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文档介绍:该【2024年上海市普陀区初三中考一模数学试卷及答案 Word含解析 】是由【小屁孩】上传分享,文档一共【10】页,该文档可以免费在线阅读,需要了解更多关于【2024年上海市普陀区初三中考一模数学试卷及答案 Word含解析 】的内容,可以使用淘豆网的站内搜索功能,选择自己适合的文档,以下文字是截取该文章内的部分文字,如需要获得完整电子版,请下载此文档到您的设备,方便您编辑和打印。:..2020年上海市普陀区初三一模数学试卷一、?,那么下列等式中,不一定正确的是()y5x?y8xx???y?8C.?D.?y5yy?,如果函数图像的对称轴是y轴,那么这个函数是()?x2??x2?2x??x2???x?1?,∠C=90°,sinA?,那么下列说法中正确的是()????,正确的是()=0,a是非零向量,那么ka?,那么e??a,那么b?a或b??,如果向量b??5a,那么a//??x?m?2?ny?mx?n的图像如图1所示,那么一次函数的图像经过()、二、、三、、二、、三、,在RtVABC中,∠ACB=90°,CD⊥AB,垂足为点D,如果VADC?,AD=9,那么BC的长C2VCDB是():..二、填空题r1rrr????:2?a?b??a?b?____________?2?y??a?2?,那么的取值范围是____________f?x??3x2?2x?1x?2f?x??,如果,?ax2?2ax?c与x轴的一个交点的坐标是(1,0),?x2?2x?2的图像向下平移m(m>0)个单位后,它的顶点恰好落在x轴上,,∠C=90°,cotB?,BC=2,那么AC=,VABC的中线AD、CE交于点G,点F在边AC上,GF//BC,,在VABC与VAED中,?,要使VABC与VAED相似,还需添加一个条件,这个条AEED件可以是____________(只需填一个条件),在RtVABC中,∠C=90°,AD是三角形的角平分线,如果AB?35,AC?25,,斜坡AB长为100米,坡角∠ABC=30°,现因“改小坡度”工程的需要,将斜坡AB改造成坡度i=1:5的斜坡BD(A、D、C三点在地面的同一条垂线上),那么由点A到点D下降了____________米(结果保留根号):..,在四边形ABCD中,∠ABC=90°,对角线AC、BD交于点O,AO=CO,CD⊥BD,如果CD=3,BC=5,那么AB=,在RtVABC中,∠C=90°,AC=5,sinB?,点P为边BC上一点,PC=3,将VABC绕点13P旋转得到VA'B'C'(点A、B、C分别与点A'、B'、C'对应),使B'C'//AB,边A'C'与边AB交于点G,那么A'G的长等于____________三、解答题2sin260??cos60?:tan260??4cos45?,在VABC中,点D、E、F分别在边AB、AC、BC上,DE//BC,EF//AB,AD:AB=1:3.(1)当DE=5时,求FC的长;uuurruuurruuuruuurrr(2)设AD?a,CF?b,那么FE?______,EA?______(用向量a,b表示).:..,在VABC中,点P、D分别在边BC、AC上,PA⊥AB,垂足为点A,DP⊥BC,垂足为点P,APBP?.PDCD(1)求证:∠APD=∠C;(2)如果AB=3,DC=2,?y?A?3,k?2?与函数(m、k为不等于零的常数)的图像有一个公共点,其中正比例函xk数y的值随x的值增大而减小,:如图11,四边形ABCD的对角线AC、BD相交于点O,S?(1)求证:?;OBOACDVOAB?kS??k?1?2S.(2)设的面积为S,,求证:AB四边形ABCD?8?xOyy?ax2?a?x?c?a?0?A??3,?2?(如图12),已知抛物线??经过点,?3?:..B?0,?2?x与y轴交于点,抛物线的顶点为点C,对称轴与轴交于点D.(1)求抛物线的表达式及点C的坐标;(2)点E是x轴正半轴上的一点,如果∠AED=∠BCD,求点E的坐标;(3)在(2)的条件下,点P是位于y轴左侧抛物线上的一点,如果VPAE是以AE为直角边的直角三角形,,在梯形ABCD中,AD//BC,∠C=90°,AD=2,BC=5,DC=3,点E在边BC上,tan∠AEC=3,点M是射线DC上一个动点(不与点D、C重合),联结BM交射线AE于点N,设DM?x,AN?y.(1)求BE的长;(2)当动点M在线段DC上时,试求y与x之间的函数解析式,并写出函数的定义域;(3)当动点M运动时,直线BM与直线AE的夹角等于45°,:..一、选择题:(本大题共6题,每题4分,满分24分)1.(B);2.(C);3.(A);4.(D);5.(B);6.(C).二、填空题:(本大题共12题,每题4分,满分48分)???2;;?2b;10.(?3,0);;;1ABAC13.;14.?B??E(?等);;?103;17.;18..413三、解答题(本大题共7题,其中第19---22题每题10分,第23、24题每题12分,第25题14分,满分78分)312?()2?:原式?···································································(4分)2(3)2?4?231?22?·······················································································(3分)3?22?3?22.······················································································(3分):(1)∵DE//BC,EF//AB,∵DE?BF.····················································································(1分)∵DE?5,∵BF?5.·······································································(1分)∵DE//BC,ADDE∵?.··················································································(1分)ABBCAD151∵?,∵?.·····································································(1分)AB3BC3解得BC?15,················································································(1分)FC?10.························································································(1分)uuurruuurrrEA?1(2)FE??2a,?a?b.·······················································(2分+2分):(1)∵PA?AB,DP?PC,:..∵?BAP??CPD?90?.····································································(1分)在Rt△ABP与Rt△PCD中,APBP?,PDCD∵Rt△ABP∵Rt△PCD.····································································(1分)∵?APB??PDC.············································································(1分)∵?DPB??APB??APD,?DPB??PDC??C,得?APD??C.···············································································(2分)(2)∵Rt△ABP∵Rt△PCD.∵?B??C.∵AB?AC.····················································································(1分)∵AB?3,DC?2,∵AD?1.···························································(1分)∵?APD??C,?PAD??CAP,∵△APD∵△ACP.·········································································(1分)ADAP∵?.·················································································(1分)APAC得AP?3.····················································································(1分)xA?3,k?2?y?:由点在函数的图像上,可得k3k?2?.················································································(1分)k整理,得k2?2k?3?0.··································································(1分)解得k?3,k??1.·····································································(2分)12∵正比例函数y的值随x的值增大而减小,∵k??1.························································································(2分)y??x?3,?3?得,点A.·································································(2分)mA?3,?3?y?由点在函数的图像上,可得xm??9.···················································································(1分)9∵y??.······················································································(1分)x9两个函数的解析式分别为y??x,y??.x:..:(1)过点A作AH⊥BD,垂足为点H.····················································(1分)11∵S=?DO?AH,S=?OB?AH,△AOD△AOB221?DO?AHS2DO∴?AOD??.·····························································(2分)S1OB?AOB?OB?AH2SCO同理,?BOC?.·········································································(1分)SOA?AOB∵S?S,△AOD△BOCDOCO∵?.···············································································(1分)OBOADOCO(2)∵?,?COD??AOB,OBOA∵△OCD∵△OAB.······································································(1分)CDDOCO∵???k.··································································(1分)ABBOAOSCD2???OCD????k2.··································································(1分)S?AB??OAB∵△OAB的面积为S,∴S?k2?S.············································(1分)?OCDSDO又∵?AOD??k,∵S?k?S.·············································(1分)SOB?AOD?OAB同理,S?k?S.······································································(1分)?BOC∴S?S?S?S?S四边形ABCD△AOB△BOC△COD△DOA?S?k?S?k2?S?k?S?(k2?2k?1)?S?(k?1)2S.································································(1分):(1)由抛物线y?ax2?(a?)x?c经过点A??3,?2?和点B?0,?2?,3:..?c??2,?4??a?,得?8解得?3···············································(2分)?9a?3(a?)?c??2.??3?c??∵抛物线的表达式是y?x2?4x?2.·············································(1分)33点C的坐标是(?,?5).···································································(1分)2(2)联结AB交CD于点F,过点A作AH?OD,H为垂足.∵A??3,?2?,B?0,?2?,∵AB??.·····································································(1分)2∵CD?5,∵CF?△BCF中,tan?BCF??.·················································(1分)CF2AH在Rt△AEH中,tan?AEH?,EH∵?AED??BCD,AH1∵?.∵EH?4.·····································································(1分)EH2∵OH?3,∵OE?1.∵点E的坐标是?1,0?.·······································································(1分)(3)∵△PAE是以AE为直角边的直角三角形,∵?PAE?90?或?PEA?90?.4设点P点的坐标为(m,m2?4m?2).3①当?PAE?90?时,?AH,∵PG?m?3,AG??m2?∵?GAE??AHE??AEH,?GAE??PAE??PAG,∵?PAG??AEH.∵tan?PAG?tan??31∵?.∵?.····················································(1分)AGEH42?m2?4m33解得m??3,m??.23∵m??3不合题意舍去,∵m??.23∵点P的坐标是(?,?5).·······························································(1分)2:..②当?PEA?90?时.?9?12913?129同理可得点P的坐标是(,).···································(2分):(1)过点A作AH?BC,H为垂足.∵AH?BC,∴?AHE?90?.∵?C?90?,∴?AHE??C.∴AH//DC.∵AD//BC,DC?3∴AH?DC?3.······························································(1分)同理可得HC?AD?2.···························································································(1分)AH在Rt△AEH中,?AHE?90?,tan?AEH?3,∴?∴EH?1.················································································································(1分)∵BC?5,∴BE?2.·····························································································(1分)(2)延长BM、AD交于点G.·············································································(1分)DGDM∵DG//BC,∴?.BCMC由DM?x,DC?3,BC?5,DGx5x得?,解得DG?.··········································································(1分)53?x3?x6?3x∴AG?.·········································································································(1分)3?xANAG∵AG//BC,∴?.BNBE在Rt△AEH中,?AHE?90?,EH?1,AH?3,可得AE?10.·········································································································(1分)6?3xy3?x由?,10?y2310x?610化简,得y?(0?x?3).····························································(2分)12?x1(3)①当点M在线段DC上时,DM?.·························································(2分)2②当点M在线段DC的延长线上时,点N在线段AE的延长线上,DM?13.(2分)

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