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2020年山东省济南市中考数学试题(word版)(含答案)

发布时间：2022-11-08 08:17:24   来源：文档文库   
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2020年山东省济南市中考数学试题（word版）（含答案）济南市2018年初三年级学业水平考试数学试题本卷须知：1．本试卷分第一卷〔选择题〕和第二卷〔非选择题〕两部分，总分值120分．第一卷1至2页，第二卷3至8页．考试时刻120分钟．2．答第一卷前，考生务必将自己的姓名、准考证号、考试科目用2B铅笔涂写在答题卡上，并同时将考点、姓名、准考证号、座号填写在试卷规定的地点．3．选择题选出答案后，用2B铅笔把答题卡上对应题目的正确答案标号涂黑．如需改动，用橡皮擦洁净后，再选涂其它答案，答案写在试卷上无效．4．数学考试不承诺使用运算器，考试终止后，应将本试卷和答题卡一并交回．第一卷〔选择题共48分〕一、选择题〔本大题共12个小题，每题4分，共48分．在每题给出的四个选项中，只有一项为哪一项符合题目要求的．〕1．2+〔－2〕的值是1B．C．0D．442．一组数据0、1、2、2、3、1、3、3的众数是A．0B．1C．2D．33．图中的几何体是由7个大小相同的小正方体组成的，该几何体的俯视图为第3题图D．C．B．A．4．作为历史上第一个正式提出〝低碳世博〞理念的世博会，上海世博会从一开始就确定以〝低碳、和谐、可连续进展的都市〞为主题．现在在世博场馆和周边共运行着一千多辆新能源汽车，为目前世界上规模最大的新能源汽车示范运行，估量将减少温室气体排放约28400吨．将28400吨用科学记数法表示为A．0.284×105吨B．2.84×104吨C．28.4×103吨D．284×102吨A．－4xy45．二元一次方程组的解是xy2x3A．y7x1B．y1第4题图x7C．y3x3D．y16．以下各选项的运算结果正确的选项是2015C．x6x2x3D．(ab2a2b27．在一次体育课上，体育老师对九年级一班的40名同学进行了10跳远项目的测试，测试所得分数及相应的人数如下图，那么这50试的平均分为A．(2x238x6B．5a2b2a2b3人数〔人〕立定次测6分8分10分第7题图分数
35405A．分B．分C．分D．8分4338．一次函数y2x1的图象通过哪几个象限A．一、二、三象限C．一、三、四象限B．一、二、四象限D．二、三、四象限AOMD．1BNCDN分不为OB、9．如下图，正方形ABCD中，对角线AC、BD交于点O，点M、OC的中点，那么cos∠OMN的值为1A．223B．C．22第9题图10．二次函数yx2x2的图象如下图，那么函数值y＜0时yx的取值范畴是A．x＜－1－1O2xB．x＞2C．－1＜x＜2第10题图D．x＜－1或x＞211．观看以下图形及图形所对应的算式，依照你发觉的规律运算1+8+16+24+……+8n〔n是正整数〕的结果为……⑴⑶⑵1+8+16+24=?1+8=?1+8+16=?A．(2n12B．(2n12C．(n22D．n212．如下图，矩形ABCD中，AB=4，BC=43，点E是折线段A－DAEP第11题图D－C上的一个动点〔点E与点A不重合〕，点P是点A关于BE的对称点．在点E运动的过程中，使△PCB为等腰三角形的点E的位置共有A．2个B．3个C．4个D．5个B绝密★启用前济南市2018年初三年级学业水平考试数学试题本卷须知：1．第二卷共6页．用蓝、黑色钢笔或圆珠笔直截了当答在考试卷上．2．答卷前将密封线内的项目填写清晰．第二卷〔非选择题共72分〕得分评卷人第12题图C二、填空题〔本大题共5个小题，每题3分，共15分．把答案填在题中的横线上．〕213．分解因式：x2x1=．DA14．如下图，△DEF是△ABC沿水平方向向右平BF移后的对应CE第14题图
图形，假设∠B=31°，∠C=79°，那么∠D的度数是度．15．解方程23的结果是．x12x316．如下图，点A是双曲线yO1xBC第16题图17．如下图，△ABC的三个顶点的坐标分不为A(－1，3、B(－2，－2、C(4，－2，那么△ABC外接圆半径的长度为．yA1在第二象限的分支上的任意一点，点B、C、D分不是点A关于x轴、x原点、y轴的对称点，那么四边形ABCD的面积是．y1yxDAOxCB第17题图三、解答题〔本大题共7个小题，共57分．解承诺写出文字讲明、证明过程或演算步骤．〕得分评卷人18．(本小题总分值7分x2x⑴解不等式组：2x≤4⑵如下图，在梯形ABCD中，BC∥AD，AB=DC，点M是AD的中点．求证：BM=CM．B得分评卷人152C19．(本小题总分值7分+(30AM第18题图D⑴运算：
⑵如下图，△ABC中，∠C=90°，∠B=30°，AD是△ABC的角平分线，假设AC=3．求线段AD的长．CDB得分评卷人第19题图20．(本小题总分值8分如下图，有一个能够自由转动的圆形转盘，被平均分成四个扇形，四个扇形内分不标有数字1、2、－3、－4．假设将转盘转动两次，每一次停止转动后，指针指向的扇形内的数字分不记为a、b〔假设指针恰好指在分界线上，那么该次不计，重新转动一次，直至指针落在扇形内〕．请你用列表法或树状图求a与b的乘积等于2的概率．1－4－32得分评卷人21．(本小题总分值8分第20题图如下图，某幼儿园有一道长为16米的墙，打算用32米长的围栏靠墙围成一个面积为120平方米的矩形草坪ABCD．求该矩形草坪BC边的长．16米AD草坪BC得分评卷人第21题图22．(本小题总分值9分如下图，菱形ABCD的顶点A、B在x轴上，点A在点B的左侧，点D在y轴的正半轴上，∠BAD=60°，点A的坐标为(－2，0．⑴求线段AD所在直线的函数表达式．⑵动点P从点A动身，以每秒1个单位长度的速度，按照A→D→C→B→A的顺序在菱形的边上匀速运动一周，设运动时刻为t秒．求t为何值时，以点P为圆心、以1为半径的圆与对角线AC相切？yCD得分评卷人PAAO第22题图Bx
23．(本小题总分值9分：△ABC是任意三角形．⑴如图1所示，点M、P、N分不是边AB、BC、CA的中点．求证：∠MPN=∠A．AM1AN1，，点P1、P2是边BC的三等分AB3AC3点，你认为∠MP1N+∠MP2N=∠A是否正确？请讲明你的理由．⑵如图2所示，点M、N分不在边AB、AC上，且AM1AN1，，点P1、P2、……、P2018AB2010AC2010是边BC的2018等分点，那么∠MP1N+∠MP2N+……+∠MP2018N=____________．〔请直截了当将该小咨询的答案写在横线上．〕AAAMNMNMN……BCBP1P2……P2018CCBP1P2P⑶如图3所示，点M、N分不在边AB、AC上，且第23题图1第23题图2第23题图3得分评卷人24．(本小题总分值9分如下图，抛物线yx22x3与x轴交于A、B两点，直线BD的函数表达式为y3x33，抛物线的对称轴l与直线BD交于点C、与x轴交于点E．⑴求A、B、C三个点的坐标．⑵点P为线段AB上的一个动点〔与点A、点B不重合〕，以点A为圆心、以AP为半径的圆弧与线段AC交于点M，以点B为圆心、以BP为半径的圆弧与线段BC交于点N，分不连接AN、BM、MN．①求证：AN=BM．②在点P运动的过程中，四边形AMNB的面积有最大值依旧有最小值？并求出该最大值或最小值.yDlCMNxAOEPB济南市2018年初三年级学业水平考试第24题图
数学试题参考答案及评分标准一、选择题题号答案二、填空题13.(x1214.7015.x916.417.13三、解答题x2＞x18.(1解：2x≤41C2D3C4B5D6A7B8B9B10C11A12C①②解不等式①，得x＞1，·····················································1分解不等式②，得x≥－2，·····················································2分∴不等式组的解集为x＞1.······················································3分(2证明：∵BC∥AD，AB=DC，∴∠BAM=∠CDM，·····················································1分∵点M是AD的中点，∴AM=DM，·······························································2分∴△ABM≌△DCM，····················································3分∴BM=CM.································································4分19.(1解：原式=52(52(52··············································1分(30·=5－2+1·······························································2分=5－1·································································3分(2解：∵△ABC中，∠C=90º，∠B=30º，∴∠BAC=60º，∵AD是△ABC的角平分线，∴∠CAD=30º，·····························································1分∴在Rt△ADC中，ADAC·······································2分cos302=3×······································3分3=2.·············································4分20.解：a与b的乘积的所有可能显现的结果如下表所示：ba12－3－4112－3－4224－6－8－3－3－6912－4－4－81216
··························································································6分总共有16种结果，每种结果显现的可能性相同，其中ab=2的结果有2种，··································································································7分1∴a与b的乘积等于2的概率是.··············································8分821.解：设BC边的长为x米，依照题意得········································1分x32x·······························································4分120，·2解得：x112，x220，···························································6分∵20＞16，∴x220不合题意，舍去，····················································7分答：该矩形草坪BC边的长为12米.·····································8分22.解：⑴∵点A的坐标为(－2，0，∠BAD=60°，∠AOD=90°，∴OD=OA·tan60°=23，∴点D的坐标为〔0，23〕，················································1分设直线AD的函数表达式为ykxb，2kb0k3，解得，b23b23∴直线AD的函数表达式为y3x23.·······························3分⑵∵四边形ABCD是菱形，∴∠DCB=∠BAD=60°，∴∠1=∠2=∠3=∠4=30°，AD=DC=CB=BA=4，··························································5分如下图：①点P在AD上与AC相切时，AP1=2r=2，∴t1=2.·············································································6分②点P在DC上与AC相切时，yCP2=2r=2，P2D∴AD+DP2=6，23∴t2=6.······························7分③点P在BC上与AC相切时，CP3=2r=2，∴AD+DC+CP3=10，∴t3=10.······························8分④点P在AB上与AC相切时，AP4=2r=2，P1P3C14AOP4第22题图Bx
∴AD+DC+CB+BP4=14，∴t4=14，∴当t=2、6、10、14时，以点P为圆心、以1为半径的圆与对角线AC相切.···············································9分23.⑴证明：∵点M、P、N分不是AB、BC、CA的中点，∴线段MP、PN是△ABC的中位线，∴MP∥AN，PN∥AM，···············1分∴四边形AMPN是平行四边形，····2分M∴∠MPN=∠A.······················3分12⑵∠MP1N+∠MP2N=∠A正确.···············4分如下图，连接MN，·························5分ANAMAN1，∠A=∠A，ABAC3∴△AMN∽△ABC，∵∴∠AMN=∠B，BP1P2C第23题图MN1，BC31∴MN∥BC，MN=BC，··················6分3∵点P1、P2是边BC的三等分点，∴MN与BP1平行且相等，MN与P1P2平行且相等，MN与P2C平行且相等，∴四边形MBP1N、MP1P2N、MP2CN差不多上平行四边形，∴MB∥NP1，MP1∥NP2，MP2∥AC，························································7分∴∠MP1N=∠1，∠MP2N=∠2，∠BMP2=∠A，∴∠MP1N+∠MP2N=∠1+∠2=∠BMP2=∠A.······················································8分⑶∠A.·········································9分24.解：⑴令x22x30，解得：x11,x23，∴A(－1，0，B(3，0························2分∵yx22x3=(x124，∴抛物线的对称轴为直线x=1，将x=1代入y3x33，得y=23，∴C〔1，23〕.····························3分⑵①在Rt△ACE中，tan∠CAE=∴∠CAE=60º，MCFDlyNCE3，AEAOEPBx第24题图
由抛物线的对称性可知l是线段AB的垂直平分线，∴AC=BC，∴△ABC为等边三角形，·····················································4分∴AB=BC=AC=4，∠ABC=∠ACB=60º，又∵AM=AP，BN=BP，∴BN=CM，∴△ABN≌△BCM，∴AN=BM.·········································································5分②四边形AMNB的面积有最小值．········································6分设AP=m，四边形AMNB的面积为S，3×42=43，4∴CM=BN=BP=4－m，CN=m，过M作MF⊥BC,垂足为F,由①可知AB=BC=4，BN=CM=BP，S△ABC=那么MF=MC•sin60º=3(4m，233211∴S△CMN=CNMF=m•···················7分(4m=m3m，·2422∴S=S△ABC－S△CMN=43－〔=32m3m〕43····························································8分(m2233·4∴m=2时，S取得最小值33.··············································9分

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