2023年山东省济南市东南片区中考一模数学试题（含答案）

这是一份2023年山东省济南市东南片区中考一模数学试题（含答案），共16页。试卷主要包含了填空题，解答题等内容，欢迎下载使用。

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
1

8
65
2

a
76
3

b
85
4

c
94

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．（10分）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）
1100
1400
销售价格（元）
今年的销售价格
2000
24．（10分）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
1
2
3
4
5
6
7
8
9
10
答案
B
C
C
A
B
D
C
A
D
D
二、填空题
11．4(a+1)(a-1) 12．49 13．3 14．163π-83 15．52023 16．①②④
三、解答题
17．原式=3-22-2+2×22 ······························································································4分
=1-2 ········································································································6分
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）85.5 83.6 ··························································································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， ······················································································3分
∴CE=22AC＝40，
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， ········· ························································································6分
∴AB＝AE+BE＝40+403 ··························································································7分
作BF⊥AC于点F，则∠AFB=90°
在Rt△BEC中，cos∠BAC＝BF AB =22
∴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=50000(1-20%)x， ························································································2分
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：4=k2．
∴k＝8． · ·······················································································2分
当x＝6时，y=43．
∴F(6，43)． ························································································3分

（2）作点F关于x轴的对称点G(6，-43)， 则MF=MG
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+b中，
得：2a+b=46a+b=-43，解得：a=-43b=203，
∴y=-43x+203， ························································································5分
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，
∴ AMMN=ABBC
又∵∠AMN=∠B，
∴△ABC∽△AMN，
∴AMAB=ANAC ， ∠BAC=∠MAN ··································································6分
∴∠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=12×8=4
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=DEDH=2，
∴△BDE∽△CDH， ························································································9分
∴∠DCH=∠DBC=45°，BE=2CH=6， ····························································10分
设CE=x，则CD=x+6，
∵DE=8，
∴由勾股定理得：x2+x+62=82，
解得：x=23-3或x=-23-3（舍），
∴CD=23+3， ························································································11分
在Rt△CDH中，CQ=QH=3，
∴△CDH的面积为12×23+3×3=323+92 ···················································12分

26． 解：（1）将点A（﹣1，0），点B（3，0）代入y＝ax2+bx+3得：
a-b+3=09a+3b+3=0， ····················································································1分
解得：a=-1b=2． ·····················································································2分
∴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）＝-32m2+92m ·······································6分
∵△PBD的面积为3，
∴-32m2+92m=3 ···················································································7分
∴m1=1， m2=2 ···················································································8分
（3）解法1
∵在Rt△CMP中，PM=2CM
∴tan∠MCP=2
设AC交y轴于点F，连接DF，过点C作CE⊥x轴于点E，如图，
∵A（﹣1，0），C（1，4），
∴AE=2，CE=4
∴OA＝1，OE＝1，CE＝4．
∴OA＝OE，AC＝AE2+CE2=25．
Rt△AEC中，tan∠CAE=2，tan∠ACE=12
∵tan∠MCP= tan∠CAE
∴∠MCP= tan∠CAE，∴△DAC是等腰三角形
∵FO⊥AB，CE⊥AB，
∴FO∥CE，
∴OF＝12CE＝2，F为AC的中点．
∵△DAC是等腰三角形，∴DF⊥AC．
∵FO⊥AD，
∴△AFO∽△FDO．∴AOOF=OFOD∴12=2OD
∴OD＝4．
∴D（4，0）．···································································10分
设直线CD的解析式为y＝kx+m，
∴k+m=44k+m=0，解得：k=-43m=163．
∴直线CD的解析式为y＝-43x+163．
∴y＝-43x+163y＝﹣x2+2x+3，解得：x1=1y1=4，x2=73y2=209
∴P1（73，209）． ··········································································12分

方法2：
过点A作AC的垂线，与CP的延长线交于点K，过点A作y轴的平行线l，过点K作KG⊥l于点G，过点C作CH⊥l于点H，易证△CHA∽△AGK
∵A（﹣1，0），C（1，4），
∴CH=2，AH=4，OA=1
∵tan∠MCP=2
∴
∴AG=4，KG=8，
∴K（7，-4）···································································10分
设直线CK的解析式为y＝kx+m，
∴k+m=47k+m=-4，解得：k=-43m=163．
∴直线CD的解析式为y＝-43x+163．
∴y＝-43x+163y＝﹣x2+2x+3，解得：x1=1y1=4，x2=73y2=209
∴P1（73，209）． ··········································································12分