河南省周口市鹿邑县2020-2021学年七年级下学期期中数学试题(word版含答案)

这是一份河南省周口市鹿邑县2020-2021学年七年级下学期期中数学试题(word版含答案)，共8页。试卷主要包含了下列各项中，是无理数的是，在平面直角坐标系中，点在，下列各命题中，为真命题的是，如图，，若，则的度数为等内容，欢迎下载使用。
2020-2021学年度第2学期期中考试卷七年级数学满分：120分一．选择题．（每题只有一个正确答案，请将正确答案填在下面的表格里．每题3分，共30分）1．下列各项中，是无理数的是A． B．0.23 C． D．2．在平面直角坐标系中，点在A．第一象限 B．第二象限 C．第三象限 D．第四象限3．下列各命题中，为真命题的是A．锐角大于它的余角  B．锐角大于它的补角C．钝角大于它的补角  D．锐角与锐角之和大于90°4．将点向右平移1个单位长度，再向下平移2个单位长度后得到，则点的坐标为A． B． C． D．5．如图，，若，则的度数为        5题图A．30° B．60° C．70° D．120°6．下列各项中，比3大比4小的无理数是A．3.14 B． C． D．7．关于图形平移的特征叙述，有下列两种说法：①一个图形和它经过平移所得的图形中，两组对应点的连线一定平行；②一个图形和它经过平移所得的图形中，两组对应点的连线一定相等；其中判断正确的是A．①错②对 B．①对②错 C．①②都错 D．①②都对8．如图所示，小明在操场上点B处看位于点A处小亮的位置时，下列说法正确的是    8题图A．点A在点B的北偏东40°的方向25m处B．点A在点B的南偏东50°的方向25m处C．点A在点B的南偏西40°的方向25m处D．点A在点B的南偏西50°的方向25m处9．如图直线与直线相交于点O，，若过点O作，则的度数为        9题图A．50° B．130° C．50°或90° D．50°或130°10．已知表示三个数中最小的那个数，例如当时，，当时，则x的值为A． B． C． D．二．填空题．（每题3分，共15分）11．16的算术平方根为________．12．若点在y轴上，则点P的坐标为________．13．某人在练习场上驾驶汽车，两次拐弯后的方向与原来的方向相反．则两次拐弯的角度可能是________．①第一次左拐40°，第二次向右拐140° ②第一次左拐50°，第二次向右拐130°③第一次左拐70°，第二次向右拐110° ④第一次左拐70°，第二次向左拐110°14．当与互为相反数时，的值为________．15．如图，直线，，，则________．     15题图三．解答题．（本大题8小题，共75分）16．（8分）计算：．17．（8分）已知，，．（1）在如图所示的直角坐标系中描上各点，画出三角形；（2）将向下平移2个单位长度，再向左平移2个单位长度得到三角形，画出平移后的图形并写出、、的坐标．18．（8分）将下列推理过程依据补充完整．如图，已知平分，，求证：平分证明：∵平分（已知）∴（角平分线的定义）∵（已知）∴（________________________________）∴（等量代换）∵（已知）∴________________（________________________________）∴（________________________________）∴________________（等量代换）∴平分（角平分线的定义）19．（9分）已知，．（1）已知x的值为4，求a的值及的平方根；（2）如果一个正数的两个平方根分别是x和y，求这个数．20.（10分）如图所示，已知，．（1）求证：；（2）若平分，于点F，，求的度数．21.（10分）如图，直线与直线相交于点O，．（1）的邻补角为________________（写一个即可）；（2）若，判断直线与的位置关系并说明理由；（3）若，求的度数22．（10分）在平面直角坐标系中，已知点．（1）若点P到x轴的距离为9，求m的值及此时点P的坐标；（2）若点P在过点，且与x轴平行的直线上，求点P的坐标．23．（12分）如图①，点D、E、F分别在线段、、上，与相交于点G，与互补．（1）写出的所有同位角________________；判断与的位置关系．并说明理由；（2）如图②，连接，，过点G作于点H．点N是线段上一点，且．求证：平分；（3）在（2）的条件下，若平分，请问的大小是否发生变化？若不变，请直接写出的度数；若改变，请说明理由．2020-2021学年度第二学期其中考试卷参考答案七年级数学说明：1．如果考生的解答与本参考答案提供的解法不同，可根据提供的解法的评分标准精神进行评分．2．评阅试卷，要坚持每题评阅到底，不能因考生解答中出现错误而中断对本题的评阅．如果考生的解答在某一步出现错误，影响后继部分而未改变本题的内容和难度，视影响的程度决定对后面给分的多少，但原则上不超过后继部分应得分数之半．3．评分标准中，如无特殊说明，均为累计给分．4．评分过程中，只给整数分数．一．选择题．（每题只有一个正确答案，每题3分，共30分）题号12345678910答案DBCDDCADDC二．填空题．（每题3分，共15分）11．4 12． 13．④ 14．2 15．120°三．解答题．（本大题8小题，共75分）16．解：原式···························································（8分）17．解：（1）如图·······················································（2分）（2）如图····························································（5分），，································································（8分）18．解：两直线平行，内错角相等两直线平行，内错角相等两直线平行，同位角相等（括号内每空2分，其余每空1分）19．解：（1）当时，，∴··················································（2分）∴，∴2的平方根为·····················································（4分）（2）依题意得：，即，∴···············································（7分）∴这个数为···························································（9分）20．解：（1）证明：∵，∴·················································（2分）∴∵，∴，∴··································································（5分）（2）∵，，∴，∴······························································（7分）∵，∴······························································（8分）∵平分，∴，∴，∴······························································（10分）21．解：（1）或························································（2分）（2）·······························································（3分）理由如下：∵，∴又∵，∴····························································（5分）∴··································································（6分）（3）∵，∴∵，∴······························································（8分）∴··································································（10分）22．解：（1）依题意得：···············································（1分）解得：或-5···························································（3分）点P的坐标为或························································（5分）（2）若轴，点P与点A的纵坐标相等，即·····································（6分）···································································（8分）∴点P的坐标为························································（10分）23．解：（l），·························································（2分）···································································（3分）理由如下：∵与互补，∴∵，∴∴··································································（6分）（2），，∴∴，∴······························································（8分）∵，∴∴平分······························································（9分）（3）不变····························································（10分）∵平分，∴即∵，∴∵，∴······························································（11分）∴，即······························································（12分）