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）
展开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分.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.已知反比例函数的图象经过点，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x和y都扩大为原来的2倍，那么分式的值（ ）
A.扩大为原来的4倍 B.扩大为原来的2倍
C.不变 D.缩小为原来的
7.若、都在函数的图象上，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分.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.当____________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（n） | 50 | 100 | 150 | 200 | 250 | 300 | 500 |
投中次数（m） | 28 | 60 | 78 | 104 | 123 | 152 | 251 |
投中频率（m/n） | 0.56 | 0.60 | 0.52 | 0.52 | 0.49 | 0.51 | 0.50 |
12.从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落在边上的中点处.若边，则的长等于_____________.
15.如图，点4在双曲线上，点B在双曲线上，且轴，则的面积等于__________.
16.如图，在正方形中，点E、F分别在边、上，且，，则___________°.
三、解答题（本大题共9小题，共84分.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）； （2）
18.（10分）
（1）计算：； （2）
19.（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O、A、B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E、F、G、H，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8：00点从学校出发，到了农场休息整顿30分钟后，按原路返回，13：30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在直线上.
（2）如图2，连接，则的最小值等于_________________.
25（10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P在y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
一、选择题
题号 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
答案 | B | D | B | B | C | D | C | A |
二、填空题
9.； 10.2； 11.0.5； 12.>；
13.； 14.3； 15.1； 16.61；
三、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.（1）50；12人；··································································（2分）
（2）144；·········································································（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人.·········································（7分）
20.
·················································································（2分）
（1）；···········································································（4分）
（2）、、.·········································································（7分）
21.（1）四边形为菱形.································································（1分）
法1：连接、，······································································（2分）
∵四边形为矩形，
∴.···············································································（3分）
∵点E、F、G、H，分别是四边的中点
∴，，············································································（4分）
∴，··············································································（5分）
∴四边形为菱形.·····································································（6分）
法2：∵四边形为矩形，
∴，；
.·················································································（3分）
∵点E、F、G、H，分别是四边的中点
∴，，
∴，
∴，··············································································（5分）
∴四边形为菱形.·····································································（6分）
（2）24.···········································································（10分）
22.解：设学生返回学校时的速度为x千米/时.················································（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/时.···················································（8分）
23.解：（1）将代入得，，
∴反比例函数解析式为.································································（2分）
将代入得，，
∴.
将，代入得，
解得：
∴一次函数解析式为.·································································（4分）
（2）或.···········································································（7分）
（3）设一次函数与x轴交与点C，与y轴交与点D.
则，，
∴，··············································································（8分）
∴
·················································································（10分）
24.（1）①证：∵四边形、四边形均为正方形，
∴，，.············································································（2分）
∴，
即，··············································································（3分）
∴.···············································································（4分）
②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.
在和中
∴················································································（8分）
∴.
∴，
又∵，
∴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.
点P为直线与y轴交点，
所以.·············································································（10分）
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