河南省周口市鹿邑縣2020-2021學(xué)年七年級下學(xué)期期中數(shù)學(xué)試題(word版含答案)

這是一份河南省周口市鹿邑縣2020-2021學(xué)年七年級下學(xué)期期中數(shù)學(xué)試題(word版含答案)，共8頁。試卷主要包含了下列各項(xiàng)中，是無理數(shù)的是，在平面直角坐標(biāo)系中，點(diǎn)在，下列各命題中，為真命題的是，如圖，，若，則的度數(shù)為等內(nèi)容，歡迎下載使用。
2020-2021學(xué)年度第2學(xué)期期中考試卷七年級數(shù)學(xué)滿分：120分一．選擇題．（每題只有一個(gè)正確答案，請將正確答案填在下面的表格里．每題3分，共30分）1．下列各項(xiàng)中，是無理數(shù)的是A． B．0.23 C． D．2．在平面直角坐標(biāo)系中，點(diǎn)在A．第一象限 B．第二象限 C．第三象限 D．第四象限3．下列各命題中，為真命題的是A．銳角大于它的余角  B．銳角大于它的補(bǔ)角C．鈍角大于它的補(bǔ)角  D．銳角與銳角之和大于90°4．將點(diǎn)向右平移1個(gè)單位長度，再向下平移2個(gè)單位長度后得到，則點(diǎn)的坐標(biāo)為A． B． C． D．5．如圖，，若，則的度數(shù)為        5題圖A．30° B．60° C．70° D．120°6．下列各項(xiàng)中，比3大比4小的無理數(shù)是A．3.14 B． C． D．7．關(guān)于圖形平移的特征敘述，有下列兩種說法：①一個(gè)圖形和它經(jīng)過平移所得的圖形中，兩組對應(yīng)點(diǎn)的連線一定平行；②一個(gè)圖形和它經(jīng)過平移所得的圖形中，兩組對應(yīng)點(diǎn)的連線一定相等；其中判斷正確的是A．①錯(cuò)②對 B．①對②錯(cuò) C．①②都錯(cuò) D．①②都對8．如圖所示，小明在操場上點(diǎn)B處看位于點(diǎn)A處小亮的位置時(shí)，下列說法正確的是    8題圖A．點(diǎn)A在點(diǎn)B的北偏東40°的方向25m處B．點(diǎn)A在點(diǎn)B的南偏東50°的方向25m處C．點(diǎn)A在點(diǎn)B的南偏西40°的方向25m處D．點(diǎn)A在點(diǎn)B的南偏西50°的方向25m處9．如圖直線與直線相交于點(diǎn)O，，若過點(diǎn)O作，則的度數(shù)為        9題圖A．50° B．130° C．50°或90° D．50°或130°10．已知表示三個(gè)數(shù)中最小的那個(gè)數(shù)，例如當(dāng)時(shí)，，當(dāng)時(shí)，則x的值為A． B． C． D．二．填空題．（每題3分，共15分）11．16的算術(shù)平方根為________．12．若點(diǎn)在y軸上，則點(diǎn)P的坐標(biāo)為________．13．某人在練習(xí)場上駕駛汽車，兩次拐彎后的方向與原來的方向相反．則兩次拐彎的角度可能是________．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．當(dāng)與互為相反數(shù)時(shí)，的值為________．15．如圖，直線，，，則________．     15題圖三．解答題．（本大題8小題，共75分）16．（8分）計(jì)算：．17．（8分）已知，，．（1）在如圖所示的直角坐標(biāo)系中描上各點(diǎn)，畫出三角形；（2）將向下平移2個(gè)單位長度，再向左平移2個(gè)單位長度得到三角形，畫出平移后的圖形并寫出、、的坐標(biāo)．18．（8分）將下列推理過程依據(jù)補(bǔ)充完整．如圖，已知平分，，求證：平分證明：∵平分（已知）∴（角平分線的定義）∵（已知）∴（________________________________）∴（等量代換）∵（已知）∴________________（________________________________）∴（________________________________）∴________________（等量代換）∴平分（角平分線的定義）19．（9分）已知，．（1）已知x的值為4，求a的值及的平方根；（2）如果一個(gè)正數(shù)的兩個(gè)平方根分別是x和y，求這個(gè)數(shù)．20.（10分）如圖所示，已知，．（1）求證：；（2）若平分，于點(diǎn)F，，求的度數(shù)．21.（10分）如圖，直線與直線相交于點(diǎn)O，．（1）的鄰補(bǔ)角為________________（寫一個(gè)即可）；（2）若，判斷直線與的位置關(guān)系并說明理由；（3）若，求的度數(shù)22．（10分）在平面直角坐標(biāo)系中，已知點(diǎn)．（1）若點(diǎn)P到x軸的距離為9，求m的值及此時(shí)點(diǎn)P的坐標(biāo)；（2）若點(diǎn)P在過點(diǎn)，且與x軸平行的直線上，求點(diǎn)P的坐標(biāo)．23．（12分）如圖①，點(diǎn)D、E、F分別在線段、、上，與相交于點(diǎn)G，與互補(bǔ)．（1）寫出的所有同位角________________；判斷與的位置關(guān)系．并說明理由；（2）如圖②，連接，，過點(diǎn)G作于點(diǎn)H．點(diǎn)N是線段上一點(diǎn)，且．求證：平分；（3）在（2）的條件下，若平分，請問的大小是否發(fā)生變化？若不變，請直接寫出的度數(shù)；若改變，請說明理由．2020-2021學(xué)年度第二學(xué)期其中考試卷參考答案七年級數(shù)學(xué)說明：1．如果考生的解答與本參考答案提供的解法不同，可根據(jù)提供的解法的評分標(biāo)準(zhǔn)精神進(jìn)行評分．2．評閱試卷，要堅(jiān)持每題評閱到底，不能因考生解答中出現(xiàn)錯(cuò)誤而中斷對本題的評閱．如果考生的解答在某一步出現(xiàn)錯(cuò)誤，影響后繼部分而未改變本題的內(nèi)容和難度，視影響的程度決定對后面給分的多少，但原則上不超過后繼部分應(yīng)得分?jǐn)?shù)之半．3．評分標(biāo)準(zhǔn)中，如無特殊說明，均為累計(jì)給分．4．評分過程中，只給整數(shù)分?jǐn)?shù)．一．選擇題．（每題只有一個(gè)正確答案，每題3分，共30分）題號12345678910答案DBCDDCADDC二．填空題．（每題3分，共15分）11．4 12． 13．④ 14．2 15．120°三．解答題．（本大題8小題，共75分）16．解：原式···························································（8分）17．解：（1）如圖·······················································（2分）（2）如圖····························································（5分），，································································（8分）18．解：兩直線平行，內(nèi)錯(cuò)角相等兩直線平行，內(nèi)錯(cuò)角相等兩直線平行，同位角相等（括號內(nèi)每空2分，其余每空1分）19．解：（1）當(dāng)時(shí)，，∴··················································（2分）∴，∴2的平方根為·····················································（4分）（2）依題意得：，即，∴···············································（7分）∴這個(gè)數(shù)為···························································（9分）20．解：（1）證明：∵，∴·················································（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）依題意得：···············································（1分）解得：或-5···························································（3分）點(diǎn)P的坐標(biāo)為或························································（5分）（2）若軸，點(diǎn)P與點(diǎn)A的縱坐標(biāo)相等，即·····································（6分）···································································（8分）∴點(diǎn)P的坐標(biāo)為························································（10分）23．解：（l），·························································（2分）···································································（3分）理由如下：∵與互補(bǔ)，∴∵，∴∴··································································（6分）（2），，∴∴，∴······························································（8分）∵，∴∴平分······························································（9分）（3）不變····························································（10分）∵平分，∴即∵，∴∵，∴······························································（11分）∴，即······························································（12分）