江苏省如东县2022-2023学年下学期七年级数学期中试卷

这是一份江苏省如东县2022-2023学年下学期七年级数学期中试卷，共10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。
2022-2023学年度第二学期期中学情调研初一年级数学试卷（总分150分，考试时间120分钟）一、选择题（本大题共10小题，每小题3分，共30分．在每小题给出的四个选项中，恰有一项是符合题目要求的，请将正确选项的字母代号填涂在答题卡相应位置上．）1． 下列各数中，化简结果为﹣2023的是（ ▲ ）A．   B．   C．  D．2．不等式x≤2的解集在数轴上表示正确的是（ ▲ ）A． B．C． D．3．已知a＞b，则下列不等式中不成立的是（ ▲ ）A． B． C． D．4．若是方程          的一个解，则a的值为（ ▲ ）A．1 B． C． D．5如图，已知棋子“车”的坐标为（﹣2，﹣1），棋子“炮”的坐标为（3，﹣2），则棋子“马”的坐标为（ ▲ ）A．（1，1）       B．（﹣1，1）          C．（﹣1，﹣1）       D．（1，﹣1）      6．如图，直线，分别与、交于、，，则的度数为（ ▲ ）A．105° B．115° C．125° D．135°7． 我国古代数学名著《孙子算经》中记载了一个问题，大意是：有几个人一起去买一件物品，每人出8元，多3元；每人出7元，少4元．问有多少人？若设有x人，该物品值y元，则下面所列方程组正确的是（ ▲ ）A． B． C． D． 8．若点A（a，3），B（2，b）是与y轴平行的直线上不同的两点，且到x轴的距离相等，则点M（a，b）的坐标是（ ▲ ）A．（2，3） B．（2，﹣3） C．（﹣2，3） D．（﹣2，﹣3）9．若的解集中的最大整数解为2，则a的取值范围是（ ▲ ）A．2＜x＜3 B．2≤x＜3 C．2＜x≤3         D．2≤x≤310．一副直角三角尺叠放如图所示，现将30°的三角尺ABC固定不动，将45°的三角尺BDE绕顶点B逆时针转动，点E始终在直线AB的上方，当两块三角尺至少有一组边互相平行时，则∠ABE所有符合条件的度数为（ ▲ ）A．45°，75°，120°，165° B．45°，60°，105°，135°C．15°，60°，105°，135° D．30°，60°，90°，120°二、填空题（本大题共8小题，第11～12小题每小题3分，第13～18小题每小题4分，共30分．不需写出解答过程，请把最终结果直接填写在答题卡相应位置上）11．36的平方根是  ▲  ．12．“相等的角是对顶角”是  ▲  命题．（填“真”或“假”）13．已知二元一次方程组，则4x−y的值为  ▲  ．14．在平面直角坐标系中，点P(3，x＋1)在第四象限，那么x的取值范围为  ▲  ．15．如图，直线AB，CD相交于O，OE平分∠AOC，OF⊥OE，若∠BOD＝40°，∠DOF=▲．16．某次知识竞赛共20道题，每一题答对得10分，不答得0分，答错扣5分，小聪有一道题没答，竞赛成绩超过90分．设他答对了x道题，则根据题意可列出不等式为  ▲  ．     17．如图，把一个长方形纸条ABCD沿AF折叠，已知∠ADB=32°，AE∥BD，则∠DAF= ▲ ．18．若方程组的解是，则方程组的解是 ▲ ．三、解答题（本大题共8小题，共90分．请在答题卡指定区域内作答，解答时应写出文字说明、证明过程或演算步骤）19．（本小题满分10分）（1）计算； （2）若一个正数x的两个平方根分别是2m﹣4与3m﹣1，求x的值．  20．（本小题满分10分）（1）解方程组：；  （2）解不等式： 21．（本小题满分10分）根据题意，完成推理填空：如图，AB∥CD，∠1=∠2，∠3=∠4，试说明AD∥BE．解：∵AB∥CD（已知）∴∠4=∠  ▲  （       ▲       ）又∵∠3=∠4∴∠3=∠  ▲  （       ▲       ）∵∠1=∠2（已知）∴∠1+∠CAF=∠2+∠CAF（       ▲       ）即∠  ▲  =∠  ▲  ∴∠3=∠  ▲  （       ▲       ）∴AD∥BE（       ▲       ） 22．（本小题满分10分）如图，△ABC的顶点都在格点上，已知点C的坐标为（4，﹣1）．（1）写出点A，B的坐标；（2）平移△ABC，使点A与点O重合．作出平移后的△OB′C′，并写出点B′，C′的坐标．（3）写出线段BB′与CC′的位置和大小关系．   23．（本小题满分10分）已知（a≠0）是关于x，y的二元一次方程组．（1）求方程组的解（用含a的代数式表示）；（2）若x﹣2y＞0，求a的取值范围；      24．（本小题满分12分）《中共中央国务院关于深化教育改革全面推进素质教育的决定》中明确指出：“健康体魄是青少年为祖国和人民服务的基本前提，是中华民族旺盛生命力的体现．”王老师所在的学校为加强学生的体育锻炼，需要购买若干个足球和篮球．他曾三次在某商场购买过足球和篮球，其中有一次购买时，遇到商场打折销售，其余两次均按标价购买．三次购买足球和篮球的数量和费用如下表： 足球数量（个）篮球数量（个）总费用（元）第一次65700第二次37710第三次78693（1）王老师是第  ▲  次购买足球和篮球时，遇到商场打折销售的；（2）列方程组求足球和篮球的标价；（3）如果现在商场均以标价的6折对足球和篮球进行促销，王老师决定从该商场一次性购买足球和篮球60个，且总费用不能超过2500元，那么最多可以购买多少个篮球？             25．（本小题满分14分）如图，直线，直线EF与AB、CD分别交于点G、H，．小安将一个含30°角的直角三角板PMN按如图①放置，使点N、M分别在直线AB、CD上，且在点G、H的右侧，∠P＝90°，∠PMN＝60°．(1)填空：∠PNB＋∠PMD  ▲  ∠P（填“>”“<”或“＝”）；(2)若∠MNG的平分线NO交直线CD于点O，如图②．①当，时，求的度数；②小安将三角板PMN保持并向左平移，在平移的过程中求∠MON的度数（用含的式子表示）．
26．（本小题满分14分）在平面直角坐标系中，若点P（x，y）的坐标满足x－2y+3=0，则我们称点P为“健康点”；若点Q（x，y）的坐标满足x+y－6=0，则我们称点Q为“快乐点”．（1）已知点M（﹣1,2）、N（﹣1,1）、E（5,1）、F（3，﹣4），其中是  ▲  “健康点”，  ▲  为“快乐点”（2）若点A既是“健康点”又是“快乐点”，若B是x轴上的“健康点”，C是y轴上的“快乐点”．①求△ABC的面积；②若P为坐标轴上一点，且△BPC与△ABC面积相等，求点P的坐标． 
2022-2023学年度第二学期期中学情调研初一年级数学试卷参考答案一、选择题题号12345678910答案BBDADCABCA二、填空题11. ±6；12. 假；13. 4；14. x<-1；15. 70°；16. 10x﹣5（19﹣x）＞90；17. 29°；18. 三、解答题19．（本小题满分10分）（1）解：原式=﹣2＋（）+（）···························································3分        = ······················································································5分（2）解：根据题意得：2m﹣4+3m﹣1＝0，解得m＝1 ···························································3分∴2m－4=2－4=﹣2····························································4分∴x= ··································································5分20．（本小题满分10分）（1）解：，，得，解得y=﹣3·································································································3分把代入①，得  ·········································································4分所以原方程组的解是 ·········································································5分（2）解不等式：解：去分母得：2（2x﹣1）﹣（5x﹣1）＜4，去括号得：4x﹣2﹣5x+1＜4， ·········································································7分移项、合并得：﹣x＜5， ·············································································9分系数化为1得：x＞﹣5． ·············································································10分 21．（本小题满分10分）根据题意，完成推理填空：（评分标准：每空1分）如图，AB∥CD，∠1=∠2，∠3=∠4，试说明AD∥BE．解：∵AB∥CD（已知）∴∠4=∠BAE（两直线平行，同位角相等）又∵∠3=∠4（已知）∴∠3=∠BAE（等量代换）∵∠1=∠2（已知）∴∠1+∠CAF=∠2+∠CAF（等式的性质）即∠BAE=∠DAC∴∠3=∠DAC（等量代换）∴AD∥BE（内错角相等，两直线平行 ）   22．（本小题满分10分）解：（1）根据图象知：A（3，4），B（0，1）；···············································2分（2）如图，△OB'C'即为所求，点B'的坐标为（-3，-3），C'的坐标为（1，-5）．····8分（3）根据平移的性质知：线段BB′与CC′的位置关系为平行，大小关系为相等，即BB′∥CC′，BB′=CC′．··························10分   23．（本小题满分10分）解：（1），①+②得：3x+3y＝6，∴x+y＝2③，①﹣③得：x＝1﹣2a，············································································3分②﹣③得：y＝1+2a，············································································5分∴方程组的解为；·····································································6分 （2）∵x﹣2y＞0，∴1﹣2a﹣2（1+2a）＞0，···········································································7分∴1﹣2a﹣2﹣4a＞0，∴﹣6a＞1，∴a；·····························································································10分24．（本小题满分12分）解：（1）王老师是第  三  次购买足球和篮球时，遇到商场打折销售的；··················2分（2）解：设足球的标价为元，篮球的标价为元．根据题意，得              ，····································································4分解得：       ．························································································6分答：足球的标价为50元，篮球的标价为80元；··················································7分（3）解：设购买a个篮球，依题意有  ··················································9分解得．   ············································································11分答：最多可以买38个篮球．······························································12分25．（本小题满分14分）（1）＝ ·····································································································2分（2）①∵NO∥EF，PM∥EF，∴NO∥PM，····························································································3分∴∠ONM＝∠PMN＝60°，············································································4分∵NO平分∠MNO，∴∠ANO＝∠ONM＝60°，····································································5分∵AB∥CD，∴∠NOM＝∠ANO＝60°，····································································6分∵NO∥EF∴α＝∠NOM＝60°；····················································································7分②点N在G的右侧时，如图②，∵PM∥EF，∠EHD＝α，∴∠PMD＝α，∴∠NMD＝60°+α，∵AB∥CD，∴∠ANM＝∠NMD＝60°+α，∵NO平分∠ANM，∴∠ANO＝∠ANM＝30°+ α，∵AB∥CD，∴∠MON＝∠ANO＝30°+ α；····································································10分点N在G的左侧时，如图，∵PM∥EF，∠EHD＝α，∴∠PMD＝α，∴∠NMD＝60°+α，∵AB∥CD，∴∠BNM+∠NMO＝180°，∠BNO＝∠MON，∵NO平分∠MNG，∴∠BNO＝[180°﹣（60°+α）]＝60°﹣α，∴∠MON＝60°﹣α，  ·············································································13分综上所述，∠MON的度数为30°+ α或60°﹣α．········································14分26．（本小题满分14分）解：（1）其中是N“健康点”，E为“快乐点”······················································2分（2）①过程略，A（3，3） ···········································································4分过程略，，·····················································································6分过程略，，·······················································································8分过程略求得△ABC的面积为    ； ························································10分②过程略综上所述，点P的坐标为：或或或．答对一个1分