四川省南充市2023-2024学年八年级上学期期末数学试题（含答案）

—2024学年度（上）期末教学质量监测八年级数学试卷注意事项：1.考试时间120分钟，试卷满分150分.2.答题前将姓名、座位号、准考证号填在答题卡指定位置.3.所有解答内容均需涂、写在答题卡上.4.选择题须用2B铅笔将答题卡相应题号对应选项涂黑，若需改动，须擦净另涂.5.非选择题在答题卡对应题号位置用0.5毫米黑色字迹笔书写.一、选择题（本大题共10个小题，每小题4分，共40分）每小题都有代号为A，B，C，D四个答案选项，其中只有一个是正确的，请根据正确选项的代号填涂答题卡对应位置，填涂正确记4分，不涂、涂错或多涂记0分.1.以下为2023年成都大运会奖牌“蓉光”上的部分设计元素，其中是轴对称图形的是（）A. B. C. D.2.下列运算正确的是（）A. B. C. D.3.如图，△ABE≌△ACD，若AB＝8，AE＝5，则BD的长度为（）A.2 B.3 C.4 D.54.一个多边形的内角和与它的外角和相等，则这个多边形是（）A.四边形 B.五边形 C.六边形 D.七边形5.如图，在△ABC中，分别以B，C两点为圆心，大于长为半径作弧，连接两弧交点得到直线l，l分别交AC、BC于E、F两点，连接BE，若AC＝9，AB＝5，则△ABE的周长为（）A.10 B.12 C.14 D.186.已知，则的值为（）A.0 B.1 C.2 D.37.已知a，b，c为△ABC三边，且满足，则△ABC是（）A.直角三角形 B.等边三角形 C.等腰三角形 D.不能确定8.某人从A地步行到B地时，速度为a，再从B地原路返回到A地时，速度为b，则他自A地到B地再返回A地的平均速度为（）A. B. C. D.9.如图，在直角坐标系中，点A、点B的坐标分别为（2，8）和（6，0），若点P是y轴上的一个动点，且A、B、P三点不在同一条直线上，当△ABP的周长最小时，点P的坐标为（）A.（0，4） B.（0，5） C.（0，6） D.（0，8）10.若整数m使得关于x的方程的解为非负整数，且关于y的不等式组至少有3个整数解，则所有符合条件的整数m的和为（）A.7 B.5 C.0 D.－2二、填空题（本大题共6个小题，每小题4分，共24分）请将答案填在答题卡对应题号的横线上.11.2023年8月“麒麟9000S”芯片横空出世，标志着我国14纳米以下先进工艺制程已取得突破性进展（14纳米＝0.000000014米），把0.000000014用科学记数法表示为________.12.若点A（a，4）与点B（3，b）关于y轴对称，则________.13.如图，直线l经过正方形ABCD的顶点A，分别过该正方形的顶点B、D作BE⊥l于E，DF⊥l于F.若BE＝3，DF＝6，则EF的长为________.14.分式方程的解为________.15.若，且，则的值为________.16.如图，在等腰△ABC中，AB＝BC，∠A＝30°，BD⊥AC于D，E是AB延长线上的一点，F是线段BD上的一点，EF＝CF.下列结论：①BC平分∠EBD；②∠A＝∠BEF＋∠FCD；③△EFC是等边三角形；④BC＝BE＋BF.其中正确的结论有________.（填写序号）三、解答题（本大题共9个小题，共86分）解答题应写出必要的文字说明或推演步骤.17.（8分）计算：（1）（2）18.（8分）如图，在△ABC中，∠C＝90°，∠B＝38°，AD为∠BAC的平分线，E为线段BD上一点，且∠CEA＝50°.求∠DAE的度数.19.（8分）先将化简，并从“－1，0，1，2”中选择一个适当的数作为a的值，再计算出结果.20.（10分）分解因式：（1）（2）21.（10分）在平面直角坐标系xoy中，△ABC的一个顶点为A（2，4）.（1）作△ABC关于x轴的对称图形并求出△ABC的面积；（2）若P是x轴上一点，且与△ABC的面积相等，请求出点P的坐标.22.（10分）如图，在△ABC中，D为BC的中点，DE⊥AB于E，DF⊥AC于F，且BE＝CF.（1）求证：AE＝AF；（2）若AB＝10，DE＝3，求△ABC的面积.23.（10分）2023年中国新能源汽车销量再创新高，其中油电混动汽车备受青睐，因为其既可以用纯油模式行驶，也可以切换成纯电模式行驶.若某品牌油电混动汽车从甲地行驶到乙地，当完全用油做动力行驶时，所需油费为160元；当完全用电做动力行驶时，所需电费为40元，已知汽车行驶中每千米所需的油费比电费多0.6元.（1）求汽车行驶中每千米需要的电费是多少元?（2）若汽车从甲地到乙地，部分路段使用纯电模式行驶，其余路段采用纯油驱动，若所需的油电费用合计不超过88元，则至少需要在纯电模式下行驶多少千米?24.（10分）如图，在△ABC中，∠ABC＝90°，AB＝BC，D为AB的中点，过点A作l1∥BC，过点B作l2⊥CD于F，l1与l2交于点E，连接CE、DE.（1）求证：△ABE≌△BCD；（2）试证明△BCE是等腰三角形.25.（12分）如图，在四边形ABCD中，∠BAD＝120°，∠BCD＝60°，AB＝AD，BC＝DC，在边BC、DC所在直线上分别有E、F两点，且始终有.图1图2图3（1）如图1，当E、F在BC、DC上，AE＝AF时，求证：BE＋DF＝EF；（2）如图2，当E、F在BC、DC上，AE≠AF时，（1）问中的结论是否仍成立请说理；（3）如图3，当E、F在边BC、DC的延长线上时，直接写出BE、DF、EF之间的数量关系，不必证明.八年级数学试题参考答案及评分意见说明：1．阅卷前务必认真阅读参考答案和评分意见，明确评分标准，不得随意拔高或降低标准．2．全卷满分150分，参考答案和评分意见所给分数表示考生正确完成当前步骤时应得的累加分数．3．参考答案和评分意见仅是解答的一种，如果考生的解答与参考答案不同，只要正确就应该参照评分意见给分．合理精简解答步骤，其简化部分不影响评分．4．要坚持每题评阅到底．如果考生解答过程发生错误，只要不降低后继部分的难度且后继部分再无新的错误，可得不超过后继部分应得分数的一半，如果发生第二次错误，后面部分不予得分；若是相对独立的得分点，其中一处错误不影响其它得分点的评分．一、选择题（本大题共10个小题，每小题4分，共40分）1.D2.A3.B4.A5.C6.B7.C8.D9.C10.A9.【解析】作点B关于y轴的对称点，连接AB′，AB′与y轴的交点即为所求．由已知数据易得△CAB′为等腰直角三角形AC∥y轴，∠OPB′＝45°，故P（0，6）10.【解析】解分式方程得：，由分式方程的解为非负整数，可得：m＋5＝0，3，6，9，12…，解之：m＝－5，－2，1，4，7…；解不等式组：m≤y＜10，且不等式组至少有3个整数解，得到m≤7，所以m＝－5，－2，1，4，7.（因分式方程中x≠1故m＝－2舍去）．故m可取的整数值为－5，1，4，7.其和为7．二、填空题（本大题共6个小题，每小题4分，共24分）11.12.113.914.15.16.①②③④16.【解析】①∠EBC＝180°－60°－60°＝60°②连接AF可证∠BEF＝∠BAF，∠FCD＝∠FAD③由∠BEF＋∠FCD＝30°，∠AEC＋∠ACE＝150°得∠FEC＋∠FCE＝120°，所以∠EFC＝60°④截BG＝BE，△BEG为正三角形，△BEF≌△GEC（SAS）三、解答题（本大题共9个小题，共86分）17.（1）解：原式···········································（2分）（结果正确，没有上一步不扣分）······················（4分）（2）解：原式·············································（2分）···································································（4分）18.证明：∵在△ABC中，，∴····················································（2分）∵AD为∠BAC的平分线····························································（3分）∴···················································（4分）∵∠CEA＝∠B＋∠BAE，∠CEA＝50°··············································（6分）∴∠BAE＝50°－38°＝12°································································（7分）∴∠DAE＝∠BAD－∠BAE＝26°－12°＝14°···········································（8分）19.解：原式················································（2分）··································································（3分）·············································································（4分）·······················································································（5分）∵“±1和0”使分式中分母为0，舍去∴a＝2·····················································································（6分）∴原式············································································（8分）20.（1）解：原式·································································（2分）····························································（5分）（2）解：原式·······················································（2分）·································································（5分）21．解：（1）如图所示·····························································（2分）·········································（5分）（2）因为，可设P为（m，0）得····················（7分）解之：或····················（9分）∴点P为：或······（10分）22.证明：（1）∵DE⊥AB，DF⊥AC∴∠DEB＝90°，∠DFC＝90°∵D为BC的中点∴BD＝CD在Rt△BDE和Rt△CDF中∴Rt△BDE≌Rt△CDF（HL）·····················································（3分）∴∠B＝∠C∴AB＝AC···············································································（4分）∴AB－BE＝AC－CF∴AE＝AF···············································································（5分）（2）连接AD··················································································（6分）∵AB＝AC（已证）∴△ABC是等腰三角形································································（7分）∵AD为△ABC的中线∴S△ABD＝S△ACD··········································································（8分）∵AB＝10，DE＝3∴····································································（9分）∴··································································（10分）23.解：（1）设汽车行驶中每千米需要的电费是x元.············································································（2分）解之，得x＝0.2······································································（3分）经检验：x＝0.2是该分式方程的解且符合题意.·······························（4分）∴汽车行驶中每千米需要的电费是0.2元.·····································（5分）（2）甲地到乙地的路程为：40÷0.2＝200千米···································（6分）设汽车用电行驶了m千米，则用油行驶了（200－m）千米················（7分）由题意得0.2m＋0.8（200－m）≤88··············································（8分）解之，得m≥120··································································（9分）∴至少需要用纯电模式行驶120千米.··········································（10分）24.证明：（1）∵l1∥BC∴∠EAB＋∠ABC＝180°∵∠ABC＝90°∴∠EAB＝180°－∠ABC＝180°－90°＝90°·······································（1分）∵l2⊥CD∴∠EBC＋∠DCB＝90°···························································（2分）∵∠EBC＋∠EBA＝90°∴∠DCB＝∠EBA·································································（3分）在△AEB和△BDC中∴△AEB≌△BDC（ASA）·····················································（5分）（2）由△AEB≌△BDC（已证）得AE＝BD∵D为AB的中点，即AD＝BD∴AE＝AD·················································································（6分）∵∠ABC＝90°，AB＝AC∴△ABC是等腰直角三角形························································（7分）∴∠BAC＝45°∴∠EAC＝90°－45°＝45°∴∠EAC＝∠DAC·····································································（8分）在△EAC和△DAC中∴△EAC≌△DAC（SAS）··························································（9分）∴DC＝CE由（1）△AEB≌△BDC得BE＝CD∴BE＝CE∴△BCE是等腰三角形·····························································（10分）25.证明：（1）连接AC在△ABC和△ADC中∴△ABC≌△ADC（SSS）························································（1分）∴∠B＝∠D∵∠BAD＝120°，∠BCD＝60°∴∠B＋∠D＝360°－120°－60°＝180°∴∠B＝∠D＝90°······································································（2分）在Rt△ABE和Rt△ADF中∴Rt△ABE≌Rt△ADF（HL）·····················································（3分）∴BE＝DF，∠BAE＝∠DAF∵∠EAF＝60°∴∠BAE＋∠DAF＝120°－60°＝60°△AEF是等边三角形∴AE＝EF，∠BAE＝∠DAF＝30°·················································