2023.5龙岩二检初三数学试卷+答案
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2023年龙岩市九年级学业(升学)质量检查数学试题参考答案一、选择题:本大题共10题,每题4分,共40分。题号12345678910答案BDCDCBDCAB10.【解析】RtABC中,B30,BAC90,ACB60当AE//BC时,EACACB60,由ACD沿AC翻折得ACE可得EACDACACEACB60,即ADC和AEC都是等边三角形.在RtABC中,B30,BAC90,AB3,AC3,∵ACD沿AC翻折得ACE,EACDACACEACB60,ADC和AEC都是等边三角形.DCDADB,133∴四边形ADCE的面积=ADC的面积的2倍=ABC的面积=33.22二、填空题:本大题共6题,每题4分,共24分。811.x1,x312.3.513.4或5或614.6515.8.64×1016.①③12216.【解析】法一:∵抛物线yaxbxc(a0,c0)经过A(1,3),B(2,9)两点,abc3b2a∴,解得:,4a2bc9c52a51∵c0,∴52a0,a,∴b2a,∴abc0成立,即①正确;22abca(2a)(52a)72a,57当a时,abc72a0;227当a时,abc72a0,∴abc0不一定成立,即②不一定正确;2b2a111抛物线的对称轴x(a0),2a2a2a2111抛物线开口向上,当x时,函数值y随x的增大而增大.即③正确.22a法二:已知抛物线经过A(1,3),B(2,9),C(0,c),其中点C在y轴负半轴,可以画出该抛物线的大致图象如图,九年级数学答案第1页(共10页)
1211因为,93,∴抛物线的对称轴直线xm,m,222观察图象可知,a0,b0,c0,abc0即①正确.1抛物线开口向上,当xm时,函数值y随x的增大而增大.即③正确.2x1时,yabc,从抛物线的大致图象无法确定y值符号,∴②不一定正确..三、解答题:本大题共9题,共86分。解答应写出文字说明、证明过程或演算步骤。17.(8分)332解:原式=4422.................................................................4分22=42334..................................................................6分=3...................................................................8分18.(8分)证明:四边形ABCD是平行四边形,ADBC,AD//BC,...................................................................2分DACBCA,180DAC180BCA,即DAEBCF,......................................4分ADBC在DAE和BCF中,DAEBCF,AECFDAEBCF(SAS),.....................................................................6分DEBF.....................................................................8分证法二:连结BD交EF于O点,....................................................................1分∵四边形ABCD是平行四边形,∴OAOC,OBOD,.....................................................................3分∵E,F在直线AC上,AECF,∴OEOAAEOCCFOF,...................................................................5分∴四边形BEDF是平行四边形,..................................................................6分∴DEBF..................................................................8分19.(8分)23(x2)(x2)x2x1解:原式.................................................................1分x2x22x1x2..................................................................2分2x2x2x1(x1)(x1)x2.................................................................4分2x2(x1)x12(1),............................................6分x1x1九年级数学答案第2页(共10页)
1311x1222当x时,原式3(13)...................8分2x11111122220.(8分)证明:(1)连接OC,OB点C是AB的中点,ACBC,AOCBOC,又∵OAOBOCAB(备注:写成“∵点C是AB的中点,OCAB,”不扣分)...................1分直线EF与O相切于点C,OCEF,...............2分AB//EF;..............4分(2)ODOC,OCDD,EOCOCDD2D,...........5分由(1)知,OCEF,EOCOEC90,即EOCDEF90,DEF3D,EOC2D,5D90,D18,...............7分DCFDEFD4D72................................8分21.(8分)解:(1)根据频数分布表中A组的频数为2和扇形统计图中A组的占比为5%可得n25%40,∴a2400.05,b4400.1;(注:无过程不扣分)..................4分(2)由(1)知,A组和B组是不符合教育部规定体育锻炼时间的人群,其人数占比为0.050.10.15,∴符合教育部规定体育锻炼的占比为10.150.85,..........................6分∴该校九年级的480名学生中,达到教育部规定体育锻炼时间的人数大约为4800.85408...........8分22.(10分)解:(1)设每个乙种纪念品进价为x元,则每个甲种纪念品的进价为(x4)元,..........1分400240依题意有,...............2分x4x解得:x6,...................................3分经检验,x6是原分式方程的解,x+410,....................................4分∴每个甲种纪念品和乙种纪念品进价分别是10元和6元.................................5分(2)设进货方案是甲种纪念品进货y个,那么乙种纪念品进货(400y)个,∵进货资金不超过3000元,∴10y6(400y)3000,...........................6分解得y150,.....................................................................7分由题意,专卖店获得的销售利润为W(1310)y(86)(400y)800y,.........................................................................................................................................8分∵W随着y的增大而增大,∴当y150时,W有最大值950,...................9分∴该专卖店获销售利润最大的进货方案是甲种纪念品进货150个,乙种纪念品进货250个...................................................................10分23.(10分)解:(1)作法一:作法二:作法三:九年级数学答案第3页(共10页)
....4分(2)证明:方法一:过E作EF//AC交BC于F,................................5分则CEFB................................6分由作法得,BDBE,DABEAB,DABABC,BAEABC,AE//BC,四边形ACFE是平行四边形,....................7分AEBCBE180,ACEF,.......8分ACBD,EFEB,EBCEFB,..................................................................................9分CCBE,AEBC180................................10分方法二:过A,E分别作AG^BC,EH^BC,垂足分别为G,H,......................5分由作法得,BDBE,DABEAB,DABABC,BAEABC,AE//BC,................................6分AGBH,.................................7分BDBE,BDAC,ACBE,RtACGRtEBH,CEBH,................................8分AE//BC,AEBCBE180,..................................9分AEBC180..................................10分24.(12分)解:(1)C90,ABCCAB90,.......................................1分AD平分CAB,CAB2CAD2BAD,.....................................2分ADBD,BADABC30,....................................3分ADB180BADABC120;...................................4分(2)方法一:连接DE,过点B作BHDE于点H,则DBHBDH90,九年级数学答案第4页(共10页)
1由旋转得BDBE,∴DBHEBHDBE,..............................5分2C90,∴CBACAB90,∵ABE90,∴CBADBE90∴CABDBE11AD平分CAB,CADCABDBEDBH,...........6分22∴CDABDH,................7分CDAADB180,BDEADB180,ADE180,A,D,E三点在同一直线上...........................................................................8分方法二:连接DE,在BDE中,BDBE,180DBE1BDE90DBE,.................................................5分22C90,ABE90,ABCCAB90,ABCDBE90,CABDBE,..................................................6分RtACD中,CDA90CAD,AD平分CAB,1CADCAB,21CDA90CAD90CAB,2CDABDE,.................................7分CDAADB180,BDEADB180,ADE180,A,D,E三点在同一直线上.................................8分方法三:连接AE,过点D作DGAB于G,C90,即DCAC,AD平分CAB,DCDG,CADDAG,ACDAGD(AAS),ACAG,.................................5分ABCDBG,CDGB90,BCABGD,DGAC6分BDAB,..........................................BDBE,ACAG,DGAG,.....................................7分BEAB九年级数学答案第5页(共10页)
AGDABE90,AGDABE,GADBAE,A,D,E三点在同一直线上.............................................8分(3)解法一:过点D作DGAB于G,...........................................9分C90,即DCAC,AD平分CAB,DCDG,724290,47,BGDEFB90,BEBD,BEFDBG(AAS),BFDG,BFCD3,...............................................10分EFBC,EFBEFC90,22BEEFBF5,BDBE5,则DF2,22DEDFEF25,CEFC90,56,ACDEFD,....................................................11分ADCDAD3,即EDDF252,AD35,则AE55,22ABAEBE10.......................................12分解法二:过点D作DGAB于G,........................................9分C90,即DCAC,AD平分CAB,DCDG,13724290,47,BGDEFB90,BEBD,BEFDBG(AAS),BFDG,CD3,BF3,........................................................10分2222在RtBEF中,BEBFEF345,BDBE5,DFBDBF2,EFDC90,56,DEF13,EFDABE90,EDFAEB,...................................................11分九年级数学答案第6页(共10页)
DFBEEFAB,25即,解得:AB10.....................................................12分4AB解法三:过点D作DGAB于G,过B作BHAE于H,...................9分则BGDEHB90,C90,即DCAC,AD平分CAB,DCDG,72DBE290,DBE7,BGDEFB90,BEBD,BEFDBG(AAS),BFDG,BFCD3,.............................................................10分EFBC,EF4EFBEFC90,22BFEFBF5,BDBE5,则DF2,22DEDFEF25,1EHDE52,22BHBEHE25,EHBABE90,BEHAEB,EHBEBA,.....................................11分HEBE55,即BEAB25AB,AB10.................................12分解法四:过点D作DGAB于G,过B作BHAE于H,.................9分C90,即DCAC,AD平分CAB,DCDG,72DBE290,DBE7,BGDEFB90,BEBD,BEFDBG(AAS),BFDG,BFCD3,.............................................................10分EFBC,EF4EFBEFC90,22BEEFBF5,BDBE5,则DF2,九年级数学答案第7页(共10页)
22DEDFEF25,CEFC90,56,ACDEFD,ADCDAD3,即EDDF252,AD35,∴AE=55,∵BDBE,BHAE,1∴DH=DE5,AHADDH45,.......................................11分2在RtABE和RtAHB中,AHAB2cosBAE,即ABAHAE4555100,ABAEAB0,AB10............................................................12分2222【在RtABE中,ABAEBE(55)510】25.(14分)11解:(1)直线的解析式为yx2,当x0时,y2;当yx20时,x4,22B(4,0),C(0,2),.....................................................................................1分12∵抛物线yxbxc经过B,C两点,21244bc0∴2,.....................................2分c23解得:b,c2,............................................................................................3分2123∴抛物线的解析式为yxx2;....................................4分22(2)∵APCPABPBA,APEABC,当PE平分APC时,APC2APE,∴APEPABPBA,∴PE//AB,PAPB,....................5分3∴点P在抛物线的对称轴x上,点P,E的纵坐标相同,231135∵当x时,yx22,2222435∴P(,),................................................................................................6分2412352∴yxx2,整理得:2x6x30,224九年级数学答案第8页(共10页)
315解得:x,.........................................7分2∵点E在AP上方的抛物线上,3155∴点E的坐标为(,),2433151PE15;.......................................8分222123(3)由y0,xx20,得x1,x4,1222A(1,0),B(4,0),AB5,设ODm,则CDBD4m,222在RtOCD中,根据勾股定理,(4m)m2,335解得:m,D(,0),CDBD4m,.................................9分222CF333,CFCD,FD7104∵APCAPEEPCPABPBA,又∵APEABP,∴EPCPAB,∵DCDB,∴PBADCB,∴ABP∽PCF,CFPB,PBPCCFAB,..................................................10分CPAB由B(4,0),C(0,2),可知BC25,法一:设BPt,则PC25t,代入上式,3215t(25t)5,整理得,t25t0,4413解得:t5,t5,122213BP5或5,...................................11分22过点P作PMAB于点M,则BPM∽BCO,BMMPBP,...................................12分BOOCBC151BMMP21①当BP5时,,2422541解得:BM1,MP,OMOBBM3,21点P的坐标为(3,);.....................................13分2九年级数学答案第9页(共10页)
353BMMP23②当BP5时,,2422543解得:BM3,MP,OMOBBM1,23点P的坐标为(1,);213综上所述,点P的坐标为(3,)或(1,)..........................................14分22法二:设BP5t,过点P作PMAB于点M,则BPM∽BCO,BMMPBPBMMP5t,即..........................................11分BOOCBC4225∴PMt,BM2t,OM42t,PC255t,∴P(42t,t).................................................................................12分CFPBCPAB345t3,t(2t),即(2t3)(2t1)0,255t5431解得:t,t,..........................................................................................13分122231P(1,),P(3,)..................................................................................14分1222九年级数学答案第10页(共10页)
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