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2018-2019学年七年级数学上学期期末检测卷 (新版)华东师大版

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  • 2021-03-31 13:48:45
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1期末检测卷一、选择题(每小题4分,共40分)1.-2017的绝对值是().A.2017B.-2017C.20171D.201712.当3x时,代数式x210的值是().A.1B.2C.3D.43.下面不是同类项的是().A.2与12B.ba22与ba2C.m2与n2D.22xy-与2212yx4.下列式子中计算正确的是().A.22550xyxyB.22523aaC.22243xyxyxyD.235abab5.下列各数中,比3大的数是().A.B.1.3C.4D.26.下列物体中,主视图是圆的是().ABCD7.中国药学家屠呦呦发明的青蒿素为保护人类健康做出了重大贡献,荣获2015年诺贝尔生理学或医学奖,奖金约为3020000元人民币.将3020000用科学记数法表示为().A.41002.3B.410302C.61002.3D.6103028.如图,锯木板前,在木板两端固定两个点,用墨盒弹一根墨线然后再锯,这样做的数学道理是().A.两点确定一条直线B.两点之间线段最短C.在同一平面内,过直线外或直线上一点,有且只有一条直线垂直于已知直线D.经过已知直线外一点,有且只有一条直线与已知直线平行9.下面图形中,射线OP是表示北偏东60°方向的是().210.一组数据:2,1,3,x,7,-9,„,满足“从第三个数起,若前两个数依次为a、b,则紧随其后的数就是2ab”,例如这组数中的第三个数“3”是由“221”得到,那么该组数据中的x为().A.-2B.-1C.1D.2二、填空题(每小题4分,共24分).11.在有理数5.0、-5、35中,属于分数的共有个.12.把多项式xx229按字母x降幂排列是.13.若50A,则A的补角为.14.在数轴上,点A表示的数是5,若点B与A点之间距离是8,则点B表示的数是.15.如图,直线a∥b,将三角尺的直角顶点放在直线b上,若∠1=35°,则∠2=.16.观察下列数字:第1层12第2层456第3层9101112第4层1617181920„„„„在上述数字宝塔中,第4层的第二个数是17,请问2510为第层第个数.三、解答题(共86分).17.(8分)计算:5×(-2)+(-8)÷(-2)18.(8分)计算:5497332(第15题图)319.(8分)先化简,再求值:yxxyyxxyyx22252223,其中1x,1y.20.(8分)如图,已知A、B、C、D是正方形网格纸上的四个格点,根据要求在网格中画图并标注相关字母.①画线段AB;②画直线AC;③过点B画AD的平行线BE;④过点D画AC的垂线,垂足为F.21.(8分)如图,点B是线段AC上一点,且20AB,8BC.(1)试求出线段AC的长;(2)如果点O是线段AC的中点.请求线段OB的长.22.(10分)根据解答过程填空(写出推理理由或根据):如图,已知∠DAF=∠F,∠B=∠D,试说明AB∥DC证明∵∠DAF=∠F(已知)∴AD∥BF()∴∠D=∠DCF()BCDA4∵∠B=∠D()∴∠=∠DCF(等量代换)∴AB∥DC()23.(10分)某水泥仓库一周7天内进出水泥的吨数如下(“+”表示进库,“-”表示出库):+30、-25、-30、+28、-29、-16、-15、(1)经过这7天,仓库里的水泥是增多还是减少了?增多或减少了多少吨?(2)经过这7天,仓库管理员结算发现库里还存200吨水泥,那么7天前,仓库里存有水泥多少吨?(3)如果进仓库的水泥装卸费是每吨a元、出仓库的水泥装卸费是每吨b元,求这7天要付多少元装卸费?24.(12分)下列是某初一数学兴趣小组探究三角形内角和的过程,请根据他们的探究过程,结合所学知识,解答下列问题.兴趣小组将图1△ABC三个内角剪拼成图2,由此得△ABC三个内角的和为180度.(1)请利用图3证明上述结论.(2)三角形的一条边与另一条边的反向延长线组成的角,叫做三角形的外角.如图4,点D为BC延长线上一点,则∠ACD为△ABC的一个外角.①请探究出∠ACD与∠A、∠B的关系,并直接填空:∠ACD=.②如图5是一个五角星,请利用上述结论求∠A+∠B+∠C+∠D+∠E的值.525.(14分)我们知道:对边平行且相等,四个角都是直角的四边形是长方形.你可以利用这一结论解答问题.(1)如图1是某直三棱柱的表面展开图.①请指出图中哪三个字母表示多面体的同一点;②如果沿BC、GH将其表面展开图.....剪成三块,恰好拼成一个长方形,那么△BMC应满足什么条件?(直接写出所有满足条件......,不必说明理由)(2)将图2中边长都是20cm的等边三角形纸片剪拼成一个底面是等边三角形的直三棱柱模型,使它的表面积与原等边三角形的面积相等;请按要求设计一种剪拼方法(用虚线表示你的设计方案,把剪拼线段用粗黑实线,在图中标注出必要的符号和数据).6答案一、1.A;2.D;3.C;4.C;5.D;6.C;7.C;8.A;9.C;10.B.二、11.2;12.229xx;13.130°;14.3或13;(每对一个得两分)15.55°;16.50、11.三、17.(本题8分)解:原式=-10+4„„„„„„„„„„„6分(化简正确每个2分)=-6„„„„„„„„„„„„„„„„„„„„„„„8分18.(本题8分)解:原式=45293„„„„„„„„„4分(化简正确每个2分)=4589„„„„„„„„„„„„„„„„6分=109„„„„„„„„„„„„„„„„„„„7分19„„„„„„„„„„„„„„„„„„„„„8分19.(本题8分)解:原式=yxxyyxxyyx22254263„4分(化简正确每个2分)xy2„„„„„„„„„„„„„„„„„„„„„5分当1,1yx时,原式=112„„„„„„„„„„„„„7分2„„„„8分(没化简直接代入求值且答案正确得3分)20.(本题8分)每画对一条得2分(点E、点F没标注各扣1分)721.(本题8分)解:(1)∵BCABAC„„„„„„„„„„„„„„„2分又∵AB=20,BC=8∴AC820„„„„„„„„„„„„„„„„„3分28„„„„„„„„„„„„„„„„„„4分(2)∵O是AC的中点,∴ACCO21„„„„„„„„„„„„„„„„„5分14„„„„„„„„„„„„„„„„„6分∴BCCOOB„„„„„„„„„„„„„„„7分8146„„„„„„„„„„„„„„„„„8分22.(本题10分)证明:∵∠DAF=∠F(已知)∴AD∥BF(内错角相等,两直线平行)„„„„2分∴∠D=∠DCF(两直线平行,内错角相等)„„„4分∵∠B=∠D(已知)„„„„„„„„„„„„6分∴∠B=∠DCF(等量代换)„„„„„„„„„8分∴AB∥DC(同位角相等,两直线平行).„„„„„10分23.(本题10分)解:(1)∵+30-25-30+28-29-16-15=-57„„„„„„„„„2分∴经过这7天,仓库里的水泥减少了57吨„„„„„„„„3分(2)∵200+57=257„„„„„„„„„„„„„„„„„„4分∴那么7天前,仓库里存有水泥257吨„„„„„„„„6分(3)依题意:进库的装卸费为:aa582830;„„„„„„„„„„„7分出库的装卸费为:bb1151516293025„„„„8分∴这7天要付多少元装卸费58115ab„10分(直接列式求得答案且正确不扣分)824.证明:(1)过点C作ABCM//„„„„„„„„„„„„„„1分ABCM//(已作)2A(两直线平行,同位角相等)„„„„2分1B(两直线平行,内错角相等)„„„„„3分018021BCA„„„„„„„„„4分0180BABCA„„„„„„„„„5分(2)①∠A+∠B,„„„„„„„„„„„„„8分②对于△BDN,∠MNA=∠B+∠D,„„„„„9分对于△CEM,∠NMA=∠C+∠E,„„„„10分对于△ANM,∠A+∠MNA+∠NMA=180o,„„11分∴∠A+∠B+∠D+∠C+∠E=180o,„„„„„„„„12分25.解:(1)点A、M、D三个字母表示多面体的同一点.„„„„„3分(2)△BMC应满足的条件是:a、∠BMC=90°,且BM=DH,或CM=DH;„„„„„„5分b、∠MBC=90°,且BM=DH,或BC=DH;„„„„„7分c、∠BCM=90°,且BC=DH,或CM=DH;„„„„„„9分(3)如上图,沿黑线剪开,把剪下的三部分拼成一个正三角形,再沿虚线折叠即可.MD21ABC92018-2019学年七年级数学上学期期末检测卷 (新版)华东师大版

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