四川省资阳市安岳县2021—2022学年度学业质量检测七年级（下）数学期末试题(含答案)

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这是一份四川省资阳市安岳县2021—2022学年度学业质量检测七年级（下）数学期末试题(含答案)，共11页。试卷主要包含了选择题，填空题，解答题等内容，欢迎下载使用。
安岳县2021—2022学年度学业质量检测七年级·数学（本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，第Ⅰ卷1至2页，第Ⅱ卷3至8页，全卷满分150分，考试时间120分钟。）题号ⅠⅡ总分总分人一二三171819202122232425得分第Ⅰ卷（选择题共40分）得分评卷人一、选择题（本大题10个小题，每小题4分，共40分。请在每小题给出的4个选项中，将唯一正确的答案序号填在题后括号里．） 1．方程的解是（）A．=5 B．=4 C．=3.5 D．=22．下列图形中，既是中心对称图形又是轴对称图形的是（）A． B． C． D．3．已知一个正多边形的每个外角都为，则这个多边形的边数为（）A．12 B．10 C．8 D．64．若是关于、的二元一次方程，则、的值分别是（）A． B． C． D．5．下列说法正确的是（）A．若，则 B．若，则C．若，则 D．若，则 6．如图1，在△ABC中，已知点D、E、F分别为边BC、AD、CE的中点，且，则△BEF的面积为（）A．2 B．4 C．6 D．87．王实同学在解关于的方程时，误将“”看作“”，得到方程的解为，那么原方程的解为（）A． B． C． D． 8．如图2，在直角△ABC中，，点D在AB边上，将△ABC沿CD折叠，使点B恰好落在AC边上的点E处，若，则的度数是（）A． B． C． D．9．小明在拼图时，发现8个大小一样的长方形，恰好可以拼成一个大的长方形（如图3-1所示），小红看见了说：“我也来试一试” ．结果小红七拼八凑，拼成了一个正方形（如图3-2所示），中间还留下了一个小洞，恰好是边长为2的小正方形，则每个小长方形的面积为（）A．60 B．72 C．54 D．4810．如图4，一个运算程序，若输入x的值需要经过两次才能输出结果，则x的取值范围是（）A． B． C． D．第Ⅱ卷（非选择题共110分）得分评卷人二、填空题（本大题6个小题，每小题4分，共24分．请把答案直接填在题中的横线上．） 11．若是关于的方程的解，则的值为 .12．如图5，△ABC沿线段BA方向平移得到△DEF，若AB=9，AE=3，则平移的距离为 .13．若关于的不等式组的解集为，则的值为 .14．如图6，△OAD≌△OBC，且，，则 .15．定义新运算“”：对于任意有理数、都有，等式右边是通常的加法、减法及乘法运算.比如，若，则= .16．下列说法：①若线段AP，BP，AB满足AP+BP＞AB，则P点一定在线段AB外；②用两种正多边形铺满地面，正八边形不能与正方形匹配；③已知一个等腰三角形两边的长分别为4与6，则该三角形的周长一定为16；④如图7所示的图形绕着中心旋转或或后能与自身重合．其中正确的有 . （填序号）三、解答题（本大题共9个小题，共86分，解答应写出必要的文字说明、证明过程或演算步骤．）得分评卷人 17．（本小题满分10分）解下列方程（组）：（1）（2）得分评卷人 18．（本小题满分9分）解不等式组，并求不等式组的整数解. 得分评卷人 19．（本小题满分9分）如图8，△ABC的三个顶点和点O都在正方形网格的格点上，每个小正方形的边长都为1.（1）将△ABC先向右平移4个单位，再向上平移2个单位得到△A1B1C1，请画出△A1B1C1；
（2）请画出△A2B2C2，使△A2B2C2与△ABC关于点O成中心对称；（3）在（1）、（2）中所得到的△A1B1C1与△A2B2C2成轴对称吗？若成轴对称，请画出对称轴；若不成轴对称，请说明理由．得分评卷人 20．（本小题满分9分）如图9，在△ABC中，点D在BC上，点E在AC上，AD交BE于点F．已知EG//AD交BC于点G，EH⊥BE，交BC于点H，∠HEG=52°.（1）求∠BFD的度数；（2）若∠BAD=∠EBC，∠C=46°，求∠BAC的度数；得分评卷人 21．（本小题满分9分）已知关于x的方程组和有相同的解，求、的值. 得分评卷人 22．（本小题满分10分）如图10，在四边形ABCD中，，CE平分∠BCD交AB于点E，连结DE.（1）若，，求的度数；（2）若，试说明. 得分评卷人 23．（本小题满分10分）类型价格A型B型进价（元/盏）3565标价（元/盏）50100某商场用2700元购进A、B两种新型节能日光灯共60盏，这两种日光灯的进价、标价如下表：（1）求这两种日光灯各购进多少盏？（2）若A型日光灯按标价的9折出售，要使这批日光灯全部售出后商场获得不少于700元的利润，则B型日光灯应按标价的至少几折出售？得分评卷人 24．（本小题满分10分）我们定义：如果两个一元一次不等式有公共整数解，那么称这两个不等式互为“云不等式”，其中一个不等式称为另一个不等式的“云不等式”.（1）在不等式：①，②，③中，不等式的“云不等式”是（填序号）；（2）若关于的不等式不是的“云不等式”，求的取值范围；（3）若，关于的不等式与不等式互为“云不等式”，求的取值范围. 得分评卷人 25．（本小题满分10分）如图11-1，在直角△ABC与直角△BCD中，∠ACB=∠DCB=90°，∠A=30°，∠D=45°，固定△BCD，将△ABC绕点C按顺时针方向旋转一个大小为的角（）得到△ACB，.（1）在旋转过程中，当B，C⊥BD时，= °；（2）如图11-2，旋转过程中，若边AB，与边BC相交于点E，与BD相交于点F，连结AD，设∠DAB，=，∠BCB，=，∠ADB=，试探究的值是否发生变化，若不变化请求出这个值；若变化，请说明理由；安岳县2021—2022学年度学业质量检测七年级·数学答案一、选择题（共10小题，每小题4分）题号12345678910答案DBBCDBCCAC二、填空题（共6小题，每小题4分）11．1 　 12．6 13．5 14． 15．3 16．①④三、解答题（共9小题）17．解：（1）·························································5分·······················（2） ································································10分18．解：解不等式，得 ··················································6分·······················所以原不等式组的解为. ···················································7分·······················故此不等式组的整数解为：－３，－２，－１，０·······························9分·······················19．解：（1）如答图1所示：ΔA1B1C1，即为所求······························3分·······················（2）如图所示：ΔA2B2C2，即为所求········································6分·······················（3）如图所示：ΔA1B1C1与ΔA2B2C2成轴对称，································7分·······················直线、即为所求·························································9分·······················20．解：（1）∵EH⊥BE，∴∠BEH=90°，···································1分∵∠HEG=52°，∴∠BEG=38°，···········································2分又∵EG//AD，∴∠BFD=∠BEG=38°；·······································4分（2）∵∠BFD=∠BAD+∠ABE，∠BAD=∠EBC，·································5分∴∠BFD=∠EBC+∠ABE=∠ABC=38°，········································7分∵∠C=46°，∴∠BAC=180°-∠ABC-∠C=180°-38°-46°=96°.·······················9分21．解：由题意得：，解得，··············································4分将代入得，解得，·······················································9分22．解：（1）∵∠B+∠ADC=180°，∠A+∠B+∠BCD+∠ADC=360°，∴∠A+∠BCD=180°，······················································1分∵∠A=50°，∴∠BCD=130°，················································2分∵CE平分∠BCD，∴∠BCE=∠BCD=65°，·······································3分∵∠B=80°，∴∠BEC=180°-∠BCE-∠B=180°-65°-80°=35°；··························5分（2）由（1）知，∠A+∠BCD=180°，∴∠A+∠BCE+∠DCE=180°，················································6分∵∠CDE+∠DCE+∠1=180°，∠1=∠A，∴∠BCE=∠CDE，························································8分∵CE平分∠BCD，∴∠BCE=∠DCE，··········································9分∴∠CDE=∠DCE.·························································10分23.解：（1）设购A型日光灯x盏，购B型日光灯y盏，由题意得·····································································3分·······················解得：，答：购A型日光灯40盏，购B型日光灯20盏.·····································5分（2）设B型日光灯应按标价的m折出售，由题意得，···································································8分得m≥8. 答：B型日光灯应按标价的至少8折出售·······································10分·······················24．解：（1）①②；······················································2分（2）解不等式可得，解不等式可得，·························································4分∵关于的不等式不是的“云不等式”，∴，解得；·····························································6分（3）当时，即时，不等式的解集为，不等式的解集为，···························8分∵关于的不等式与不等式互为“云不等式”，∴，即，∴；····························································10分25．解：（1）45；······················································3分·······················（2）结论：的值不变，理由如下：··········································4分·······················如图11-2中，在直角ΔABC与直角ΔBCD中，∠ACB=∠DCB=90°，∠A=30°，∠D=45°，∴∠B=45°，∴∠B，=60°，··················································5分·······················∵∠EFB是ΔDFA的一个外角，∴∠EFB=∠DAB，+∠ADB，∠EFB=①，·······································6分·······················又∵∠BEF是ΔCB，E的一个外角，∴∠BEF=∠BCB，+∠B，，∴∠BEF=②，······································7分·······················∴①+②得：∠EFB+∠BEF=·················································8分·······················又在ΔEFB中，∠B=45°，∴∠EFB+∠BEF=180°-45°=135°，············································9分·······················∴=135°，∴····························································10分·······················