2023年山西省朔州市朔城区中考一模数学试题（含答案）

这是一份2023年山西省朔州市朔城区中考一模数学试题（含答案），共14页。试卷主要包含了本试卷分第Ⅰ卷和第Ⅱ卷两部分，4，众数是10B等内容，欢迎下载使用。
2023年山西省初中学业水平测试信息卷数学注意事项：1.本试卷分第Ⅰ卷和第Ⅱ卷两部分。全卷共6页，满分120分，考试时间120分钟。2.答卷前，考生务必将自己的姓名、准考证号填写在本试卷相应的位置。3.答案全部在答题卡上完成，答在本试卷上无效。4.考试结束后，将本试卷与答题卡一并交回。第Ⅰ卷  选择题（共30分）一、选择题（本大题共10小题，每小题3分，共30分）在每个小题给出的四个选项中，只有一项符合题目要求，请选出并在答题卡上将该项涂黑.1.下列各数中，是负数的是（    ）A. B.0 C.-1 D.2.小颖在研究无盖的正方体盒子的展开图时，画出下面4个展开图，其中符合要求的共有（    ）A.1个 B.2个 C.3个 D.4个3.下列事件中，是必然事件的是（    ）A.疫情期间，对从疫情高风险区归来的人员进行核酸检测，检测结果为阳性B.任意画一个三角形，其内角和为180°C.某校开展“喜迎二十大，筑梦向未来”主题学习活动中，抽到A同学分享发言D.打开电视机，正在播放“天宫课堂”4.大型电视专题片《领航》自2022年10月8日在中央广播电视总台央视开播以来，引发社会各界广泛关注.截止10月11日，专题片《领航》相关视频内容及宣传报道跨媒体总触达人次超7.56亿次.数据7.56亿用科学记数法表示为（    ）A. B. C. D.5.不等式组的解集在数轴上表示正确的是（    ）A. B.C. D.6.把1~9这9个数填入方格中，使其任意一行，任意一列及两条对角线上的数之和都相等，这样便构成了一个“九宫格”，它源于我国古代的“洛书”（图①），是世界上最早的“幻方”.图②是仅可以看到部分数值的“九宫格”，则其中x的值为（    ）A.1 B.3 C.4 D.67.如图是某市连续20天的平均气温折线统计图，则下列说法正确的是（    ）A.平均数是9.4，众数是10 B.中位数是9，平均数是10C.中位数是9.4，众数是9 D.中位数是9.5，众数是98.已知线段，，，，下列说法中正确的是（    ）A.b，d，c，a成比例  B..d，b，a，c成比例C.b，c，d，a成比例  D.b，d，a，c成比例9.将等腰直角三角板与量角器按如图所示的方式摆放，使三角板的直角顶点与量角器的中心O重合，且两条直角边分别与量角器边缘所在的弧交于A，B两点.若厘米，则的长度为（    ）A.厘米 B.厘米 C.厘米 D.厘米10.如图，菱形的边长为8，，点E，F分别是，边上的动点，且，过点B作于点G，连接，则长的最小值是（    ）A. B. C. D.第Ⅱ卷  非选择题（共90分）二、填空题（本大题共5个小题，每小题3分，共15分）11.________________.12.我市举办的“喜迎二十大·奋进新征程——乡村振兴成果展”吸引了众多市民前来参观，如图是该展览馆出入口示意图.小颖和母亲从同一人口进入分别参观，参观结束后，她们恰好从同一出口走出的概率是________.13.已知点，，在函数的图像上，那么，，的大小关系是________.14.如图，将一根细长的绳子沿中间对折，再沿对折后的绳子的中间对折1次，这样连续对折n次，最后用剪刀沿对折n次后的绳子的中间将绳子剪断，此时绳子将被剪成________段.15.如图，在平面直角坐标系中，有7个半径为1的小圆拼在一起，下面一行的4个小圆都与x轴相切，上面一行的3个小圆都在下一行右边3个小圆的正上方，且相邻两个小圆只有一个公共点，从左往右数，y轴过第2列两个小圆的圆心，点P是第3列两个小圆的公共点.若过点P有一条直线平分这7个小圆的面积，则该直线的函数表达式是____________.三、解答题（本大题共8个小题，共75分）解答应写出文字说明，证明过程或演算步骤.16.（每小题4分，共8分）（1）计算：；（2）解方程：.17.（本题7分）如图，在四边形中，，在上取两点E，F，使，连接，.（1）若，试说明；（2）在（1）的条件下，连接，，试判断与有怎样的数量关系，并说明理由.18.（本题8分）为了解学生掌握垃圾分类知识的情况，增强学生环保意识，某学校举行了“垃圾分类人人有资”的知识测试活动.现从该校七、八年级中各随机抽取20名学生的测试成绩（满分10分，6分及6分以上为合格）进行整理、描述和分析，下面给出了部分信息.七年级20名学生的测试成绩为：7，8，7，9，7，6，5，9，10，9，8，5，8、7，6，7，9，7，10，6、八年级20名学生的测试成绩条形统计图如图：七、八年级抽取的学生的测试成绩的平均数、众数、中位数、8分及以上人数所占百分比如下表所示：年级 平均数 众数 中位数 8分及以上人数所占百分比七年级7.5a745%八年级7.58bc根据以上信息，解答下列问题：（1）直接写出上述表格中的a、b，c的值：（2）根据以上数据，你认为该校七、八年级中哪个年级学生掌握垃圾分类知识较好？请说明理由（写出一条理由即可）；（3）该校七、八年级共1200名学生参加了此次测试活动，估计参加此次测试活动成绩合格的学生人数是多少？19.（本题9分）阅读下列材料，并完成相应的任务.西姆松定理是一个平面几何定理，其表述为：过三角形外接圆上异于三角形顶点的任意一点作三边或其延长线的垂线，则三垂足共线（此线常称为西姆松线）.某数学兴趣小组的同学们尝试证明该定理.如图（1），已知内接于，点P在上（不与点A，B，C重合），过点P分别作，，的垂线，垂足分别为点D，E，F.求证：点D，E，F在同一条直线上.如下是他们的证明过程（不完整）：如图（1），连接，，，，取的中点Q，连接，，则，（依据1）∴点E，F，P，C四点共圆，∴.（依据2）又∵，∴.同上可得点B，D，P，E四点共圆，……任务：（1）填空：①依据1指的是中点的定义及________________________；②依据2指的是________________（2）请将证明过程补充完整.（3）善于思考的小虎发现当点P是的中点时，，请你利用图（2）证明该结论的正确性.20.（本题9分）如图是小开家所在居民楼，楼底C点的左侧30米处有一个山坡，坡角为30°，E点处有一个图书馆，山坡坡底到图书馆的距离为40米，在图书馆E点处测得小开家的窗户B点的仰角为45°，居民楼与山坡的剖面在同一平面内.（1）求的高度；（结果精确到个位，参考数据：）（2）某天，小开到家后发现有资料落在图书馆，此时离图书馆闭馆仅剩5分钟，若小开在平地的速度为6，上坡速度为4，电梯速度为1.25，等候电梯及上、下乘客所耽误的时间共3分钟，请问小开能否在闭馆前赶到图书馆？21.（本题10分）今年植树节期间，某景观园林公司购进一批成捆的A，B两种树苗，每捆A种树苗比每捆B种树苗多10棵，每捆A种树苗和每捆B种树苗的价格分别是630元和600元，而每棵A种树苗和每棵B种树苗的价格分别是这一批树苗平均每棵价格的0.9倍和1.2倍.（1）求这一批树苗平均每棵的价格是多少元？（2）如果购进的这批树苗共5500棵，A种树苗最多购进3500棵，为了使购进的这批树苗的费用最低，应购进A种树苗和B种树苗各多少棵？并求出最低费用.22.（本题11分）综合与探究在矩形的边上取一点E，将沿翻折，使点C恰好落在边上的点F处.（1）如图①，若，求的度数；（2）如图②，当，且时，求的长；（3）如图③，延长，与的角平分线交于点M，交于点N，当时，请直接写出的值.23.（本题13分）综合与实践如图，抛物线与x轴交于A，B两点，且，与y轴交于点C，连接，抛物线的对称轴为直线，D为第一象限内抛物线上一动点，过点D作于点E，与交于点F，设点D的横坐标为m.（1）求抛物线的表达式；（2）当线段的长度最大时，求D点的坐标：（3）抛物线上是否存在点D，使得以点O，D，E为顶点的三角形与相似？若存在，求出m的值；若不存在，请说明理由.  2023年山西省初中学业水平测试信息卷数学参考答案（三）一、（每小题3分，共30分）1—5 C C B D C 6—10 A A D B C二、（每小题3分，共15分）11.5 12. 13.14. 15.三、（共75分）16、解：（1）原式································································2分；·············································································4分（2），，·············································································5分即，···········································································6分∴，···········································································7分解得：，.·······································································8分17.解：（1）∵，∴，···········································································1分∵，∴，···········································································2分在和中，，······································································3分∴.；···········································································4分（2），理由如下：································································5分∵，∴，···········································································6分又∵，∴四边形是平行四边形，∴.·············································································7分18.解：（1），，；································································3分（2）八年级学生掌握垃圾分类知识较好，理由如下（写出其中一条即可）；①七、八年级学生成绩的平均数相同，八年级学生成绩的中位数7.5高于七年级学生成绩的中位数7；②七、八年级学生成绩的平均数相同，八年级学生成绩的众数8高于七年级学生成绩的众数7；③七、八年级学生成绩的平均数相同，八年级8分及以上人数所占比例50%，高于七年级8分及以上人数所占比例45%.······5分（3）∵七年级20名学生中，成绩在6分及6分以上的有18人，八年级20名学生中，成绩在6分及6分以上的有18人.设七年级有m名学生参加了此次活动，则八年级有名学生参加了此次活动，根据题意得，（人）.答：估计参加此次测试活动成绩合格的学生人数是1080人.···································8分19.解：（1）①直角三角形斜边上的中线等于斜边的一半②圆内接四边形对角互补····························································2分（2）如图（1），连接，，，，取的中点Q，连接，，则，∴点E，F，P，C四点共圆，∴.又∵，∴，同上可得点B，D，P，E四点共圆，∴，···········································································3分∴，∴.·············································································4分∴点D，E，F在同一条直线上；······················································5分（3）如图（2），连接，，.·························································6分∵点P是的中点，∴，∴，.···········································································7分又∵，，∴，∴，···········································································8分∴.·············································································9分20.解：（1）如图，作于点F，作，交的延长线于点G，得矩形，∴，，··········································································2分根据题意可知：米，，米，，∴米，··········································································3分∴米，∴米，∴米，··········································································4分∴（米），······································································5分答：的高度约为85米；·····························································6分（2）根据题意得：（秒），·························································8分5分秒，∵，∴小开能在闭馆前赶到图书馆.························································9分21.解：（1）设这一批树苗平均每棵的价格是x元，根据题意得，····································································2分解得.···········································································3分经检验，是原分式方程的根，且符合题意.···············································4分答：这一批树苗平均每棵的价格是20元.·················································5分（2）由（1）可知A种树苗每棵价格为（元），B种树苗每棵价格为（元），设购进A种树苗t棵，这批树苗的费用为w元，则.··············································································7分∵.∴w随t的增大而减小，又∵，∴当时，w最小，····························································8分此时，B种树苗有（棵），.··············································································9分答：为了使购进的这批树苗的费用最低，应购进A种树苗3500棵，B种树苗2000棵，最低费用为111000元.10分22.解：（1）由折叠可知，，..··············································································1分∵，∴，∴.∵，∴，········································································2分∴；···········································································3分（2）∵，∴.又∵，∴.∵，∴，········································································5分∴，∴.∵，，∴，······································································7分∴，∴；········································································8分（3）.提示：····································································11分如图，过点N作于点G，∴.∵，∴.∵，∴，∴.∵.∴，∴.∵为的平分线，∴.∵，，∴，∴，，∴，.∵，∴，∴.23.解：（1）设点A的坐标为，∵，∴点B的坐标为，··································································1分∵抛物线的对称轴为直线，∴，解得，∴，∴A点的坐标为，B点的坐标为，······················································2分将A，B两点的坐标代入抛物线，得，解得.∴抛物线的表达式为；·····························································3分（2）∵抛物线的表达式，∴点C的坐标为，·································································4分设直线的解析式为，将，代入，得，解得，∴直线的解析式为.································································5分∴点，则点，∴，···········································································6分∵，∴时，的长度最大，·······························································7分∴点D的坐标为；·································································8分（3）存在，m的值为1或.····························································9分∵于点E，，∴要使与相似，只需有一个锐角相等，················································10分当时，，设直线的解析式为，将，代入，得，解得.∴直线的解析式为，∴直线的解析式为，∴解方程组，解得或.∵点D是第一象限内抛物线上一动点，∴点D的坐标为，∴；当时，∵，，∴.∵，，，，∴，解得（舍去）或.·································································12分综上所述，m的值为1或.····························································13分