2014北京丰台中考一模数学(含解析)

1/14PDCBAEDCBA132412013“”350.18350.18A35.01810B23.501810C33.501810D20.3501810212A12B12C2D23164A23B12C13D164ABC90ADACDEBC1145BA35B25C45D555PACDCABBDCDBD1.2AB1.8BP12PDA6B8C18D246ABCD7122451/585560482/14xyB3B2B1A4A3A2A1OFCEDBA112A60B60C12D578ABCD4cmAB23cmADECDPAAEECCQAABBCC1cm/sPQ(s)tAPQ2(cm)yyt二、填空题（本题共16分，每小题4分）9324xxy___________________101x__________________11ABC90ACBFAC12CFACDEABC1302DEAFED______________12l33yx1A(0,1)1Ayl1BO1OBy2A2Ay2BO2OBy3A4A_____________nA__________________QPEDCBA3/14PFEDCBA三、解答题（本题共30分，每小题5分）13ECDAABDFEDABECPDABCDEF[Com][14113tan303215316122xxxx1623210xx2(3)2(2+)7xxx[ZXXK]4/14GFEDCBA17列方程或方程组解应用题：“”400060004018x2(1)210mxmxm12m四、解答题（本题共20分，每小题5分）19ABCDEFABCDBDAAGDBCBG1DEBF290G60C=2BCDEBF5/14OFEDCBA20RtABC90ACBDABBDOACEDEBCF1BDFF21CF3sin5AO211E2__________________23400380160170x/cmA155xB155160xC160165xD165170xE170x6/14221AFACBBEACDCEEDFAFAF2ABOCD(3,4)A(0,2)B(0,0)O(4,0)C(4,2)DAABOCD五、解答题（本题共22分，第23题7分，第24题7分，第25题8分）231L22yxbxcx(1,0)A(3,0)B2L243ykxkxk0kP1b___________c___________290APBk315yk2LEFEFEFxy1234432112344321OFEDCBA11yxODCBA217/14ABCEDFGHCHFGEPBDA24ABC90BACABAC11DEABACAFBEBCFEFCDHEFCD22ADAEAFBEGBCFFFPCDBEPBPFPAF[XK]25xOy2yaxcx(2,0)ABy(0,23)CACPA1CABQB2At12APQAOC?t3y(0,)Mm(0,1)NmAMMNNPmtAMMNNP8/141B2A3C4D5B6D7A8B9(2)(2)xxyxy1022yxx111612(0,8)1(0,2)n13ABDFBCPDAEDFECPDEBABCDEFBEDEABBACEDFABCDEF(ASA)143=23133=115316xx1x122xx13x113x1622=69427xxxx2=322xx23210xx2321xx=12117x(40)x4000600040xx80x80x408040120x9/14801201812(1)210mxmxm1m2=(2m)4(m1)(m1)4022212(1)1mmxmm11x2121+11mxmmm11m12m2m3m2m3m191ABCDABCDABCD12DFCD12BEABDEBEDEBEDEBF2AGDBADBC90GADBG90DBGDBC60C2BC23BD12232DEBFDFBBCDSSSBCBDDEBF23201OEOACE90OEC90ACBOEBCFOEDODOEOEDODEBDFF21=32OEBFx6BFx3sin5OEAAO5AOx10/148ABAOBOx3sin5BCAAB245BCx2466155CFBFBCxxx56x532OExO52211137.5%17.5%25%15%5%25%=40402=802401086412310+18400+380(25%+15%)=3324022AOACBBMAOxMDDNxNMNFAFAO43yxBM423yx3(,0)2MAC416yxDN418yx9(,0)2NMN(1.5,0)FAF843yx23122(1)(3)286yxxxx8b6c222243(43)(2)ykxkxkkxxkxk(2,)Pk90APB(2,1)P(2,1)P1k3243=15kxkxkk24120xx16x22x11/148EFEFEF824CCMABAFMABEACDABACBAECADAEADABEACD(SAS)ABEACDBCDCBE90BACAFBE90ABGBAGCAMBAGABECAMCMAB90BAEACMABEACMABECAMABACBAEACMBAEACM(ASA)ACCECMECFMCF45CECMECFMCFCFCFECFMCF(SAS)EFCMFCBFG90CBGBFG90BCDCFEEFCD2AOFOABEACDABEACDBCDCBEOBOC45BAOCAOABOCAFABOCAFABCABAOACFABOCAF(ASA)BOAFAOCF12/14AFCOAOFCFOAOCFAFCOOFOFAOFCFOAFOCOFPOHPFEPOFPFOPOPFBPBOOPAFPF2512yaxc(2,0)A(0,23)C3223ac23232yx23232yx2APt2BQt4AB42AQtAQPAOC12AQAOAPAC85tAPQAOC12APAOAQAC1t1t85t3A(2,1)EEy(2,1)FFFHACHyNPHAMMNNP1AE4EF60CAO33EG233AG343FG13226PGFG312322PFMF322tAGPGPF3231+33yx23(0,1)3N23(0,)3M233mAMMNNP112312322PFMN13/141.B350.1823.501810B2.A1212A3.C456421=63C4.D114535EDCDEBC=35CEDC90BC55BD5.BABPCDP1.21.8ABCDBPPD12PD8CDB6.DD7.A60125755A8.B4AEAB04tAPQ234yx46t1111(24)234(4)23(4)(6)(423)2222ytttt6623t1(623)42ytB9.(2)(2)xxyxy32224(4)(2)(2)xxyxxyxxyxy(2)(2)xxyxy14/1410.22yxx1x12ba22yxx11.16CDDEDEAC1=22DEAC12CFACDECFDCFE1302DECF4EFDCACADAFED161612.(0,8)(0,2)n1130ABO11OA122OBOA234OBOA348OBOA12nnOA(0,8)(0,2)n.