山东省烟台市2021-2022学年八年级下学期期中数学试题(word版含答案)

这是一份山东省烟台市2021-2022学年八年级下学期期中数学试题(word版含答案)，共6页。试卷主要包含了答题前务必用0，非选择题必须用0，当时，化简的结果是等内容，欢迎下载使用。
2021—2022学年度第二学期期中学业水平检测初三数学温馨提示：1.本试卷共6页，共120分；考试时间120分钟.2.答题前务必用0.5毫米黑色签字笔将自己的姓名、准考证号填写在答题卡规定的位置上.3.选择题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用橡皮擦干净后，再选涂其他答案标号.4.非选择题必须用0.5毫米黑色签字笔作答，答案必须写在答题卡指定区域内相应的位置；如需改动，先划掉原来的答案，然后再写上新的答案；不能使用涂改液、胶带纸修正带.5.写在试卷上或答题卡指定区域外的答案无效.一、卷面书写（满分3分）二、选择题（本题共12个小题，每小题3分，满分36分）每小题都给出标号为A，B，C，D四个备选答案，其中有且只有一个是正确的.1.下列关于x的方程是一元二次方程的是（    ）A. B. C. D.2.王师傅应客户要求加工4个长为4、宽为3的矩形零件.在交付客户之前，王师傅需要对4个零件进行检测.根据零件的检测结果，图中有可能不合格的零件是（    ）A. B. C. D.3.下列二次根式中，不能与合并的是（    ）A. B. C. D.4.下列等式成立的是（    ）A. B. C. D.5.已知关于x的方程有两个不相等的实数根，则a的取值范围是（    ）A. B.且 C. D.且6.当时，化简的结果是（    ）A. B. C. D.7.如图，四边形为菱形，点A、B、C、D在坐标轴上，点A的坐标为，长为3，则菱形的面积等于（    ）A.20 B. C. D.8.如图，在矩形中，点E是上一点，且，，垂足为点F，下列结论中，不一定正确的是（    ）A. B. C. D.9.如图，某小区有一块长为30m，宽为24m的矩形空地，计划在这块空地上划出四分之一的区域修建绿地，小明的设计方案如图所示，若设绿地的宽度为m，可列方程为（    ）A. B.C. D.10.如图，四边形，都是正方形，点E，G分别在边，上，连接，过点E作交于点H.若，，则的长为（    ）A.1 B.2 C.3 D.11.若，则等于（    ）A.-1 B.-5 C.5 D.112.下列说法：①关于x的一元二次方程（）无实数根；②无论m为何值，关于x的一元二次方程都有两个不相等的实数根；③若关于x的一元二次方程有两个不相等的实数根，则m的取值范围是且.其中说法正确的是（    ）A.①②③ B.①② C.①③ D.②三、填空题（本大题共6个小题，每小题3分，满分18分）13.计算的结果是__________.14.若关于x的一元二次方程的常数项为0，则m的值为__________.15.若代数式有意义，则x的取值范围为__________.16.m是方程的根，则代数式的值是__________.17.如图，在矩形中，E为的中点，的角平分线交于点F.若，，则的长度为__________.18.如图，在正方形中，和为直角三角形，，，，则的长是__________.四、解答题（本大题共7个小题，满分63分）19.（本题满分8分）解方程：（1）（2）20.（本题满分8分）计算：（1）（2）21.（本题满分9分）已知关于x的方程是一元二次方程.（1）求m的值；（2）解这个一元二次方程.22.（本题满分9分）（1）在中，，，.求的周长和面积；（2）已知，，求的值.23.（本题满分9分）如图，在菱形中，于点E，于点F.（1）求证：；（2）分别延长和相交于点G，若，，求的值.24.（本题满分10分）如图，已知正方形中，点E是边延长线上一点，连接，过点B作，垂足为点F，与交于点G.（1）求证：；（2）若，，求的长.25.（本题满分10分）如图，点E是平行四边形对角线上一点，点F在延长线上，且，与交于点G，连接，，.（1）求证：；（2）若，点G恰好是的中点，求证：四边形是矩形.初三数学参考答案一、卷面书写（满分3分）二、选择题（每题3分，共36分）BCDAD  CCDDD  AB三、填空题（每题3分，共18分）13.  14.-3  15.且  16.-2018  17.6  18.72四、解答题（本题共7个题，满分63分）19.（本题满分8分）（1）；  （2），20.（本题满分8分）（1）；  （2）21.（本题满分9分）（1）-1；···········································································4分（2）方程为，即，···································································5分，·················································································9分22.（本题满分9分）（1）周长为，面积为；································································5分（2）9·············································································9分23.（本题满分9分）（1）证明：∵四边形是菱形，∴，·······················································1分∵于点E，于点F，∴，································································2分∵，，，∴，········································································3分∴，···············································································4分∴；···············································································5分（2）解：∵四边形是菱形，∴，∴，······················································6分∵，∴，∴，∴，····································································8分∴.················································································9分24.（本题满分10分）（1）证明：∵四边形是正方形，∴，，····················································1分∵，∴，∵，∴.······································································3分∴，···············································································4分∴；···············································································5分（2）解：由（1），，∴，·····························································6分∵，，∴，，········································································8分在中，············································································10分25.（本题满分10分）（1）证明：连接，交于点O，···························································1分∵四边形是平行四边形，∴，····························································2分∵，∴是的中位线，···································································3分∴，即；···········································································4分（2）证明：由（1）得，∴，，∵G是的中点，∴，···································································5分∴，∴，···········································································6分∴四边形是平行四边形，·······························································7分∵四边形是平行四边形，∴，∵，∴，∵，∴，∴，···············································································9分∴平行四边形是矩形.·································································10分