上海市16区2018届中考一模数学试卷分类汇编:平面向量(含答案)

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上海市16区2018届九年级上学期期末(一模)数学试卷分类汇编平面向量专题宝山区20.(本题满分10分,每小题各5分)如图,AB∥CD∥EF,而且线段AB、CD、EF的长度分别为5、3、2.(1)求AC:CE的值;(2)如果AE记作a,BF记作b,求CD(用a、b表示).长宁区20.(本题满分10分,第(1)小题5分,第(2)小题5分)如图,在ABC中,点D在边AB上,DE//BC,DF//AC,DE、DF分别交边AC、BC于点E、F,且23ECAE.(1)求BCBF的值;(2)联结EF,设aBC,bAC,用含a、b的式子表示EF.崇明区20.(本题满分10分,每小题各5分)如图,在ABC△中,BE平分ABC交AC于点E,过点E作EDBC∥交AB于点D,已知5AD,4BD.第20题图FBACDE(1)求BC的长度;(2)如果ADa,AEb,那么请用a、b表示向量CB.奉贤区20.(本题满分10分,第(1)小题满分6分,第(2)小题满分4分)已知:如图,在平行四边形ABCD中,AD=2,点E是边BC的中点,AE、BD想交于点F,过点F作FG∥BC,交边DC于点G.(1)求FG的长;(2)设ADa,DCb,用、ab的线性组合表示AF.虹口区如图,在△ABC中,点E在边AB上,点G是△ABC的重心,联结AG并延长交BC于点D.(1)若ABa,ACb,用向量、ab表示向量AG;(2)若∠B=∠ACE,AB=6,26AC,BC=9,求EG的长.ABCDE(第20题图)第20题图黄浦区嘉定区金山区如图,已知平行四边形ABCD,点M、N分别是边DC、BC的中点,设=ABa,=ADb,求向量MN关于a、b的分解式.静安区闵行区浦东新区20.(本题满分10分,每小题5分)如图,已知△ABC中,点D、E分别在边AB和AC上,DE∥BC,且DE经过△ABC的重心,设BCa.(1)DE▲(用向量a表示);(2)设ABb,在图中求作12ba.(不要求写作法,但要指出所作图中表示结论的向量.)普陀区22.(本题满分10分)下面是一位同学做的一道作图题:已知线段a、b、(如图),求作线段x,使::abcx.他的作法如下:1.以点O为端点画射线OM,ON.2.在OM上依次截取OAa,ABb.3.在ON上截取OCc.(第20题图)ABCDEbacMOABCDabcN4.联结AC,过点B作BD∥AC,交ON于点D.所以:线段____________就是所求的线段x.(1)试将结论补完整:线段▲就是所求的线段x.(2)这位同学作图的依据是▲;(3)如果4OA,5AB,ACm,试用向量m表示向量DB.松江区20.(本题满分10分,每小题各5分)如图,已知△ABC中,D、E、F分别是边AB、BC、CA上的点,且EF//AB,2CFADFADB.(1)设ABa,ACb.试用a、b表示AE(2)如果△ABC的面积是9,求四边形ADEF的面积.徐汇区19.(本题满分10分,第(1)小题满分4分,第(2)小题满分6分)如图,在△ABC中,∠ACD=∠B,AD=4,DB=5.(1)求AC的长(2)若设,CAaCBbuurruurr,试用a、b的线性组合表示向量CDuuur.杨浦区20.(本题满分10分,第(1)、(2)小题各5分)已知:如图,Rt△ABC中,∠ACB=90°,sinB=35,点D、E分别在边AB、BC上,且AD∶DB=2∶3,DE⊥BC.(1)求∠DCE的正切值;(2)如果设ABa,CDb,试用a、b表示AC.(第20题图)CEFBADABCDE(第20题图)参考答案宝山区长宁区20.(本题满分10分,第(1)小题5分,第(2)小题5分)解:(1)∵23ECAE∴52ACEC(1分)∵DE//BC∴52ACECABBD(2分)又∵DF//AC∴52ABBDBCBF(2分)(2)∵52BCBF∴53BCFC∵aBC,CF与BC方向相反∴aCF53(2分)同理:bEC52(2分)又∵CFECEF∴abEF5352(1分)崇明区20、(1)∵BE平分ABC∠∴ABECBE∠∠∵EDBC∥∴DEBCBE∠∠∴ABEDEB∠∠………………………………………………………2分∴4BDDE∵EDBC∥∴DEADBCAB……………………………………1分又∵5AD,4BD∴9AB∴459BC∴365BC………………………………………2分(2)∵EDBC∥∴5=9DEADBCAB∴95BCDE…………………………………………………………1分又∵ED与CB同向∴95CBED………………………………1分∵ADa,AEb∴EDab……………………………1分∴9955CBab…………………………………………………………2分奉贤区虹口区黄浦区金山区静安区闵行区20.(本题共2小题,第(1)小题4分,第(2)小题6分,满分10分)如图,已知向量ar、br和pur,求作:(1)向量132abrr.(2)向量pur分别在ar、br方向上的分向量.20.解:(1)作图.…………………………………………………………………………(3分)结论.…………………………………………………………………………(1分)(2)作图.…………………………………………………………………………(4分)结论.…………………………………………………………………………(2分)浦东新区20.解:(1)DE23a.……………………………(5分)(2)图正确得4分,结论:AF就是所要求作的向量.…(1分).arpur(第20题图)br(第20题图)ABCDEF普陀区22.解:(1)CD;··························································································································(2分)(2)平行线分线段成比例定理(两条直线被三条平行的直线所截,截得的对应线段成比例);或:三角形一边的平行线性质定理(平行于三角形一边的直线截其他两边所在的直线,截得的对应线段成比例).···············································································································································(2分)(3)∵BD∥AC,∴ACOABDOB.················································································(1分)∵4OA,5AB,∴49ACBD.·········································································(2分)得94BDAC.·········································································································(1分)∵94BDAC,ACm,DB与AC反向,∴94DBm.··········································································································(2分)青浦区松江区20.解:(1)∵EF//AB∴CFCEFAEB又CFADFADB∴CEADEBDB…………………………………………(1分)∴DE∥AC,………………………………………(1分)∴四边形ADEF是平行四边形………………………(1分)AEAFAD……………………………………(1分)∵2CFADFADB,ABa,ACb∴13AFb,23ADa2133AEab………………………………………(1分)(2)∵EF//AB,2CFFA∴9:4:ABCCEFSS………………………………(1分)∵△ABC的面积是9,∴4CEFS……………………………………………(1分)由(1)得DE∥AC,且2ADDB∴9:1:ABCBDESS………………………………(1分)∴1BDES…………………………………………(1分)∴四边形ADEF的面积=9-4-1=4……………………(1分)徐汇区19.(1)在△ABC中,∠ACD=∠B,∠A=∠A,∴ACDABC:V.……………………………………………………(2分)∴ADACACAB,即2ACADABg∴249AC,6.AC……………………………………………(2分)(2)49CDCAADaABuuuruuruuurruuur……………………………………………(2分)4()9aACCBruuuruur4()9aabrrr………………………………(2分)5499abrr………………………………………………………(2分)杨浦区20.(本题满分10分,第(1)、(2)小题各5分)解:(1)∵∠ACB=90°,sinB=35,∴35ACAB.-------------------------(1分)∴设AC=3a,AB=5a.则BC=4a.∵AD:DB=2:3,∴AD=2a,DB=3a.∵∠ACB=90°即AC⊥BC,又DE⊥BC,∴AC//DE.∴DEBDACAB,CEADCBAB.∴335DEaaa,245CEaaa.∴95DEa,85CEa.----------(2分)∵DE⊥BC,∴9tan8DEDCECE.-----------------------------(2分)(2)∵AD:DB=2:3,∴AD:AB=2:5.------------------------------------------------(1分)∵ABa,CDb,∴25ADa.DCb.--------------------(2分)∵ACADDC,∴25ACab.-----------------------------------(2分)

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