江苏省苏州工业园区某校2021-2022学年七年级下学期期末调研数学试卷 (word版含答案)

这是一份江苏省苏州工业园区某校2021-2022学年七年级下学期期末调研数学试卷 (word版含答案)，共8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。
2021—2022学年第二学期期末调研试卷七年级数学一、选择题（本大题共10小题，每小题2分，共20分．在每小题所给出的四个选项中，恰有一项是符合题目要求的，请将正确选项前的字母代号填涂在答题卡相应位置上）1．a3·a2的结果是A．aB．a5 C．a6D．a92．氢原子的半径约为0.000 000 000 05 m，用科学记数法表示0.000 000 000 05是A．5×10－9B．0.5×10－10C．5×10－11D．5×10－123．若a＞b，则下列不等式不成立的是A．－3a＞－3b       B．3a＞3b    C．＞       D． a＋3＞b＋34．下列因式分解正确的是（　　）A．a2+b2＝（a+b）2 B．a2+2ab+b2＝（a﹣b）2 C．2a2﹣a＝2a（a﹣1） D．a2﹣b2＝（a﹣b）（a+b）5．若△ABC≌△DEF，且∠A＝60°，∠B＝70°，则∠F的度数为（　　）A．50° B．60° C．70° D．80°6．若x2﹣mx+16是完全平方式，则m的值等于（　　）A．2 B．4或﹣4 C．2或﹣2 D．8或﹣87．如图，AB∥CD，BC平分∠ABD，若∠1=65°，则∠2的度数是A．65°B．60°C．55°D．50°         第7题                              8．一辆汽车从A地驶往B地，前路段为普通公路，其余路段为高速公路．已知汽车在普通公路上行驶的速度为60km/h，在高速公路上行驶的速度为100km/h，汽车从A地到B地一共行驶了2.2h．设普通公路长、高速公路长分别为xkm、ykm，则可列方程组为A．B．C．D．9．在△ABC中，AC＝5，中线AD＝4，那么边AB的取值范围为（　　）A．1＜AB＜9 B．3＜AB＜13 C．5＜AB＜13 D．9＜AB＜1310． 甲，乙，丙三人进行乒乓球比赛，规则是：两人比赛，另一人当裁判，输者将在下一局中担任裁判，每一局比赛没有平局．已知甲，乙各比赛了4局，丙当了3次裁判．问第2局的输者是（　　）A．甲 B．乙 C．丙 D．不能确定二、填空题（本大题共8小题，每小题2分，共16分．不需写出解答过程，请把答案直接填写在答题卡相应位置上）11．命题“两直线平行，内错角相等”的逆命题是▲．12．把方程2x＋y＝3写成用含x的代数式表示y的形式，则y＝  ▲  ．13．一个三角形的三边为2、4、x，另一个三角形的三边为y、2、5，若这两个三角形全等，则x+y＝▲ ．14．如图，直线a、b被直线c所截，∠1＝50°．当∠2＝▲ °时，a∥b．15．关于x，y的方程组的解满足x－y＝6，则m＝  ▲  ．16．如图，点D在AB上，点E在AC上，BE、CD相交于点O，∠A＝40°，∠C＝30°，∠BOD＝100°．则∠B＝ ▲ °．     17．如图，∠A＋∠B＋∠C＋∠D＋∠E＋∠F＝  ▲  °．18．若关于x的一元一次不等式组仅有2个整数解，则m的取值范围是 ▲ ．三、解答题（本大题共64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）17．（6分）计算（1）(2a2)3÷(a2)2；（2）(a＋b)(a－3b)．    18．（6分）分解因式（1）2a(x－y)＋b(y－x)；（2）4a2－16a＋16．    19.（5分）先化简，再求值：(a＋2b)(a－2b)＋(a－2b)2，其中，a＝，b＝1．    20.（4分）解方程组 21.（5分）解不等式组请结合题意，完成本题的解答.（1）解不等式①，得 ▲ .（2）解不等式③，得 ▲ .（3）把不等式①、②和③的解集在数轴上表示出来.   （4）从图中可以找出三个不等式解集的公共部分，得不等式组的解集▲ .      22.（5分）画图并填空：如图，方格纸中每个小正方形的边长都为1，△ABC的顶点都在方格纸的格点上，将△ABC经过一次平移，使点C移到点C'的位置．（1）请画出△A'B'C'；（2）连接AA'、BB'，则这两条线段的关系是▲；（3）在方格纸中，画出△ABC的中线BD和高CE；（4）线段AB在平移过程中扫过区域的面积为▲．           23．（8分）如图，△ABC中，AB＝BC，∠ABC＝90°，F为AB延长线上一点，点E在BC上，且AE＝CF．（1）求证：Rt△ABE≌Rt△CBF；（2）若∠CAE＝30°，∠BAC＝45°，求∠ACF的度数．         24．（7分）为了进一步丰富校园活动，学校准备购买一批足球和篮球．购买7个足球和5个篮球的费用相同；购买40个足球和20个篮球共需3400元.(1)求每个足球和篮球各多少元？(2)如果学校计划购买足球和篮球共80个，总费用不超过4800元，那么最多能买多少个篮球？          25．（8分）如图，∠1，∠2，∠3，∠4是四边形ABCD的四个外角．用两种方法证明∠1+∠2+∠3＋∠4＝360°．         26．（10分）数学概念百度百科这样定义凹四边形：把四边形的某些边向两方延长，其他各边有不在延长所得直线的同一旁，这样的四边形叫做凹四边形．如图①，在四边形ABCD中，画出DC所在直线MN，边BC、AD分别在直线MN的两旁，则四边形ABCD就是凹四边形．性质初探（1）在图①所示的凹四边形ABCD中，求证：∠BCD＝∠A＋∠B＋∠D．深入研究（2）如图②，在凹四边形ABCD中，AB与CD所在直线垂直，AD与BC所在直线垂直，∠B、∠D的角平分线相交于点E．①求证：∠A＋∠BCD＝180°；②随着∠A的变化，∠BED的大小会发生变化吗？如果有变化，请探索∠BED与∠A的数量关系；如果没有变化，请求出∠BED的度数．               
七年级数学参考答案一、选择题 (本大题共10小题，每小题2分，共20分12345678910BCADCDDCBC二、填空题 (本大题共8小题，每小题2分，共16分)11．内错角相等，两直线平行 12．y＝－2x＋3   13．9     14．130    15．4           16． 10      17．360            18．　3＜m≤4三、解答题（本大题共64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）17．（6分）（1）原式＝8a6÷a4·····································2分＝8a2·············································3分（2）原式＝a－3ab＋ab－3b·····································2分＝a－2ab－3b·········································3分18．（6分）（1）原式＝2a(x－y)－b(x－y)·····························1分＝(x－y)(2a－b)··································3分（2）原式＝4(a2－4a＋4)··································1分＝4(a－2)2．····································3分19.（5分）原式＝a2－4b2＋a2－4ab＋4b2·································2分＝2a2－4ab，············································3分把a＝，b＝1代入得，原式＝－··································5分20.（4分）解：由①得y＝x－3……③·································2分把③代入②得x＝2···················································3分把x＝2代入③得y＝－1················································4分所以原方程组的解为·················································5分解法二：①×3得3x－3y＝9·············································1分①－②得5y＝－5·········································2分解得y＝－1·····················································3分把y＝－1代入①得x＝2················································4分所以原方程组的解为·················································5分21.（5分）（1）x≥－3.···········································1分（2）x＜1.·················································2分（3）数轴表示略. （图形、描点各1分）··························4分（4）－2＜x＜1·············································6分      22.（5分）（1）如图······················2分（2）AA'∥BB'且AA'＝BB'··················4分（3）略································6分（4）12································8分21．（本题8分）证明：（1）HL（2）60°24．（7分）（1）解：设每个足球为x元，每个篮球为y元．·····························································2分解得：答：每个足球为50元，每个篮球为70元．······························4分（2）解：设买篮球m个，则买足球（80－m）个．70m＋50(80－m)≤4800············································6分m≤40························································7分∵m为整数∴m最大取40答：最多能买40个篮球．·······················8分 25．（8分）证法1：∵∠1＋∠BAD＝180°，∠2＋∠ABC＝180°，∠3＋∠BCD＝180°，∠4＋∠CDA＝180° 1分∴∠1＋∠BAD＋∠2＋∠ABC＋∠3＋∠BCD＋∠4＋∠CDA＝180°×4＝720°．·····2分∵∠BAD＋∠ABC＋∠BCD＋∠CDA＝360°，······························3分∴∠1+∠2+∠3＋∠4＝360°．··········································4分证法2：连接BD，···················································1分∵∠1＝∠ABD+∠ADB，∠3＝∠CBD+∠CDB，····························2分∴∠1+∠2+∠3＋∠4＝∠ABD+∠ADB＋∠2+∠CBD+∠CDB＋∠4，·············3分＝180°×2＝360°．···································4分    
26．（本题10分）（1）证明：如图①，延长DC交AB于点E．∵∠BEC是△AED的一个外角，∴∠A＋∠D＝∠BEC．·········································1分同理，∠B＋∠BEC＝∠BCD．···································2分∴∠BCD＝∠A＋∠D＋∠B．····································3分       （2）①证明：如图②，延长BC、DC分别交AD、AB于点F、G．   由题意可知∠AFC＝∠AGC＝90°．····························4分                   又∵在四边形AFCG中，∠AFC＋∠AGC＋∠A＋∠FCG＝360°，∴∠A＋∠FCG＝180°．····································5分∵∠FCG＝∠BCD，∴∠A＋∠BCD＝180°．····································6分②解：由（1）可知，在凹四边形ABED中，∠A＋∠ABE＋∠ADE＝∠BED．①·····························7分同理，在凹四边形EBCD中，∠BED＋∠EBC＋∠EDC＝∠BCD．②···························8分∵BE平分∠ABC，∴∠ABE＝∠EBC．同理，∠ADE＝∠EDC．①－②，得∠A＋∠BCD＝2∠BED．····························9分由（2）①可知，在凹四边形ABCD中，∠A＋∠BCD＝180°．∴2∠BED＝180°，∴∠BED＝90°．··········································10分