2023年山东省济南市东南片区中考一模数学试题（含答案）

2023年九年级学业水平模拟测试一数学试题一、选择题，本大题共10个小题，每小题4分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的．）1．的相反数是（）A． B．2 C． D．2．如图所示的几何体，其俯视图是（）正面A． B． C． D．3．为完善城市轨道交通建设，提升城市公共交通服务水平，济南市城市轨道交通2020~2025年第二期建设规划地铁总里程约为159600米．把数字“159600”用科学记数法表示为（）A． B． C． D．4．如图，平行线，被直线所截，平分，若，则的度数是（）A．39° B．51° C．78° D．102°5．下列图案中，既是中心对称图形又是轴对称图形的是（）A． B． C． D．6．已知实数，在数轴上对应点的位置如图所示，则下列判断正确的是（）A． B． C． D．7．“二十四节气”是中华农耕文明与天文学智慧的结晶，被国际气象界誉为“中国第五大发明”．小明购买了“二十四节气”主题邮票，他要将“立春”“立夏”“秋分”三张邮票中的两张送给好朋友小亮．小明将它们背面朝上放在桌面上（邮票背面完全相同），让小亮从中随机抽取一张（不放回），再从中随机抽取一张，则小亮抽到的两张邮票恰好是“立春”和“秋分”的概率是（）A． B． C． D．8．函数与在同一坐标系中的图象如图所示，则函数的大致图象为（）A． B． C． D．9．如图，已知锐角，按如下步骤作图：（1）在射线上取一点，以点为圆心，长为半径作，交射线于点，连接；（2）分别以点，为圆心，长为半径作弧，交于点，；③连接，，．根据以上作图过程及所作图形，下列结论中错误的是（）A． B．若，则C． D．10．已知二次函数，将其图象在直线左侧部分沿轴翻折，其余部分保持不变，组成图形．在图形上任取一点，点的纵坐标的取值满足或，其中．令，则的取值范围是（）A． B． C． D．二、填空题（本大题共6个小题，每小题4分，共24分．）11．因式分解：______．12．如图，一个可以自由转动的转盘，被分成了9个相同的扇形，转动转盘，转盘停止时，指针落在阴影区域的概率等于______．13．比大的最小整数是______．14．如图，扇形纸片的半径为4，沿折叠扇形纸片，点恰好落在上的点处，图中阴影部分的面积为______．15．如图（1），已知小正方形的面积为1，把它的各边延长一倍得到新正方形；把正方形边长按原法延长一倍得到正方形（如图（2））…；以此下去，则正方形的面积为______．16．正方形的边长为8，点、分别在边、上，将四边形沿折叠，使点落在处，点落在点处，交于．以下结论：①当为中点时，三边之比为；②连接，则；③当三边之比为时，为中点；④当在上移动时，周长不变．其中正确的有______（写出所有正确结论的序号）．三、解答题（本大题共10个小题，共86分．解答应写出文字说明、证明过程或演算步骤．）17．（6分）计算：18．（6分）解不等式组：，并写出它的所有非负整数解．19．（6分）在中，点，在对角线上，且，连接，．求证：．20．（8分）为深入学习贯彻党的二十大精神，某校开展了以“学习二十大，永远跟党走，奋进新征程”为主题的知识竞赛．发现该校全体学生的竞赛成绩（百分制）均不低于60分，现从中随机抽取名学生的竞赛成绩进行整理和分析（成绩得分用表示，共分成四组），并绘制成如下的竞赛成绩分组统计表和扇形统计图，其中“”这组的数据如下：82，83，83，84，84，85，85，86，86，86，87，89．竞赛成绩分组统计表组别竞赛成绩分组频数平均分18652a763b854c94请根据以上信息，解答下列问题：（1）______．（2）“”这组数据的众数是______分，方差是______；（3）随机抽取的这名学生竞赛成绩的中位数是______分，平均分是______分；（4）若学生竞赛成绩达到85分以上（含85分）为优秀，请你估计全校1200名学生中优秀学生的人数．21．（8分）如图，一艘游轮在处测得北偏东45°的方向上有一灯塔B，游轮以海里/时的速度向正东方向航行2小时到达处，此时测得灯塔在处北偏东15°的方向上．（1）求到直线的距离；（2）求游轮继续向正东方向航行过程中与灯塔的最小距离是多少海里?（结果精确到1海里，参考数据：，，，，）22．（8分）如图，是的直径，，是上两点，且，过点的切线交的延长线于点，交的延长线于点，连结，交于点．（1）求证：；（2）若，的半径为2，求的长．23．山地自行车越来越受到中学生的喜爱，各品牌相继投放市场．某车行经营的A型车去年销售总额为50000元，今年每辆销售价比去年降低400元，若卖出的数量相同，销售总额将比去年减少20%．（1）今年A型车每辆售价多少元?（2）该车行计划新进一批A型车和新款B型车共60辆，且B型车的进货数量不超过A型车数量的两倍，应如何进货才能使这批车获利最多?A，B两种型号车的进货和销售价格如下表：A型车B型车进货价格（元）11001400销售价格（元）今年的销售价格200024．如图，在矩形中，，，分别以，所在的直线为轴和轴建立平面直角坐标系．反比例函数的图象交于点，交于点，．（1）求的值与点的坐标；（2）在轴上找一点，使的周长最小，请求出点的坐标；（3）在（2）的条件下，若点是轴上的一个动点，点是平面内的任意一点，试判断是否存在这样的点，，使得以点，，，为顶点的四边形是菱形．若存在，请直接写出符合条件的点坐标；若不存在，请说明理由．25．（12分）某校数学兴趣学习小组在一次活动中，对一些特殊几何图形具有的性质进行了如下探究：（1）发现问题：如图1，在等腰中，，点是边上任意一点，连接，以为腰作等腰，使，，连接．求证：．（2）类比探究：如图2，在等腰中，，，，点是边上任意一点，以为腰作等腰，使，．在点运动过程中，是否存在最小值？若存在，求出最小值，若不存在，请说明理由．（3）拓展应用：如图3，在正方形中，点是边上一点，以为边作正方形，是正方形的中心，连接．若正方形的边长为8，，求的面积．26．（12分）抛物线过点，点，顶点为，与轴相交于点．点是该抛物线上一动点，设点的横坐标为．（1）求抛物线的表达式及点的坐标；（2）如图1，连接，，，若的面积为3，求的值；（3）连接，过点作于点，是否存在点，使得．如果存在，请求出点的坐标；如果不存在，请说明理由．九年级数学试卷（2023.4）参考答案及评分标准一、选择题题号12345678910答案BCCABDCADD二、填空题11．4(a+1)(a-1)12．4913．314．163π-8三、解答题17．原式=3-22-2+2×22······························································································4=1-18．解：解不等式①，得：x＜5，··················································································2分解不等式②，得：x＜4，······························································································4分原不等式组的解集是x＜4．··························································································5分非负整数解为0，1，2，3·························································································6分19．证明：∵四边形ABCD是平行四边形，∴AD＝BC，AD∥BC，····································································································2分∴∠DAE＝∠BCF，··································································································3分∵AE=FC，∠DAE＝∠BCF，AD＝BC，∴△ADE≌△CBF（SAS），·······························································································4分∴∠DEA=∠BFC············································································································5分∴∠DEC=∠BFA∴DE∥BF·····················································································································6分20．解：（1）20························································································2分（2）86，3.5································································································4分（3）85.583.6··························································································6分（4）1200×2750=648答：获奖的人数是648人．································································································8分21．解：（1）如图，由题意可得，∠CAB＝45°，过点C作CE⊥AB于点E，·······································1分在△ABC中，∠BAC＝45°，∴△ACE是等腰直角三角形，·································2分由题意得：AC＝2×202=402，∴CE=22即点C到线段AB的距离为40海里；····················································································4分（2）由题意可得，∠DCB＝15°，则∠ACB＝105°，∵∠ACE＝45°，∴∠CBE＝30°，·································································································5分∵在Rt△BEC中，AE＝CE＝40，∴BE=3CE＝403∴AB＝AE+BE＝40+403··························································································7分作BF⊥AC于点F，则∠AFB=90°在Rt△BEC中，cos∠BAC＝BFAB=∴BF=202+206≈77答：与灯塔B的最小距离是77海里．········································································8分22．解：（1）证明：如图，连接OD，∵DE是⊙O的切线，∴DE⊥OD，∴∠ODF=90°···········································································1分∵BD=CD，∴∠CAD=∠DAB，·····················································································2分∵OA=OD，∴∠DAB=∠ODA，∴∠CAD=∠ODA，∴OD∥AE，························································································3分∴∠AEF=∠ODF=90°∴AE⊥EF························································································4分（2）解：∵∠CAD=∠ODA，∠AGE=∠OGD，∴△OGD∽△EGA，∴，················································································5分∵∠AEF=∠ODF，∠F=∠F∴△ODF∽△AEF∴·································································································6分∴·····································································································7分∴BF=2························································································8分23．解：（1）设今年A型车每辆售价x元，则去年售价每辆为（x+400）元，····················1分由题意，得50000x+400=解得：x＝1600．························································································3分经检验，x＝1600是原方程的根，且符合题意，······························································4分答：今年A型车每辆售价1600元；··········································································5分（2）设今年新进A型车a辆，则B型车（60﹣a）辆，获利y元，由题意，得················6分y＝（1600﹣1100）a+（2000﹣1400）（60﹣a），y＝﹣100a+36000．························································································7分∵B型车的进货数量不超过A型车数量的两倍，∴60﹣a≤2a，∴a≥20．························································································8分∵y＝﹣100a+36000．∴k＝﹣100＜0，∴y随a的增大而减小．························································································9分∴a＝20时，y最大＝34000元．∴B型车的数量为：60﹣20＝40辆．∴当新进A型车20辆，B型车40辆时，这批车获利最大．···········································10分24．解：（1）∵在矩形OABC中，OA＝6，OC＝4，∴AB=4，BC=6∵BE=4∴点E（2，4）························································································1分把E（2，4）代入y=kx中，得：∴k＝8．························································································2分当x＝6时，y=4∴F(6，4（2）作点F关于x轴的对称点G(6，-43)，则连接GE与x轴交于点M，连接EF，此时△EMF的周长最小．·······································4分设EG的函数关系式为y＝ax+b把E（2，4），G(6，-43)代入y＝ax得：2a+b=46a+b=-43∴y=-4当y＝0时，x＝5，∴M（5，0）．························································································6分（3）点P的坐标为（0，0）或(-1，0)或(10，0)或·································10分25．(1)证明：∵∠BAC=∠MAN，∴∠BAC－∠CAM=∠MAN－∠CAM，即∠BAM=∠CAN，········································1分∵AB=AC，AM=AN，∴△ABM≌△ACN，························································································3分∴∠ACN=∠ABM．························································································4分(2)解：AN存在最小值，理由如下：∵AM=MN，AB=BC，∴又∵∠AMN=∠B，∴△ABC∽△AMN，∴AMAB=ANAC，∠BAC∴∠BAC-∠MAC=∠MAN-∠MAC即∠BAM=∠CAN∴△ABM∽△ACN∴∠ACN=∠B=30°························································································7分过点A作AH⊥CN交CN延长线于点H，此时AN最小，最小值为AH，Rt△ACH中，∠ACN=30°AH=12AC=1故AN存在最小值，最小值为4···················································································8分(3)解：连接BD，EH，过H作HQ⊥CD于Q，如图所示，∵H为正方形DEFG的中心，∴DH=EH，∠DHE=90°，∵四边形ABCD为正方形，∴BC=CD，∠BCD=90°，∴∠BDE+∠CDE=∠CDH+∠CDE=45°，∴∠BDE=∠CDH，∵BDCD∴△BDE∽△CDH，························································································9分∴∠DCH=∠DBC=45°，BE=2CH=6设CE=x，则CD=x+6，∵DE=8，∴由勾股定理得：x2解得：x=23-3或x=-∴CD=23+3，在Rt△CDH中，CQ=QH=3，∴△CDH的面积为1226．解：（1）将点A（﹣1，0），点B（3，0）代入y＝ax2+bx+3得：a-b+3=0解得：a=-1b=2．·····················································································∴抛物线的表达式为y＝﹣x2+2x+3．···································································3分∵y＝﹣x2+2x+3＝﹣（x﹣1）2+4，∴顶点C（1，4）．·····································································4分(2)∵点D（0，3），点B（3，0），∴直线BC解析式为y＝﹣x+3，··········································································5分过点P作PQ∥y轴交BD于点Q，设点P（m，﹣m2+2m+3），点Q（m，﹣m+3），∴S△PBD＝12×PQ×OB＝12×3（﹣m2+2m+3+m-3）＝-32∵△PBD的面积为3，