八年級(jí)下數(shù)學(xué)期末幾何計(jì)算題專題復(fù)習(xí)和訓(xùn)練

D.八年級(jí)數(shù)學(xué)下冊(cè)期末專題復(fù)習(xí)和訓(xùn)練：幾何計(jì)算題、證明題AD（（；.FBCE交于E（（（.DCMAAFBE:EG例ABCD于交求FCBD.DFCFCD..EFDBCFCFCBA:CDAB4cm,CBAD7cm故DF743cmFDEAB例CCBDA.F角形又可以給我們提供CB,“SASE例:PBCDDCABABCDACBD.ADCFDECFCFDE;為平行四邊形可以證.ABCDOFCDDEF30EDB30FCDEDB,ADE60ADB603090,ADBC;ACBD.RtRt和OOAC、BDPO這.RtRtA和AC2POABCD.DBD2PO故ACBDOBC例，求P是、DEF沿B點(diǎn)沿C.POAPO和DPO.C'ADFPADB'ABCDF11E∴BAD90,AC,,ACBDBCE22O例于H交于于形1∴A2BCRt在BDABADBD23242，222..D又BD0BD=5.E11∴BD=5=5B212SSSS1≌≌HFC11∵PE,343過2244AEFD.)SS+S1111∴PE+=即553例4,，ABCD2222E、FBC、CDE、F、C、D5∴.E、F在BC、CDDBECF?E、F在BC、CDAECF.、、POD的面積以及作為底邊的、OD.FEDAABACABC、ACDBAC60，ACAB≌ACFBECFABDEFS=S=SSS≌ACFOAECACF四邊形AECFBCS=SSABC.AECBAEC.ABCDBAD120∴1,2∴12BA∵BAD120∴ABC60∴ABC和ACD1E2FDDB∴460，ACABE12HADF例.3和ACFABACACCFEABC3若O∴≌ACFB∴BECF.CDE,,所SS≌ACF.=11=S=SDEA.SSSS故.ABCAECACFAECBAE22AECF作AHBC于H2]RtACBD在AOF中，,1S＝S11ABCDABC=AB2243AF2OF2根據(jù)勾股定理可以求出2223ACBD23221123236又AB0.AB6.222ABCD其ADE例為ABCD）A求P:EP'E'E為交于DBFEE'E'BCBCC'交P''E'P''AC4130.AEDDC8CECDDE826CE'6ABCDACOAD、AC、BCE、、FAFCE.'CD2E'C28262即BCFFAD:F上EBC1CEBE又ECDF.D為2EBDHADCG=4,2,.3S四邊形F交HFCD（.BABC≌DCB15；由EF，EFACACBD則BOC90在BOC15454,4.1為，S（（.ADAEFGABCD點(diǎn)ABE、GD.FEFGBEGDECBAEFGABCD點(diǎn)ABE、GDBEGD.DC.ABDACBPABCD,B與CAD，BC，B60例E、DAC.DCA.6