九年级数学中考模拟试卷含答案

754298631学生数分数100.590.580.570.560.550.540.5图12018年九年级数学中考模拟试卷（一）数学试卷2018.4（满分150分，100分钟完成）考生注意：1．本试卷含三个大题，共25题．答题时，考生务必按答题要求在答题纸规定的位置上作答，在草稿纸、本调研卷上答题一律无效．2．除第一、二大题外,其余各题如无特别说明，都必须在答题纸的相应位置上写出证明或计算的主要步骤．一、选择题：（本大题共6题，每题4分，满分24分）[每小题只有一个正确选项，在答题纸相应题号的选项上用2B铅笔正确填涂]1．下列实数中，有理数是（▲）（A）2；（B）2.1；（C）；（D）135．2．下列方程有实数根的是（▲）（A）4+2=0x；（B）22=1x；（C）2+21=0xx；（D）111xxx．3．已知反比例函数1yx，下列结论正确的是（▲）（A）图像经过点（-1，1）；（B）图像在第一、三象限；（C）y随着x的增大而减小；（D）当1x时，1y．4．用配方法解方程241=0xx，配方后所得的方程是（▲）（A）2(2)=3x；（B）2(+2)=3x；（C）2(2)=3x；（D）2(+2)=3x．5．“a是实数，20a”这一事件是（▲）（A）不可能事件；（B）不确定事件；（C）随机事件；（D）必然事件．6．某校40名学生参加科普知识竞赛（竞赛分数都是整数），竞赛成绩的频数分布直方图如图1所示，成绩的中位数落在（▲）（A）50.5~60.5分；（B）60.5~70.5分；（C）70.5~80.5分；（D）80.5~90.5分．二、填空题：（本大题共12题，每题4分，满分48分）[在答题纸相应题号后的空格内直接填写答案]7．计算：32()=aa▲．8．因式分解：24=aa▲．9．函数=3yx的定义域是▲．010．不等式组1020.xx，的整数解是▲．11．关于x的方程=2(1)axxa的解是▲．12．抛物线2(3)+1yx的顶点坐标是▲．13．掷一枚材质均匀的骰子，掷得的点数为合数的概率是▲．14．如果点1P（2，1y）、2P（3，2y）在抛物线2+2yxx上，那么1y▲2y．（填“”、“”或“=”）15．如图2，已知在平行四边形ABCD中，E是边AB的中点，F在边AD上，且AF︰FD=2︰1，如果ABa，BCb，那么EF▲．16．如图3，如果两个相似多边形任意一组对应顶点P、P所在的直线都经过同一点O，且有(0)OPkOPk，那么我们把这样的两个多边形叫位似多边形，点O叫做位似中心．已知ABC与ABC是关于点O的位似三角形，3OAOA，则ABC与ABC的周长之比是▲．17．如图4，在△ABC中，BC=7，AC=32，tan1C，点P为AB边上一动点（点P不与点B重合），以点P为圆心，PB为半径画圆，如果点C在圆外，那么PB的取值范围是▲．18．已知，在Rt△ABC中，∠C=90°，AC=9，BC=12，点D、E分别在边AC、BC上，且CD︰CE=3︰4．将△CDE绕点D顺时针旋转，当点C落在线段DE上的点F处时，BF恰好是∠ABC的平分线，此时线段CD的长是▲．图3ABCDEF图2图4POP'三、解答题：（本大题共7题，满分78分）[将下列各题的解答过程，做在答题纸的相应位置上]19．(本题满分10分)计算：10121552(3)2（）．20．(本题满分10分)先化简，再求值：25+3222xxxx（），其中3x．21.(本题满分10分，第（1）、（2）小题，每小题5分）如图5，在Rt△ABC中，∠C=90°，AC=3，BC=4，∠ABC的平分线交边AC于点D，延长BD至点E，且BD=2DE，联结AE．（1）求线段CD的长；（2）求△ADE的面积.22．（本题满分10分）如图6，海中有一个小岛A，该岛四周11海里范围内有暗礁．有一货轮在海面上由西向正东方向航行，到达B处时它在小岛南偏西60°的方向上，再往正东方向行驶10海里后恰好到达小岛南偏西45°方向上的点C处．问：如果货轮继续向正东方向航行，是否会有触礁的危险？（参考数据：21.41，31.73）23.（本题满分12分，第（1）、（2）小题，每小题6分）如图7，在梯形ABCD中，AD∥BC，对角线AC、BD交于点M，点E在边BC上，且DAEDCB，联结AE，AE与BD交于点F．（1）求证：2DMMFMB；（2）联结DE，如果3BFFM，求证：四边形ABED是平行四边形.MFEDCBA图7东北ABC图6EDCBA图524.（本题满分12分，第（1）、（2）、（3）小题，每小题4分）已知：如图8，在平面直角坐标系xOy中，抛物线23yaxbx的图像与x轴交于点A（3，0），与y轴交于点B，顶点C在直线2x上，将抛物线沿射线AC的方向平移，当顶点C恰好落在y轴上的点D处时，点B落在点E处．（1）求这个抛物线的解析式；（2）求平移过程中线段BC所扫过的面积；（3）已知点F在x轴上，点G在坐标平面内，且以点C、E、F、G为顶点的四边形是矩形，求点F的坐标．．25.（本题满分14分，第（1）小题4分，第（2）小题6分，第（3）小题4分）如图9-1，已知扇形MON的半径为2，∠MON=90，点B在弧MN上移动，联结BM，作ODBM，垂足为点D，C为线段OD上一点，且OC=BM，联结BC并延长交半径OM于点A，设OA=x，∠COM的正切值为y．（1）如图9-2，当ABOM时，求证：AM=AC；（2）求y关于x的函数关系式，并写出定义域；（3）当△OAC为等腰三角形时，求x的值.OMNDCBA图9-1OMNDCBA图9-2NMO备用图ABOxy备用图ABOxy图82018年九年级数学中考模拟试卷评分参考一、选择题：1．B；2．C；3．B；4．A；5．D；6．C．二、填空题：7．a；8．4aa；9．3x；10．101、、；11．21a；12．（3，1）；13．13；14．；15．2132ba；16．1︰3；17．3508PB；18．6．三、解答题：19．解：原式=5+5212．································································（8分）=251．·············································································（2分）20．解：原式=2245223xxxx，·····························································（5分）=233223xxxxx，·······················································（1分）=33xx.··················································································（1分）当3x时，原式=3333=32.············································（3分）21．解：（1）过点D作DH⊥AB，垂足为点H．················································（1分）∵BD平分∠ABC，∠C=90°，∴DH=DC=x，········································································（1分）则AD=3x．∵∠C=90°，AC=3，BC=4，∴AB=5.···········································（1分）∵sinHDBCBACADAB，∴435xx，··········································································（1分）∴43x.················································································（1分）（2）1141052233ABDSABDH.·············································（1分）∵BD=2DE，∴2ABDADESBDSDE，································································（3分）∴1015323ADES.·······························································（1分）22．解：过点A作AH⊥BC，垂足为点H．······················································（1分）由题意，得∠BAH=60°，∠CAH=45°，BC=10．·····································（1分）设AH=x，则CH=x．·······································································（1分）在Rt△ABH中，∵tanBHBAHAH，∴10tan60xx，········································（3分）∴310xx，解得53513.65x，·······································（2分）∵13.6511，················································································（1分）∴货轮继续向正东方向航行，不会有触礁的危险．·································（1分）答：货轮继续向正东方向航行，不会有触礁的危险．23．证明：（1）∵AD//BC，∴DAEAEB，··············································（1分）∵DCBDAE，∴DCBAEB，································（1分）∴AE//DC，···········································································（1分）∴FMAMMDMC．···································································（1分）∵AD//BC，∴AMDMMCMB，···················································（1分）∴FMDMMDMB，···································································（1分）即2MDMFMB．（2）设=FMa，则=3BFa，=4BMa．···········································（1分）由2MDMFMB，得24MDaa，∴2MDa，································································