山东省烟台市2022-2023学年八年级上学期期末数学试题 (含答案)

这是一份山东省烟台市2022-2023学年八年级上学期期末数学试题 (含答案)，共11页。试卷主要包含了保证答题卡清洁、完整等内容，欢迎下载使用。
2022—2023学年第一学期期末阶段性测试初三数学试题（120分钟）注意事项：1．答题前，请务必将自己的学校、姓名、准考证号填写在答题卡和试卷规定的位置上。2．答选择题时，必须使用2B铅笔填涂答题卡上相应题目的正确答案字母代号，如需改动，用橡皮擦干净后，再选涂其他答案。3．答非选择题时，必须使用0.5毫米黑色签字笔书写；做图、添加辅助线时，必须用2B铅笔。4．保证答题卡清洁、完整。严禁折叠、严禁在答题卡上做任何标记，严禁使用涂改液、胶带纸、修正带。5．请在题号所指示的答题区域内作答，写在试卷上或答题卡指定区域外的答案无效。一、书写与卷面（3分）书写规范    卷面整洁二、选择题（本题共10个小题，每小题3分，满分30分）每小题有且只有一个正确答案，请把正确答案的字母代号涂在答题卡上。1．下列分式中，是最简分式的是（    ）A． B． C． D．2．七巧板是我国的一种传统智力玩具，下列用七巧板拼成的图形是中心对称图形的是（    ）A． B． C． D．3．在中，若，则的度数是（    ）A．140° B．120° C．100° D．40°4．如图，菱形的对角线AC与BD相交于点O，若，，则BD的长为（    ）A．4 B．6 C．7 D．85．为了落实“作业、睡眠、手机、读物、体质”等五项管理要求，了解学生的睡眠状况，某校调查了一个班50名学生每天的睡眠时间，绘成睡眠时间条形统计图如图所示，则所调查学生睡眠时间的中位数为（    ）A．6h B．7h C．7.5h D．8h6．如图，将三角形纸片剪掉一角得四边形，设与四边形的外角和的度数分别为，，则正确的是（    ）A． B． C． D．无法比较与的大小7．如图所示的扇形统计图描述了某校学生对课后延时服务的打分情况（满分5分），则所打分数的众数为（    ）A．5分 B．4分 C．3分 D．45%8．当m为自然数时，一定能被下列哪个数整除（    ）A．5 B．6 C．7 D．89．如图，四边形是正方形，E为边CD上一点，绕着点A顺时针旋转90°后到达的位置，连接EF，则的形状是（    ）A．等腰三角形 B．直角三角形 C．等腰直角三角形 D．等边三角形10．如图，等腰直角三角形中，，，将BC绕点B顺时针旋转（），得到BP，连接CP，过点A作交CP的延长线于点H，连接AP，则的度数（    ）A．随着的增大而增大   B．随着的增大而减小C．保持定值45°不变   D．随着的增大，先增大后减小三、填空题（本大题共6个小题，每小题3分，满分18分）11．如果关于x的方程有增根，那么m的值为________．12．若关于x的二次三项式是完全平方式，则k的值是________．13．已知一组数据，，的平均数和方差分别为5和2，则数据，，的平均数和标准差分别是________．14．如图，在中，，，的平分线AE交BC于E点，则EC的长为________．15．如图，将长为5 cm，宽为3 cm的矩形先向右平移2 cm，再向下平移1 cm，得到矩形，则阴影部分的面积为________．16．如图，在平面直角坐标系中，三个顶点坐标分别为，，，则顶点B的坐标为________．四、解答题（本大题共9个小题，满分69分）17．（本题满分6分）分解因式：（1）．（2）18．（本题满分5分）解方程：19．（本题满分6分）先化简，然后从的范围内选择一个合适的整数作为x的值代入求值．20．（本题满分6分）如图，已知的三个顶点的坐标分别为、、．（1）画出关于原点O成中心对称的图形；（2）将绕原点O顺时针旋转90°，画出对应的，并写出点的坐标．21．（本题满分8分）核酸检测时采集的样本必须在4小时内送达检测中心，超过时间，样本就会失效．A、B两个采样点到检测中心的路程分别为30 km、36 km．A、B两个采样点的送检车有如下信息：信息一：B采样点送检车的平均速度是A采样点送检车的1.2倍；信息二：A、B两个采样点送检车行驶的时间之和为2小时．若B采样点从开始采集样本到送检车出发用了2.6小时，则B采样点采集的样本会不会失效？22．（本题满分8分）在学校组织的“文明出行”知识竞赛中，8（1）和8（2）班参赛人数相同，成绩分为A、B、C三个等级，其中相应等级的得分依次记为A级100分、B级90分、C级80分，其中8（2）班有2人达到A级，将两个班的成绩整理并绘制成如下的统计图．请解答下列问题：（1）求各班参赛人数，并补全条形统计图；（2）此次竞赛中8（2）班成绩的中位数a为________分；（3）小明同学根据以上信息制作了如下统计表： 平均数（分）中位数（分）方差8（1）班m90n8（2）班91a29请分别求出m和n的值，并从稳定性方面比较两个班的成绩．23．（本题满分8分）如图，在矩形中，对角线AC，BD相交于点O，，交BC于F，垂足为E，求的度数．24．（本题满分10分）如图，在中，，M、N分别是AD、BC的中点．（1）求证：四边形是平行四边形；（2）若，，求BD的长．25．（本题满分12分）如图①，中，，点M、N分别是AB、AC上的点，且．连接MN、CM、BN，点D、E、F、G分别是BC、MN、BN、CM的中点，连接E、F、D、G．（1）判断四边形的形状是_________（不必证明）；（2）现将绕点A旋转一定的角度，其他条件不变（如图②），四边形的形状是否发生变化？证明你的结论；（3）如图②，在（2）的情况下，请将在原有的条件下添加一个条件，使四边形是正方形．请写出你添加的条件，并在添加条件的基础上证明四边形是正方形． 2022-2023学年第一学期期末阶段性测试初三数学参考答案及评分意见一、书写与卷面（3分）评分标准：分别赋分3，2，1，0．二、选择题（每小题3分，共30分）题号12345678910答案BDADCABDCC三、填空题（每小题3分，共18分）11．，    12．，    13．6，，    14．2，    15．18，    16．．四、解答题（17题每小题3分，18题5分，19-20题每小题6分，21-23题每小题8分，24题10分，25题12分，共69分）17．解：（1）原式．································································3分（2）原式．·······································································3分18．解：原方程可变为方程两边同乘以，得，解得，··························································4分检验：当时，，所以原分式方程的解为．·····························································5分19．解：原式．···············································································5分由分式有意义的条件可知，，∴当时，∴原式．（答案不唯一，如0，3）······················································6分20．解：（1）如图所示，即为所求；····················································3分（2）如图所示，即为所求，···························································5分其中点．··········································································6分21．解：设A采样点送检车的平均速度是，················································1分根据题意，得，····································································4分解得，···········································································5分经检验，是分式方程的根，····························································6分∴B采样点送检车的平均速度为，∴B采样点送检车的行驶时间为，∵，∴B采样点采集的样本不会失效．·······················································8分22．解：（1）∵8（2）班有2人达到A级，且A等级人数占被调查的人数为20%，∴8（2）班参赛的人数为（人）························································1分∵8（1）和8（2）班参赛人数相同，∴8（1）班参赛人数也是10人，························································2分故8（1）班C等级人数为（人），补全图形如图：····································································3分（2）90；·········································································4分（3）（分），·····································································5分，···············································································7分∵8（1）班的方差大于8（2）班的方差，∴从稳定性看8（2）班的成绩更稳定．···················································8分23．解：∵四边形是矩形，∴．∵，∴．·············································································3分∵，∴．∴．·············································································5分∵，，，∴，∴，∴．·············································································8分24．（1）证明：∵是平行四边形，∴，．···········································································1分∵M、N分别是AD、BC的中点，∴．∵，∴四边形是平行四边形；·····························································3分（2）如图，连接ND，·······························································4分∵是平行四边形，∴．∵N是BC的中点，∴．∵，∴．·············································································6分∵，∴是等边三角形，∴，，···········································································7分∵是的外角，∴，∵，∴，∴，·············································································9分∴，∴．············································································10分25．解：（1）菱形；································································2分（2）不变，·······································································3分证明：由旋转得，∴，∵，，∴（），∴．·············································································5分∵点E、F分别是MN、BN的中点，∴，，同理，，，，∴，，∴四边形是平行四边形．·····························································6分且．∴四边形是菱形；···································································7分（3）添加条件：，··································································8分证明：如图，设BM与CN交于点P，DF与BM交于点Q，由（2）得，∵，∴，∴，即，∴．············································································10分∵，∴．············································································11分∵，∴，············································································12分∴菱形是正方形．