上海市宝山区2020年八年级上学期期末数学试题

20一、选择题（本大题共小题，每题分，满分分）63182x8xx22；2；xC.2x2x.的A22AxxxB2x2Bx2xC842222CxxxxD22xD.x2x.1122a；；y；ax22xxC.AAB2aBa02axC02y20yCDy3D..Bah.AABBCDah.1C2A；A；A；A；A30.1C90,2A30.(1x%)；1；；；2D.axa(1x%)2aa，ax%)2ax1x%x%xx.2a.41xA作yB作x3的x()8642AB△．．k12题，每题3分，满分36分）123a2aa..123232aa.aaa.0．x12或xx．12xx．120242xxx22x22.4x2(x4x2(x((x2x2.x2222(x2x2.0.a()2aa2x1y1x2.2x101x212：x.632．x(2)2x2.由fx662622(2)32.f22232..2yx．x2x0xy1x2y120.x2x0.：.．..和CBDCBD.ABCB,ABECBD60,BEBDABECBD(SAS)CBD..．AB.段ABAB.．ABBCACECEADEACAE∴EC∵EBCBCAEBEBCECBEBCC10EBC∴ABABAC6ABACBC10616C.G，．°CE,DGBEAF，GM，且GMMDFMGMDFMGAEAC,DEDBGAFCE,DGBEAFD△AGD和点M111AD,AD,222FMGM,GMMD又点HMH//AG,GMHDMHDMHMAE25GMHDMH25GMD50FMG180AMFGMD180405090FMGM245..A．l2xyOAADx交，BCADA作ABBAC交AOD点By.BOAADx交，为A作ABAD，BC交(3,1)OD3,AD1AOBABOA45OAB90OBA45OAABADxBCAD交CODAACB90AODOAD90OADBAC180OAB90AODBACBAC在△AOD和中ODAACBAODBACOAABAODBAC(AAS)BCADACOD3ODBC2,ACAD4B(2,4)yB(2,4)，4k2k2xy.2xy..三、解答题（本大题共题，其中:7每题分，第、题各分，满分共分）6242584619-2386531026.866538(53)(53)(53)22(53)6535610661026.，x3x123x2x6x260x2x(x2)(x3)0，x3x.12.1(mxm0xx2m241m1m2x0xxm2241m2xm0xx2241（m2m，24．．c0220kO3．x绕OB3）y）BxABB.k3kx3得1k3y；3xBx作xDBD作AACB3知在RtAOC1，中，根据勾股定理得2322，12∴AOC30AOB30,OBOA2BODAOBAOC60OBD301在RtBOD1BDOB2OD2323(1,3)∴Byx∴B.是中A于于.AEAFADCDFAC，A是中BADCAD，∴∵AEDAFD90，ADAEDAFD∴AD，AEAFDEFA∴.tykytBxC是xB点C12B1,2【答案】（1）yx，y，；（2）2xC2190C2190C2190C21901234tABxC，CABBCABAC，BCACx.12）得12kk2，，tt1得tx212xyy，2x1yxx2x2y12或B(2,1)；2yx1yC）设Cm)m4m5，20(2)(1222AB2(2)1(1)m45，22m22m2BC①20245m4m150m219，mmAB2(2(2或；CAC②20219，ABm245m4m150m2m或；Cm00C③mBCAC4545m2mm2mC.C2190C2190C2190点C或或123C2190或.4DMCD作于．x51xx122）S，S22CDNADMCDAMMNBNCN，BNCMCNMAM2xCDNADMMDND，（2）易知CM，CNxx5CDCM2xRtACDx）CD，ABCACB90,BCAC2,AB45点D是11AD,,22NCDADNDM,CDABNDM90,CDA90NDCCDM90,ADMCDM90NDCADM在和ADMNCDACDADNDCADMCDNADMASACNCM∴AMBNCNM是直角三角形，即以CMAMMNBN的2xCNAMx）CM在RtCNMx)x2x4x4，22222CDNADMMDND在RtDNMMDNDMN2x4x42222即2MD2x4x4MDx2x22222111xx1S22222当ADM22.5CMDAADM67.5，CDMCDAADM9022.567.5CMDCDMCDCM2x，在RtCDA中，根据勾股定理得CDADAC4222ADCD2CD422CD2x2x2211将22Sxx1S2得1222.x22.x51xx122）S，S22CDNADMCDAMMNBNCN，BNCMCNMAM2xCDNADMMDND，（2）易知CM，CNxx5CDCM2xRtACDx）CD，ABCACB90,BCAC2,AB45点D是11AD,,22NCDADNDM,CDABNDM90,CDA90NDCCDM90,ADMCDM90NDCADM在和ADMNCDACDADNDCADMCDNADMASACNCM∴AMBNCNM是直角三角形，即以CMAMMNBN的2xCNAMx）CM在RtCNMx)x2x4x4，22222CDNADMMDND在RtDNMMDNDMN2x4x42222即2MD2x4x4MDx2x22222111x