2）．······································13分把x＝－4代入y＝－21x2－23x＋4，得y＝－21×(－4)2－23×(－4)＋4＝2．∴点M在抛物线上．∴存在这样的点M，使以A、P、C、M为顶点的四边形为矩形．··········14分方法二（原参考答案中没有，本人添加，仅供参考）：过点作AM∥CD交抛物线于点M，连结CM．∵MA∥CD，直线CD的斜率为－2．∴设直线MA的解析式为y＝－2x＋m，把A（－3，0）代入，得m＝－6．∴直线MA的解析式为y＝－2x－6联立42321622＋xxxyy－－－－＝＝解得24＝－＝yx∴M（－4，2）．∴MA＝22234＋＋)(－＝5在Rt△COD中，∵CD＝22ODOC＋＝2224＋＝52．∴PC＝21CD＝5∴MA＝PC．∵MA∥CD，∴四边形APCM为平行四边形．又∵AP⊥CD，∴∠APC＝90°．∴四边形APCM为矩形．OCBAPDxyNM∴存在这样的点M，使以A、P、C、M为顶点的四边形为矩形．104．解：（1）方法一：∵抛物线经过原点O（0，0），∴设抛物线的解析式为y＝ax2＋bx（a≠0）．把A（1，1），B（3，1）代入上式得：·············································1分1391＝＝baba＋＋解得3431＝＝ba－·······················································3分∴经过O、A、B三点的抛物线解析式为y＝－31x2＋34x·······················4分方法二：∵A（1，1），B（3，1），∴抛物线的对称轴为直线x＝2．设抛物线解析式为y＝a(x－2)2＋h（a≠0）．······································1分把O（0，0），A（1，1）代入上式得：12102022＝＝)()(－－haha＋＋解得3431＝＝ha－·················································3分∴经过O、A、B三点的抛物线解析式为y＝－31(x－2)2＋34·················4分（2）①当0＜t≤2时，重叠部分为△OPQ，过点A作AD⊥x轴于点D，如图1．在Rt△AOD中，AD＝OD＝1，∠AOD＝45°．在Rt△OPQ中，OP＝t，∠OPQ＝∠QOP＝45°．∴OQ＝PQ＝22t．∴S＝S△OPQ＝21OQ·PQ＝21×22t×22t＝41t2（0＜t≤2）·························································6分②当2＜t≤3时，设PQ交AB于点E，重叠部分为梯形AOPE，作EF⊥x轴于点F，如图2．∵∠OPQ＝∠QOP＝45°∴四边形AOPE是等腰梯形∴AE＝DF＝t－2．∴S＝S梯形AOPE＝21(AE＋OP)·AD＝21(t－2＋t)×1＝t－1（2＜t≤3）······················································8分③当3＜t＜4时，设PQ交AB于点E，交BC于点F，重叠部分为五边形AOCFE，如图3．∵B（3，1），OP＝t，∴PC＝CF＝t－3．2OABCxy113PQ图1DP2OABCxy113图2DEFQP2OABCxy113DEFQ∵△PFC和△BEF都是等腰直角三角形∴BE＝BF＝1－(t－3)＝4－t∴S＝S五边形AOCFE＝S梯形OABC－S△BEF＝21(2＋3)×1－21(4－t)2＝－21t2＋4t－211（3＜t＜4）·················································