2024.4福州格致中学七下数学期中考试卷

2023—2024学年第二学期期中考试七年级数学试卷满分：150分；考试时间：120分钟一、选择题（本题共10小题，每小题4分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的）1．下列个数，是无理数的是3A．9B．271πC．D．322．在平面直角坐标系中，点（2，－3）所在的象限是A．第一象限B．第二象限C．第三象限D．第四象限3．如果a＞b，那么下列不等式成立的是A．a＋1＜b＋1B．－a＜－bC．1a＜1bD．a2＞b2224．将y＝2x－1代入3x－y＝4，去括号后，可得A．3x－2x＋1＝4B．3x＋2x＋1＝4C．3x＋2x－1＝4D．3x－2x－1＝45．大多数计算器都有“”键，用它可以求出一个正有理数的算术平方根（或其近似值）．用计算器依此按键“”，“1”，“4”，“4”，“＝”，最终显示的结果是A．12B．122C．±12D．±1226．已知直线a，b，c的位置如图所示，且a⊥c，b⊥c，下列关于a∥b的证明过程正确的是A．∵∠1＝∠4＝90°，∴a∥bB．∵∠2＝∠3＝90°，∴a∥bC．∵∠1＋∠3＝180°，∴a∥bD．∵∠2＝∠4＝90°，∴a∥b3x4y16，①7．利用加减消元法解方程，下列做法正确的是5x6y33②A．要消去x，可以将①×(－5)＋②×3B．要消去x，可以将①×5－②×(－3)C．要消去y，可以将①×(－3)＋②×2D．要消去y，可以将①×6－②×438．下列关于3读法正确的是A．负的三次方根负3B．负的负3的立方根C．负3的立方根的相反数D．负的3的相反数的立方根七年级数学第1页（共4页）{#{QQABCQSUogAoAIAAARhCQQngCgMQkACCAIoOwAAAoAABSRNABAA=}#} 9．如图，10块形状，大小相同的小长方形墙砖拼成一个大长方形，设小长方形墙砖的长和宽分别为x厘米和y厘米，则下列方程中，不符合依题意的是xy50xy50A．B．2xx4yx4y5y50x4y50C．D．x4yx4y10．在平面直角坐标系中，线段AB两端点的坐标分别为A（1，0），B（2，1）．将线段AB平移后，A，B的对应点分别是C，D，则C，D的坐标可以是A．（1，－1），（0，－2）B．（1，1），（2，2）C．（2，0），（4，2）D．（3，2），（0，3）二、填空题（本题共6小题，每小题4分，共24分）x1，11．已知是方程2mx－y＝2的一组解，则m的值y2是．12．若2＜a＜3，则整数a的值可以是（写出一个满足题意的a即可）．13．如图是小九同学在体育课上跳远后留下的脚印，他的跳远成绩是线段的长度．14．如图是某机器零件的设计图纸，根据图中信息，零件长度的合格尺寸L的取值范围是．15．在平面直角坐标系xOy中，已知点A（a，1），B（2，3－b），C（4，5）．若AB∥x轴，AC∥y轴，则a＋b＝．16．某校七年级举办五子棋比赛，分为若干组，其中甲组有A，B，C，D，E五名同学，这五位同学要进行单循环赛，即每两人之间要进行一场比赛，每场比赛采用三局两胜制，即三局中胜两局就获胜（若平局则重新开始一局，直到有三局分出胜负），最终得分最高的一人出线，进入下一轮．若每场比赛胜负双方根据比分不同会获得相应的积分（0∶2，1∶2，2∶1，2∶0的积分各不相同，获胜情况越好，积分越高），且积分均为正整数，则C同学总积分y的所有可能值是．甲组ABCDE获胜场数总积分A0∶22∶11∶22∶129B2∶02∶12∶01∶2x13C1∶21∶2m0∶20yD2∶10∶2n1∶22tE1∶22∶12∶02∶1312此处的“2∶1”表示E同学与D同学的比赛中，E同学赢两局，E同学以2∶1的比分战胜D同学七年级数学第2页（共4页）{#{QQABCQSUogAoAIAAARhCQQngCgMQkACCAIoOwAAAoAABSRNABAA=}#} 三、解答题（本题共9小题，共86分．解答应写出文字说明、证明过程或演算步骤）17．（本小题满分8分）计算：3（1）2|23|；（2）2(28)．18．（本小题满分10分）解下列方程：3x4y2，①xy1，①（1）（2）42xy5．②x3y．②19．（本小题满分8分）x12x5解不等式1，并在数轴上表示解集．6420．（本小题满分8分）五子连珠棋和象棋，围棋一样，深受广大棋友的喜爱，其规则是：在正方形棋盘中，由黑方先行，轮流弈子，在任一方向（横向，竖向或者是斜着的方向）上连成五子者为胜．如图是两个五子棋爱好者甲和乙的对弈图，甲执黑子先行，乙执白子后走．若白①的位置是（0，3），白②的位置是（3，1）．（1）请根据题意，画出平面直角坐标系xOy；（2）若甲的下一步落子可以在某个方向上连成四子，请写出符合题意得的其中两个落子处的坐标．21．（本小题满分8分）据说，我国著名数学家华罗庚在一次出国访问途中，看到飞机上邻座的乘客阅读的杂志上有一道智力题：一个数是59319，希望求它的立方根．华罗庚脱口说出：39．邻座的乘客十分惊奇，忙问计算的奥妙．（1）以下步骤是华罗庚迅速准确地计算出结果的过程，请补充完整：第一步：由103＝1000，1003＝1000000，可以确定359319是①位数；第二步：由59319的个位上的数是9，可以确定359319的个位上的数是②；第三步：如果划去59319后面的三位319得到数59，而③，④，由于27＜59＜64，可以确定359319的十位上的数是3；第四步：由此求得359319＝39．（2）已知287496也是一个整数的立方，请利用上面的方法求出它的立方根．答：它的立方根是⑤位数；它的立方根的个位数是⑥；它的立方根的十位数是⑦；故287496的立方根是⑧．七年级数学第3页（共4页）{#{QQABCQSUogAoAIAAARhCQQngCgMQkACCAIoOwAAAoAABSRNABAA=}#} 22．（本小题满分10分）如图，在方格纸中平移三角形ABC至三角形DEF，使点A移动到点D，点B的对应点是点E．（1）画出平移后的三角形DEF；（2）写出BC与EF的位置关系；（3）连接BE，CF，求证：∠CBE＝∠EFC．23．（本小题满分10分）魏晋时期的数学家张丘建在古算书《张丘建算经》中提出著名的百鸡问题，即“今有鸡翁一，值钱伍；鸡母一，值线三；鸡雏三，值钱一．凡百钱买鸡百只，问鸡翁，母，雏各几何？”其大意是：公鸡5文钱1只，母鸡3文钱1只，小鸡3只1文钱，用100文钱买100只鸡，其中公鸡，母鸡和小鸡都必须要有，问公鸡，母鸡和小鸡各多少只？设公鸡，母鸡和小鸡各有x，y，z只，请完成下列问题．（1）请列出满足题意的方程组，并求出y与z（用含x的代数式表示）；（2）由于x，y，z均为小于100的正整数，请写出所有满足条件的x的值．24．（本小题满分12分）如图，三角形ABC中，过点A作直线DE∥BC．（1）求证：∠BAC＋∠B＋∠C＝180°（在下面的括号内，填上推理的依据）；证明：∵DE∥BC（已知），∴∠①＝∠B，∠EAC＝∠C（两直线平行，内错角相等）．∵∠DAB，∠BAC，∠EAC组成平角，∴②（平角定义），∴∠BAC＋∠B＋∠C＝180°（③）．在此问中，∠BAC，∠B，∠C是三角形ABC的三个内角，通过（1）的证明，我们可以得到结论：④．（2）若∠ABC和∠BAE的平分线交于点F，求∠AFB的度数；（3）在（2）的条件下，过点C作CG⊥DE，垂足为点G，连接FG，若∠CGF＝∠BAF，求证：B，F，G三点共线．25．（本小题满分12分）在平面直角坐标系xOy中，点A（a，0），B（0，b）分别在x轴负半轴和y轴正半轴上，且实数a，b满足ab44b4．（1）直接写出a，b的值以及线段AB的长；（2）若P（m，0）为线段OA上一点（不含端点），过点P作x轴垂线交直线AB于点Q．①求点Q的坐标（用含m的代数式表示）；②记点P到AB的距离为d，若0＜d＜2，判断AP与OP的大小关系，并证明．七年级数学第4页（共4页）{#{QQABCQSUogAoAIAAARhCQQngCgMQkACCAIoOwAAAoAABSRNABAA=}#} 2023—2024学年第二学期期中考试七年级数学答卷一、选择题1．D2．D3．B4．A5．A6．D7．A8．C9．C10．B二、填空题11．212．5（或6，7，8）13．PB14．9.98≤L≤10.0215．616．6或7三、解答题17．（本小题满分8分）解：（1）原式＝2＋3－2········································································2分＝(2－2)＋3·····································································3分＝3．····················································································4分（2）原式＝2(2＋2)···········································································1分＝2×2＋22······································································3分＝2＋22．··············································································4分18．（本小题满分10分）解：（1）解法一：由②，得y＝2x－5．③··························································1分把③代入①，得3x＋4(2x－5)＝2．··········································2分解这个方程，得x＝2．··························································3分把x＝2代入③，得y＝－1．···················································4分x=2，∴这个方程组的解是···················································5分y=−1．解法二：②×4，得8x－4y＝20．③·····················································1分①＋③，得11x＝22，····························································2分x＝2．················································································3分把x＝2代入②，得2×2－y＝5，－y＝1，y＝－1．·············································································4分x=2，∴这个方程组的解是···················································5分y=−1．3（2）解法一：把②代入①，得y－y＝－1．·················································2分4解这个方程，得y＝4．··························································3分把y＝4代入②，得x＝12．····················································4分x=12，∴这个方程组的解是···················································5分y=4．解法二：①×4，得x－4y＝－4．③·····················································1分把②代入③，得3y－4y＝－4，················································2分解这个方程，得y＝4．··························································3分把y＝4代入②，得x＝12．····················································4分x=12，∴这个方程组的解是···················································5分y=4．解法三：①×4，得x－4y＝－4．③·····················································1分②－③，得4y＝3y＋4，·························································2分y＝4．················································································3分{#{QQABCQSUogAoAIAAARhCQQngCgMQkACCAIoOwAAAoAABSRNABAA=}#} 把y＝4代入②，得x＝12．····················································4分x=12，∴这个方程组的解是···················································5分y=4．19．（本小题满分8分）解：去分母，得2(x＋1)＞3(2x－5)＋12．··························································2分去括号，得2x＋2＞6x－15＋12．·································································3分移项，得2x－6x＞－15＋12－2．·································································4分合并同类项，得－4x＞－5．·······································································5分5系数化为1，得x＜．·············································································6分4这个不等式的解集在数轴上的表示如图所示．054··············································8分20．（本小题满分8分）解：（1）y121-1O12x-1··············································4分（2）（4，4），（5，4）．·······································································8分【或（4，5），（6，5），（2，－1），（1，－2），（6，2），（7，2）】21．（本小题满分8分）33（1）①两；②9；③3＝27；④4＝64；·····························································4分（2）⑤两；⑥6；⑦6；⑧66．··········································································8分22．（本小题满分10分）解：（1）ACDFBE··············································3分如图所示，三角形DEF为所要画的三角形；···········································4分{#{QQABCQSUogAoAIAAARhCQQngCgMQkACCAIoOwAAAoAABSRNABAA=}#} （2）BC∥EF；························································································6分AC（3）证明：∵三角形ABC平移得到三角形DEF，点B的对应点是点E，点C的对应点是点F，DF∴BE∥CF，············································································7分B∴∠BEF＋∠EFC＝180°．··························································8分由（2），得BC∥EF，∠CBE＋∠BEF＝180°．·····························································9分E∴∠CBE＝∠EFC．·································································10分23．（本小题满分10分）xyz++=100，①解：（1）根据题意，得1···················································3分53xyz++=100．②3②×3，得15x＋9y＋z＝300．③③－①，得14x＋8y＝200，7整理，得y＝25－x．·······································································5分477将y＝25－x代入①，得x＋25－x＋z＝100，443整理，得z＝75＋x．·······································································7分4（2）满足条件的x的值有4，8，12．·······················································10分24．（本小题满分12分）（1）①DAB；·······························································································1分②∠DAB＋∠BAC＋∠EAC＝180°；····························································2分③等量代换；·························································································3分④三角形三个内角的和等于180°．······························································4分（2）解：∵BC∥DE，∴∠ABC＋∠BAE＝180°．·································································5分∵∠ABC和∠BAE的平分线交于点F，11∴∠BAF＝∠GAF＝∠BAE，∠ABF＝∠ABC，·································6分2211∴∠BAF＋∠ABF＝∠BAE＋∠ABC＝90°．·····································7分22由（1），得在三角形ABF中，∠BAF＋∠ABF＋∠AFB＝180°，∴∠AFB＝90°．··············································································8分（3）证明：∵CG⊥DE，AGDE∴∠CGA＝90°，∴∠CGF＋∠AGF＝90°．·······························································9分由（2），得∠GAF＝∠BAF．F又∠CGF＝∠BAF，BC∴∠CGF＝∠GAF，····································································10分∴∠GAF＋∠AGF＝90°．由（1），得在三角形AFG中，∠GAF＋∠AGF＋∠AFG＝180°，∴∠AFG＝90°，·········································································11分∴∠AFG＋∠AFB＝180°，∴B，F，G三点共线．·································································12分{#{QQABCQSUogAoAIAAARhCQQngCgMQkACCAIoOwAAAoAABSRNABAA=}#} 25．（本小题满分12分）（1）a＝－4，b＝4，AB＝42；······························································y········3分（2）解：①由（1），得A（－4，0），B（0，4），B∴OA＝OB＝4，且∠AOB＝90°．·······················································4分∵P（m，0）为线段OA上一点（不含端点），Q∴－4＜m＜0，∴OP＝－m，AP＝m＋4．································································5分APOx∵PQ⊥x轴交AB于点Q，∴三角形AOB的面积＝三角形APQ的面积＋梯形OPQB的面积，···········6分111∴×4×4＝×(m＋4)×PQ＋×(PQ＋4)×(－m)，222化简，得16＝m·PQ＋4·PQ－m·PQ－4m，∴PQ＝m＋4．···············································································7分∵点Q在第二象限，∴Q（m，m＋4）．·········································································8分②AP＜OP．·····················································································9分理由如下：∵三角形AOB的面积＝三角形APB的面积＋三角形OPB的面积，111∴×4×4＝×42×d＋×OP×4，222∴OP＝4－2d．··························································10y分∵0＜d＜2，B∴－2＜－2d＜0，∴2＜4－2d＜4，即OP＞2．···································································11分APOx又∴AP＝4－OP＝2d＜2，∴AP＜OP．··································································12分{#{QQABCQSUogAoAIAAARhCQQngCgMQkACCAIoOwAAAoAABSRNABAA=}#}