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2019福州质检(彭雪林制)第1页共5页2019年福州市九年级质量检测数学试题一、选择题:本题共10小题,每小题4分,共40分1.下列天气预报的图标中既是轴对称图形又是中心对称图形的是().2.地球绕太阳公转的速度约为110000千米/时,将110000用科学记数法表示正确是().A.1.1×106B.1.1×105C.11×104D.11×1063.已知△ABC∽△DEF,若面积比为4:9,则它们对应高的比是().A.4:9B.16:81C.3:5D.2:34.若正数x的平方等于7,则下列对x的估算正确的是().A.1x2B.2x3C.3x4D.4x55.已知a∥b,将等腰直角三角形ABC按如图所示的方式放置,其中锐角顶点B,直角顶点C分别落在直线a,b上,若∠1=15°,则∠2的度数是().A.15°B.22.5°C.30°D.45°6.下列各式的运算或变形中,用到分配律的是().A.23×32=66B.(ab)2=a2b2C.由x+2=5得x=5-2D.3a+2a=5a7.不透明的袋子中装有除颜色外完全相同的a个白球、b个红球、c个黄球,则任意摸出一个球是红球的概率是().A.cabB.cbacaC.cbabD.bca8.如图,等边三角形ABC边长为5、D、E分别是边AB、AC上的点,将△ADE沿DE折叠,点A恰好落在BC边上的点F处,若BF=2,则BD的长是().A.724B.821C.3D.29.已知Rt△ABC,∠ACB=90,AC=3,BC=4,AD平分∠BAC,则点B到射线AD的距离是().A.2B.3C.5D.310.一套数学题集共有100道题,甲、乙和丙三人分别作答,每道题至少有一人解对,且每人都解对了其中的60道.如果将其中只有1人解对的题称作难题,2人解对的题称作中档题,FCABDEA.B.C.D.(第8题)(第5题)A21CBab2019福州质检(彭雪林制)第2页共5页3人都解对的题称作容易题,那么下列判断一定正确的是().A.容易题和中档题共60道B.难题比容易题多20道C.难题比中档题多10道D.中档题比容易题多15道二、填空题:本题共6小题,每小题4分,共24分11.分解因式:m3-4m=________.12.若某几何体从某个方向观察得到的视图是正方形,则这个几何体可以是________.13.如图是甲、乙两射击运动员10次射击成城的折线统计图,则这10次射击成绩更稳定的运动员是________.14.若分式56mm的值是负整数,则整数m的值是________.15.在平面直角坐标系中,以原点为圆心,5为半径的⊙O与线y=kx+2k+3(k≠0)交于A,B两点,则弦AB长的最小值是________.16.如图,在平面直角坐标系中,O为原点,点A在第一象限,点B是x轴正半轴上一点,∠OAB=45°,双曲线y=xk过点A,交AB于点C,连接OC,若OC⊥AB,则tan∠ABO的值是________.三、解答题:本题共9小题,共86分17.(8分)计算:|-3|+3·tan30°-(3.14-)°18.(8分)如图,已知∠1=∠2,∠B=∠D,求证:CB=CD.乙甲次数12345678910O678910成绩/环21DABC(第13题)yxCOAB(第16题)2019福州质检(彭雪林制)第3页共5页19.(8分)先化简,再求值:(1-x1)÷2212xxx,其中x=3+120.(8分)如图,在Rt△ABC中,∠ACB=90°,BD平分∠ABC.求作⊙O,使得点O在边AB上,且⊙O经过B、D两点;并证明AC与⊙O相切.(尺规作图,保留作图痕迹,不写作法)21.(8分)如图,将△ABC沿射线BC平移得到△A'B'C',使得点A'落在∠ABC的平分线BD上,连接AA'、AC'.(1)判断四边形ABB'A'的形状,并证明;(2)在△ABC中,AB=6,BC=4,若AC⊥A'B',求四边形ABB'A'的面积.DCBAA'B'C'CBAD2019福州质检(彭雪林制)第4页共5页22.(10分)为了解某校九年级学生体能训练情况,该年级在3月份进行了一次体育测试,决定对本次测试的成绩进行抽样分析.已知九年级共有学生480人,请按要求回答下列问题:(1)把全年级同学的测试成绩分别写在没有明显差别的小纸片上,揉成小球,放到一个不透明的袋子中,充分搅拌后,随意抽取30个,展开小球,记录这30张纸片中所写的成绩得到一个样本,你觉得上面的抽取过程是简单随机抽样吗?答:________(填“是”或“不是”)(2)下表是用简单随机抽样方法抽取的30名同学的体育测试成绩(单位:分):596977737262797866918584838486878885868990979198909596939299若成绩为x分,当x≥90时记为A等级,80≤x90时记为B等级,70≤x80时记为C等级,x70时记为D等级,根据表格信息,解答下列问题:①本次抽样调查获取的样本数据的中位数是________;估计全年级本次体育测试成绩在A、B两个等级的人数是________;②经过一个多月的强化训练发现D等级的同学平均成绩提高15分,C等级的同学平均成绩提高10分,B等级的同学平均成绩提高5分,A等级的同学平均成绩没有变化,请估计强化训练后全年级学生的平均成绩提高多少分?23.(10分)某汽车销售公司销售某厂家的某款汽车,该款汽车现在的售价为每辆27万元,每月可售出两辆.市场调查反映:在一定范国内调整价格,每辆降低0.1万元,每月能多卖一辆.已知该款汽车的进价为每辆25万元.另外,月底厂家根据销售量一次性返利给销售公司,销售量在10辆以内(含10辆),每辆返利0.5万元:销售量在10辆以上,超过的部分每辆返利1万元.设该公司当月售出x辆该款汽车.(总利润=销售利润十返利)(1)设每辆汽车的销售利润为y万元,求y与x之间的函数关系式;(2)当x10时,该公司当月销售这款汽车所获得的总利润为20.6万元,求x的值2019福州质检(彭雪林制)第5页共5页24.(13分)在正边形ABCD中,E是对角线AC上一点(不与点A、C重合),以AD、AE为邻边作平行四边形AEGD,GE交CD于点M,连接CG.(1)如图1,当AE21AC时,过点E作EF⊥BE交CD于点F,连接GF并延长交AC于点H.①求证:EB=EF;②判断GH与AC的位置关系,并证明.(2)过点A作AP⊥直线CG于点P,连接BP,若BP=10,当点E不与AC中点重合时,求PA与PC的数量关系.25.(13分)已知抛物线y=-21(x+5)(x-m)(m0)与x轴交于点A、B(点A在点B的左边),与y轴交于点C.(1)直接写出点B、C的坐标;(用含m的式子表示)(2)若抛物线与直线y=21x交于点E、F,且点E、F关于原点对称,求抛物线的解析式;(3)若点P是线段AB上一点,过点P作x轴的垂线交抛物线于点M,交直线AC于点N,当线段MN长的最大值为825时,求m的取值范围.MHFGDABCEDABC2019福州质检(彭雪林制)第6页共5页参考答案一、ABDBCDCBCB二、11.m(m+2)(m-2)12.正方体13.甲14.415.4316.215三、17.解:原式33313·······························································6分311·······································································7分3.···········································································8分18.证明:∵∠1∠2,∴∠ACB∠ACD.································3分在△ABC和△ADC中,BDACBACDACAC,,,∴△ABC≌△ADC(AAS),···················································6分∴CBCD.·····································································8分注:在全等的获得过程中,∠B=∠D,AC=AC,△ABC≌△ADC,各有1分.19.解:原式22121xxxxx···························································1分221(1)xxxx································································3分1xx,·······································································5分当31x时,原式31311···············································6分313333.···············································8分20.解:······································3分如图,⊙O就是所求作的圆.···················································4分证明:连接OD.∵BD平分∠ABC,∴∠CBD∠ABD.······················································5分∵OBOD,∴∠OBD∠ODB,∴∠CBD∠ODB,·····················································6分BCADO21CABD2019福州质检(彭雪林制)第7页共5页∴OD∥BC,∴∠ODA∠ACB又∠ACB90°,∴∠ODA90°,即OD⊥AC.·······························································7分∵点D是半径OD的外端点,∴AC与⊙O相切.························································8分注:垂直平分线画对得1分,标注点O得1分,画出⊙O得1分;结论1分.21.(1)四边形ABB′A′是菱形.··························································1分证明如下:由平移得AA′∥BB′,AA′BB′,∴四边形ABB′A′是平行四边形,∠AA′B∠A′BC.·········2分∵BA′平分∠ABC,∴∠ABA′∠A′BC,∴∠AA′B∠A′BA,···············································3分∴ABAA′,∴□ABB′A′是菱形.················································4分(2)解:过点A作AF⊥BC于点F.由(1)得BB′BA6.由平移得△A′B′C′≌△ABC,∴B′C′BC4,∴BC′10.·························5分∵AC′⊥A′B′,∴∠B′EC′90°,∵AB∥A′B′,∴∠BAC′∠B′EC′90°.在Rt△ABC′中,AC′228BCAB.·······························6分∵S△ABC′1122ABACBCAF,∴AF245ABACBC,··
本文标题:福建省福州市2019年九年级质检数学卷及答案
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