陕西省延安中学2021-2022学年八年级上学期期中考试数学试题

这是一份陕西省延安中学2021-2022学年八年级上学期期中考试数学试题，共9页。试卷主要包含了本试卷共4页，总分120分，领到试卷和答题卡后，请用0，在中，，，则的度数为等内容，欢迎下载使用。
试卷类型：A2021~2022学年度第一学期第二次阶段性作业八年级数学注意事项：1.本试卷共4页，总分120分。考试时间120分钟。2.领到试卷和答题卡后，请用0.5毫米黑色墨水签字笔，分别在试卷和答题卡上填写姓名、班级和准考证号，同时用2B铅笔在答题卡上填涂对应的试卷类型信息点（A或B）。3.请在答题卡上各题的指定区域内作答，否则作答无效。4.作图时，先用铅笔作图，再用规定签字笔描黑。5.考试结束，本试卷和答题卡一并交回。第一部分（选择题  共24分）一、选择题（共8小题，每小题3分，计24分.每小题只有一个选项是符合题意的）1.下列图形中，轴对称图形的个数是（    ）A.1个 B.2个 C.3个 D.4个2.若三角形的三边长分别为5、x、15，则x的值可以是（    ）A.2 B.3 C.8 D.113.在中，，则的度数为A.30° B.45° C.90° D.120°4.如图，在中，，，是边上的中线.若的周长为35，则的周长是（    ）A.20 B.24 C.26 D.295.如图，直线l是的对称轴，如果，那么的度数等于（    ）A.66° B.48° C.58° D.24°6.如图，，，，则能直接判断的理由是（    ）A. B. C. D.7.若等腰三角形的一个内角比另一个内角大30°，则这个等腰三角形的底角度数是（    ）A.50° B.80° C.50°或70° D.80°或40°8.如图，在与中，，，，，交于点D，连接.下列结论：①；②；③；④；⑤，一定正确的个数为（    ）A.1个 B.2个 C.3个 D.4个第二部分（非选择题  共96分）二、填空题（共5小题，每小题3分，计15分）9.点关于x轴的对称点的坐标为____________.10.在中，，，则的度数为11.如图，在中，平分，于点E，于点F，，，，则的长为____________.12.如图，是的平分线，垂直平分交的延长线于点F，若，则的度数为____________°.13.如图，，，，、交于点H，连接，则的度数为___________°.三、解答题（共13小题，计81分.解答应写出过程）14.（5分）在一个各内角都相等的多边形中，每一个内角都比与它相邻外角的3倍还大20°，求这个多边形的边数.15.（5分）如图，正方形网格中每个小正方形边长都是1，画出关于直线l对称的.16.（5分）如图，E是的平分线上一点，，，垂足分别为C、D，连接求证：是的垂直平分线.17.（5分）如图，在中，利用尺规在边上找一点E，使（不写作法，保留作图痕迹）.18.（5分）如图，点C，F在上，，，.求证：.19.（5分）用一条长为20的细绳围成一个等腰三角形.（1）如果底边长是腰长的一半，那么各边的长是多少？（2）能围成底边长是10的等腰三角形吗？为什么？20.（5分）如图，在中，M是的中点，求证：.21.（6分）已知：如图，D是的边的中点，，，垂足分别为E，F，且.求证：是等腰三角形.22.（7分）某段河流的两岸是平行的，数学兴趣小组在老师带领下不用涉水过河就测得河流的宽度，他们是这样做的：①在河流的一条岸边B点，选对岸正对的一棵树A；②从B处沿河岸直走20米有一棵树C，继续前行20米到达D处；③从D处沿河岸垂直的方向行走，当到达A树正好被C树遮挡住的E处停止行走；④测得的长为5米.（1）河流的宽度是____________米；（2）请你说明他们做法的正确性.23.（7分）如图，直线l与m分别是边和的垂直平分线，l与m分别交边于点D和点E.（1）若，则的周长是多少？为什么？（2）若，求的度数.24.（8分）如图，在中，，垂直平分，交边于点D，交边于点E，垂直平分，交于点F，连接.（1）求证：；（2）求的度数.25.（8分）如图，在中，，过点A作的平行线交的角平分线于点D，连接.（1）求证：为等腰三角形；（2）若，求的度数.26.（10分）如图，在中，是边的高，为的角平分线，且.为的中线，延长到点F，使得，连接，交于点G，交于点H.（1）求证：；（2）若，求证：.     试卷类型：A2021~2022学年度第一学期第二次阶段性作业八年级数学参考答案及评分标准一、选择题（共8小题，每小题3分，计24分.每小题只有一个选项是符合题意的）1.B  2.D  3.A  4.D  5.B  6.A  7.C  8.C二、填空题（共5小题，每小题3分，计15分）9.  10.52  11.3  12.68  13.130三、解答题（共13小题，计81分.解答应写出过程）14.解：设这个多边形的一个外角为，则与其相邻的内角等于，由题意，得，·······································································（2分）解得，即多边形的每个外角为40°，·······················································（3分）又∵多边形的外角和为360°，∴多边形的外角个数，∴这个多边形的边数是9.·······························································（5分）15.解：如图，为所作.·················································································（5分）16.证明：∵E是的平分线上一点，，，∴，··············································································（1分）在和中，，，∴，··············································································（2分）∴，··············································································（3分）又∵，∴是的垂直平分线.···································································（5分）17.解：如图，点E为所作.·················································································（5分）18.证明：∵，∴，即，··········································································（1分）在和中，∴.···············································································（4分）∴.···············································································（5分）19.解：（1）设底边长为x，则腰长为2x，由题意得：，·······································································（2分）解得：，∴，∴各边长为：8，8，4.································································（3分）（2）不能围成底边长是10的等腰三角形，理由如下：当底边长为10时，腰长，∵，∴不能构成三角形，故不能围成底边长是10的等腰三角形.·····················································（5分）20.证明：∵M是中点，∴，··············································································（1分）∵，，∴，··············································································（4分）∴.···············································································（5分）21.证明：∵，，∴，∵D是的中点，∴，··············································································（2分）在与中，∴，··············································································（4分）∴，∴，∴是等腰三角形.·····································································（6分）22.解：（1）5.······································································（2分）（2）由题意知，在和中，，∴，··············································································（5分）∴.即他们的做法是正确的.································································（7分）23.解：（1）的周长为10.理由：∵直线l与m分别是边和的垂直平分线，∴，，············································································（2分）∴的周长.··········································································（3分）（2）由（1）知，，，∴，，············································································（5分）又∵，∴，··············································································（6分）∴.∴.···············································································（7分）24.（1）证明：∵垂直平分，∴.···············································································（1分）∴，∵，∴，··············································································（3分）∵垂直平分，∴，··············································································（4分）∴，∴.···············································································（5分）（2）解：设，由（1）得出：，∵，∴，··············································································（6分）∵，∴，··············································································（7分）解得：，即.········································································（8分）25.（1）证明：∵平分，∴，··············································································（1分）∵，∴，∴.···············································································（2分）∴.···············································································（3分）∵，∴，∴，∴为等腰三角形.·····································································（4分）（2）解：由（1）知，，∵，，∴，··············································································（5分）LABD=△CBD=LADB=，（180°-LBAD）=20°，∴，∵，∴，由（1）知，，∴，∴.···············································································（8分）26.证明：（1）∵，∴，，············································································（1分）∵为的中线，∴，···································································（2分）在与中，.∴，··············································································（3分）∴，又∵，∴.···············································································（4分）（2）∵垂直平分，∴，∴，··············································································（5分）又∵，∴，在与F中，∴，··············································································（7分）∴，∴，∵为的角平分线，∴，··············································································（8分）在与中，，∴，··············································································（9分）∴.···············································································（10分）