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    江苏省徐州市2022-2023学年八年级下学期期末数学试题(含答案)

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    江苏省徐州市2022-2023学年八年级下学期期末数学试题(含答案)

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    这是一份江苏省徐州市2022-2023学年八年级下学期期末数学试题(含答案),共10页。试卷主要包含了选择题,三象限B.第一,解答题等内容,欢迎下载使用。
    2022~2023学年度第二学期期末测试一八年级数学试题一、选择题(本大题共8小题,每小题3分,共24.在每小题所给出的四个选项中,恰有一项是符合题目要求的)1.下列图形是我国国产品牌汽车的标识,在这些汽车标识中,是中心对称图形的是    A. B. C. D.2.为了了解一批电视机的使用寿命,从中抽取100台电视机进行试验,这个问题的样本是(    A.这批电视机  B.这批电视机的使用寿命C.抽取的100台电视机  D.抽取的100台电视机的使用寿命3.已知反比例函数图象经过点,则这个函数的图象位于(    A.第二、三象限 B.第一、三象限 C.第三、四象限 D.第二、四象限4.下列各式中属于最简二次根式的是    A. B. C. D.5.下列事件中,是必然事件的是(    A.抛掷2枚骰子,都是6点朝上 B.任意画一个三角形,其内角和是360°C.13人中至少有2人的生日在同一个月 D.两直线被第三条直线所截,内错角相等6.如果把分式中的xy都扩大为原来的2倍,那么分式的值(    A.扩大为原来的4  B.扩大为原来的2C.不变  D.缩小为原来的7.都在函数图象上,且,则(    A. B. C. D.8.如图,在平面直角坐标系中,函数)与的图像交于点则代数式的值为    A. B. C.-2 D.2二、填空题(本大题共8小题,每小题4分,共32.不需写出解答过程,直接写出答案)9.若二次根式有意义,则x的取值范围是_____________.10.____________时,分式的值为零.11.如表记录了一名球员在罚球线上投篮的结果.那么,这名球员投篮一次,投中的概率约为___________(精确到0.1.投篮次数n50100150200250300500投中次数m286078104123152251投中频率m/n0.560.600.520.520.490.510.5012.25名男生和20名女生中,随机抽取一名学生做代表,则男生做代表的可能性___________女生做代表的可能性(填写“>”、“<”、“=”)13.分式方程的解为_______________.14.如图,将正方形纸片沿折叠,使点B落在边上的中点.若边,则的长等于_____________.15.如图,点4在双曲线上,点B在双曲线上,且的面积等于__________.16.如图,在正方形中,点EF分别在边上,且,则___________°.三、解答题(本大题共9小题,共84.解答时应写出文字说明、证明过程或演算步骤)17.10分)计算:1 218.10分)1)计算: 219.7分)某中学为了解学生每天参加户外活动的情况,对部分学生每天参加户外活动的时进行了抽样调查,并将调查结果绘制作成如下两幅不完整的统计图,请根据图中信息解答下列问题:1)本次调查一共抽取了_____________名学生,并补全频数分布直方图;2_________3)若该中学共有1000名学生,请估计该校每天参加户外活动的时间为2小时的学生人数.20.7分)如图,已知,顶点.1)请画出绕坐标原点O顺时针旋转90°后得到的,并写出点B的对应点的坐标_______2)请直接写出:以OAB为顶点的平行四边形的第四个顶点C的坐标____________.21.10分)如图,在矩形中,点EFGH,分别是四边的中点;1)判断四边形的形状,并给出理由;2)当时,四边形的面积等于____________.22.8分)某学组织学生去离学校12千米的农场,早上800点从学校出发,到了农场休息整顿30分钟后,按原路返回,1330到达学校,其中去农场时的速度是返回学校时速度的1.2倍,问去农场时的速度多少?23.10分)如图,已知一次函数与反比例函数相交于点和点.1)求一次函数和反比例函数的解析式2)观察图像,直接写出关于x的不等式的解集3)求的面积.24.12分)如图,已知四边形为正方形,,点E为平面内一动点(不与点D重合),连接,以为边作正方形连接.1)如图1,当点E在对角线上移动时:①求证:②探究:的值是否为定值?若是,请求出这个定值;若不是,请说明理由;③求证:点F在直线.2)如图2,连接的最小值等于_________________.2510分)如图,一次函数的图像与反比例函数)的图像相交于点A,与x轴交于点B,与y轴交于点C轴于点D,点C关于直线的对称点为点E且点E在反比例函数的图像上.1)求b的值;2)连接,求证四边形为正方形;3)若点Py轴上,当最小时,求点P的坐标.  2022~2023学年度八下期末数学参考答案选择题题号12345678答案BDBBCDCA填空题9. 10.2 11.0.5 12.>13. 14.3 15.1 16.61解答题:17.1)解:原式··································································4分).·················································································5分)2)解:原式····································································3分).·················································································5分)18.1)解:原式··································································3分).·················································································5分)2)解:········································································2分)················································································4分).·················································································5分)19.15012人;··································································2分)2144·········································································4分)3)解:(人····································································6分)答:每天参加户外活动的时间为2小时的学生有160.·········································7分)20.·················································································2分)1···········································································4分)2.·········································································7分)21.1)四边形为菱形.································································1分)1:连接······································································2分)∵四边形为矩形,.···············································································3分)∵点EFGH,分别是四边的中点············································································4分)··············································································5分)∴四边形为菱形.·····································································6分)2四边形为矩形,.·················································································3分)∵点EFGH,分别是四边的中点··············································································5分)∴四边形为菱形.·····································································6分)224.···········································································10分)22.解:设学生返回学校时的速度为x千米/.················································1分)(小时),30分钟小时,(小时)·······················································2分)根据题意得,·················································································5分)解这个方程得:.检验:当时,所以是方程的解且符合实际意义.·························································7分)所以(千米/时).答:学生去农场时的速度为5.28千米/.···················································8分)23.解:(1)将代入得,∴反比例函数解析式为.································································2分)代入得,.代入得,解得:∴一次函数解析式为.·································································4分)2.···········································································7分)3)设一次函数x轴交与C,与y轴交与D.··············································································8分)·················································································10分)24.1)①证:四边形、四边形均为正方形,.············································································2分)··············································································3分).···············································································4分)的值为定值...,∴.···············································································6分)③过点E分别做,垂足分别为点P、点Q,连接.················································································8分).∴点F在直线.·······································································10分)3.·············································································12分)25.解(1)因为点A、点C在一次函数的图像上,所以设.因为点C、点E关于对称,所以.·············································································1分)因为点A、点E在反比例函数)的图像上,所以所以所以因为所以,所以.···································································3分)2)由(1)可求.······························································5分)易证四边形为正方形.·································································6分)3)由(1)可求点且点B、点D关于y轴对称.设直线BE的表达式为),将点代入)得,解得············································································8分)所以直线的表达式为.P为直线y轴交点,所以.·············································································10分)

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