福建省泉州市德化县2022-2023学年八年级下学期期末数学试题（含答案）

这是一份福建省泉州市德化县2022-2023学年八年级下学期期末数学试题（含答案），共11页。试卷主要包含了选择题，填空题，解答题等内容，欢迎下载使用。
2023年春八年级数学达标测试（满分：150分；考试时间：120分钟）友情提示：请把所有答案填写（涂）在答题卡上，不要错位、越界答题。一、选择题（每小题4分，满分40分）1．下列式子变形正确的是（    ）A． B． C． D．2．清代诗人袁枚创作了一首诗《苔》：“白日不到处，青春恰自来．苔花如米小，也学忔丹开．”歌颂了苔在恶劣环境下仍能绽放属于自己的美丽．若苔花的花粉粒直径约为0.0000084米，用科学记数法表示0.0000084为（    ）A． B． C． D．3．下列命题的逆命题正确的是（    ）A．平行四边形的两组对边分别平行 B．对顶角相等C．矩形是平行四边形 D．全等三角形的对应角相等4．在平行四边形中，的平分线交于点，则的长为（    ）A．4 B．3 C．2 D．15．在1000米中长跑考试中，小明开始慢慢加速，当达到某一速度后保持匀速，最后200米时奋力冲刺跑完全程，下列最符合小明跑步时的速度y（单位：米/分）与时间x（单位：分）之间的大致图象的是（    ）A． B． C． D．6．平行四边形的对角线交于点，下列条件中，不能判定平行四边形是菱形的是（    ）A． B． C． D．平分7．一次函数和的图象如图，甲、乙两位同学给出的下列结论：甲说：方程的解是；乙说：当时，．其中正确的结论的是（    ）A．甲乙都正确 C．乙正确，甲错误B．甲正确，乙错误 D．甲乙都错误8．无论实数为何值，直线与直线的交点都不可能出现任平面直角坐标系中的（    ）A．第一象限 B．第二象限 C．第三象限 D．第四象限9．若平行四边形的一边长为5，则它的对角线长可能是（    ）A．4和6 B．2和12 C．4利8 D．4和310．在反比例函数中，当时，的最大值与最小值之差为4，则值为（    ）A．8 B．6或 C．6 D．5二、填空题（每小题4分，满分24分）11．两组数据与的平均数都是7，若将这两组数据合并成一组数据，则这组新数据的中位数为_________．12．已知菱形的两条对角线的长分别是和，那么菱形的边长等于_________．13．如图，在平面直角坐标系中，若将点向右平移后，其对应点恰好落在反比例函数的图象上，已知点，则的面积为_________．14．若关于的分式方程有增根，则的值为_________．15．已知为大于1的正整数，且代数式的值也是整数，则可取的最大整数值是_________．16．平面直角坐标系中，若直线经过和两点，则代数式的值为_________．三、解答题（共9小题，满分86分）17．（8分）计算：18．（8分）解分式方程：19．（8分）先化简再求值：，其中．20．（8分）一客车和一出租车分别从甲、乙两地相向而行，同时出发，设客车离甲地距离为千米，出租车离甲地距离为千米，两车行驶的时间为小时，关于的函数图象如图所示：（1）根据图象，直接写出关于的关系式；（2）求经过多少小时，两车之间的距离为100千米？21．（8分）如图1，已知平行四边形，点为边的中点，连结并延长交的延长线与点，连结．（1）求证：四边形是平行四边形．（2）如图2，当时，求的面积．22．（10分）为了从甲、乙两位同学中选拔一人参加法制知识竞赛，举行了6次对战赛，根据两位同学6次对战赛的成绩，分别绘制了如下统计图．（1）填写下列表格（将数字写在横线上） 平均数（分）中位数（分）众数（分）甲_________9192乙90__________________（2）已知乙同学6次成绩的方差为（平方分），求出甲同学6次成绩的方差；方差公式：（3）你认为选择哪一位同学参加知识竞赛比较好？请说明理由．23．（10分）某企业员工感冒后，到药店买了一种新型感冒药，按使用说明书服用后，血液中的约物浓度（微克/亳升）与服药后时间（小时）之间的函数关系如下图所示，其中，当时，满足的关系式；当时，与成反比例．（1）求的值，并求当时，与的函数关系式；（2）若血液中药物浓度不低于2.5微克/敦升的持续时间超过5.5小时，则称药物治疗有效，请通过计算说明用这种新药治疗是否有效吗？24．（12分）如图，为正方形对角线的交点，点为线段上一动点（不与两点重合），连结，将绕点逆时针旋转后得到，过点作交于点，连结．（1）试证：四边形为正方形．（2）若点恰好是边的中点，正方形的边长，求线段的长．25．（14分）直线和交于点为常数，且，且两直线分别与轴交于两点．（1）试说明的面积为定值．（2）当的周长最小时，求点的坐标．（3）当点恰好在轴上时，将直线绕点逆时针旋转后交轴于点，求的面积．
2023年春八年级数学达标测试参考答案一、选择题(每小题4分，满分40分)1．C 2．D 3．A 4．D              5．A              6．A              7．B              8．C              9．C              10．B二、填空题(每小题4分，满分24分)11．6  12．13               13．3                            14．3              15．8                            16．4三、解答题(共9小题，共86分)17.解:原式=·········································································6分=············································································8分18.解：去分母得：,····································································4分解得：,·······································································6分经检验：是原方程的解.原方程的解为···································································8分19.解：原式==············································································4分===············································································6分当时，原式=···································································8分20.解：(1) ···········································································2分·············································································4分(2) 两车相遇前，两车之间的距离为100千米，解得：········································································6分两车相遇后，两车之间的距离为100千米，，解得：综上所述，经过小时或小时，两车之间的距离为100千米··································8分21.(1)证明：四边形ABCD是平行四边形,又点E为AD的中点，AE=DE,在AEF与DEC中，≌,，又四边形ACDF是平行四边形·························································4分(2)四边形ABCD是平行四边形AD=BC，AB=CD,CF=BC，AD=CF，由(1)证得四边形ACDF是平行四边形四边形ACDF是矩形，FAC=90°，CAB=180°FAC=90°，是直角三角形，·································································6分四边形ACDF是矩形，AF=CD=AB=1，．···········································································8分22． (1) ①90    ②87．5      ③85·····················································3分(2) 甲同学6次成绩的方差：＝＝···········································································6分(3)∵甲、乙两同学成绩平均数相同，但甲的中位数比乙的大，且甲的方差比乙的方差小∴选择甲同学参加知识竞赛比较好．················································10分23．解：(1)由图象可知，将(3，6)代入函数，得6=3t，解得t=2，·······························································2分当时，设与的函数关系式为将(3，6)代入,得，, ············································································5分(2)将代入函数得，解得，将代入函数得，解得，············································································10分这种新药治疗有效．24．(1)证明：过点E作ENAD于N，EN交BC于H，过点E作EMAB于M，则四边形AMEN为矩形，MEN＝90°···················································2分平分BAD, EM=EN,于E,BEG=90°,1=NEM-GEM=90°-GEM,2=BEG-MEG=90°-GEM,1=2.ENG=EMB,(ASA).EG=EB········································································4分由旋转知，FB=EB，EBF=90°,FB=GE，EBF+BEG=180°,  ,四边形BEGF为平行四边形,又FB=EB，FBE=90°,四边形BEGF为正方形.····························································6分(2)连结ED，由正方形对称性可知ED=EB,EB=EG，ED=EG，又.G为AD的中点，BC=4aGN=ND=，HC=ND=·····························································10分又EHC为等腰直角三角形，EC=HC=······································································12分25．解：(1)在中，令=0，得,  在中，令=0，得．·······························································2分    ，  .，即点A为直线上的一动点．·······················································4分·············································································5分(2)作点B关于直线的对称点，则点(4,3)，连结交直线于点A，设, 则,    ,令，得，·············································································9分(3)过点B作BEAB交AD于E，过点E作EH轴于H点,过点B作BG⊥EH于G，过点A作AF⊥GB于F，,,,,,.在和中，,.，············································································12分设直线则，，  .令,得， ,.············································································14分提示：本题也过点B作BEAD于E，再过点E作EH轴.