圆试卷2(答案)

试卷答案一.解答题(本题包括64小题,共0分。)1．24222CE81-1)(2OE-OCCE2x:x-)3(212-)1(xDE-CDOE-OCDE-CDCE:DECRtOE-OCCE:OCERt1,xOCx,EDD,,ACDAC,OEOE,:222222222222222222CEACACOEOC解得中在中在可得则连结设延长线必经过点中点可知为弧因延长解2．12)10(6)10(6)10(6:)10(6:)10(://)(68,10816.,,:22xxxxBFAExxBFxxAExxOMAEAEOMxOPcmMDODOMMDODMDCDCDMODMCDOM同理设中点为则连结于作解3．解:因直角三角形的垂心为直角顶点A,外心为斜边BC的中点,所以外心和垂心的距离即为斜边上中线长,∴两心距离为BC/2=2.54．解:1.2=12分米,0.5=5分米,1.3m=13分米∵132=122+52∴此三角形铁板是一直角三角形铁板.∴内切圆半径为:12+5-13/2=2(分米)(1)分别作出两锐角(任两角)平分线,标出交点O.(2)以O为圆心,2分米为半径作⊙O.(3)沿⊙O即可剪出面积最大的圆形铁板.5．解:如图(1)CF是⊙O的切线CF与直线AB不相交,证明如下:∵CF是⊙O的切线∴∠BCF=∠A∵△ABC是等边三角形∴∠ABC=∠A∴∠BCF=∠ABC∴CF∥AB∴CF与直线AB不相交(2)连结BO并延长交AC于H∵⊙O是边△ABC的外接圆∴∠BHC=90°∵点P是弧BC的中点∴∠BCE=30°又∵∠ACB=60°∴∠HCE=90°∵∠BEC=90°∴∠HBE=90°∴BE是⊙O的切线在△ACD中∵∠ACD=90°∠A=60°∴∠D=30°∴BD=BC∴DE=CE∴s△BDE=s△BCE在矩形BGCE中∴S△BCE=s△BCH=S/2∴S△BCE=S/2∴S△BDE=S/26．.,,45221025OAOMsinMlOMO,OA,OAl:相离为圆直线相交与圆直线相切与圆直线于作过夹角为与相切且与圆过点设直线解OADOABOAC7．4334223222OAFOtgOBAFOBDADctgA)2(3532rrx),(2,322222)(,,AFE.O(1):11112112221111BDADctgArrrrrxRtBOABABDrrrOOFOrxrxrrrxxrxOAEOFOFEAOFEFOxrxrxrEFxFCrOOAFEOAEO于又是是直径连结舍去不合题意解得相似又设于切圆连结解8．)21,23(),21,233(),1,3(),0,0()1,0(),2,3(),1,3(),0,0(DCBADCBA或9．606021cos6055,5.2:.),(5.221,)1(:22AOBBAOACOAACOACAOBAOBcmOBACOAOCcmOAcmACAOCRtOBOAcmABACCABOC或为正三角形中在、连结则于作解10．证明:连结AB,设PO1交⊙O1于M,连结AM∵PM为直径∴∠PAM=90°∵四边形ABCD内接于⊙O2∴∠ABP=∠C∵∠PBA=∠PMA∴∠C=∠PMA∵∠PMA+∠APM=90°∴∠C+∠CPE=90°∴∠PEC=90°∴PE⊥CD11．解:作OC⊥AB,交AB于C,交弧AB于D则CD=6cm,连结OA,则OA=12cm,OC=OD-CD=6cm(cm)3122)(3661222ACABcmAC弦12．)715(4)715()26(21715:,2)157(4)75)(62(21715734:1514:4,13//.,:,1:2222梯形由上面解出的值可得同测时如图在圆心和当梯形中在中在则、连结于交于交作直径在圆心两侧时和当如图解SONOMMNOCDABSMNOMOCMRtONOCNRtOBOCOBOCNDCNMBAMCDEFABDCNCDMABABEFCDAB13．)(64326424,8:34436:862,,,2,)(341632426:)(2426:,,.1,:2222222222cmCDADACcmCDcmADADCRtcmODOCCDODCRtcmADcmDAcmODODBCADABCOcmADBDACcmADACDRtcmODOBBDOBDRtBCODDBCOBOAABCO中在中在且则必过于作如图内时在当外心中在中在则于交、连结如图外时在当外心解14．)3,33(),6,0(),0,0(15．2210510)6060(//6032123tan24343,3102b-4a310,)3(,333242)(54,2,)2(2222.,90//044)0(,2,20)0(bb2ax-x.,)1(:222222222222222222222222222222222的半径为即圆弧弧弧又是正三角形即连结时即当即要使又的直径为圆则连结且即的两根的方程的长是关于又于的直径是圆解OaaADCDBCABADCDABaADCDABBCADaOBABAOBABCaaBEAEABCaaaabaaBEECBEbabkAOAEADkbbbbaBECECEBECEBEBECECEBEBECECEBEkbabbaAEADbaAFDFADbAEAFaDEODFDFDAFBCADbaECBEbabbAEaBCbECBEaECBExECBEECBEAEEBCAFOBC16．36y6234)61(440)61(444:6,2ab-x)2()120(261-y1223:9090,)1(,:222OSabacyxxxxxyxACAEADABACDABEECADCABEABEAEBEEOAEA圆最大时即当即相似为直径连结于交圆作直径过如图解17．)(8422,)(45.25.65.65.295.26)9(:)(6122121x-9ODOAx,,:222222cmBFDECDEBFBCBECFDFcmBFOAOBxxxOFDFODODFRtcmCDDFCDOFOFOD的中位线为解得即中在则设连结解18．3244)332(1)32(3323232360sin1,6060,,:BEABAEBEOBFBOBOFOBOFOCACOCCOBBEOBEOEOBABFBFFABFBFABCABOCFBOACOE为切点是等边三角形是等边三角形连结交于、延长连结解19．连接OH，设EH＝x，则HR＝2x．∵ND＝4，OA＝OD，∠AOD＝60°，∴AO＝2ON，OD＝2DN＝8，OR＝x＋4，222)()()(OHORHR＝＋，64)4()2(22xx＋＋，解得512＝x．20．x=-3,y=4,A(-3,-3),B(3,-3)21．7cm或17cm22．51015155.7233530cos,3530cos1010290906030602:CDACOAAMMADOMOAODADABDOAODADOAADOAA则于作解23．3412213264)131(445,tg60AH,,,)2(756045180454,22604,290BCBE.CD:(1)BCAHSAHAHBCCHBHtgAHCHBHABCRtHBCAHAAECBBCECDBCBCBDBECBDCABC同理中在为垂足作过点又为直径、连结解24．)(334323232)(3223460sin,,:cmADRcmABADDBCAD外接圆由重心定理得于作如图解25．529536//536990,:2222ABACBCCDACABCDBCBCDRABCRtCBDABCABCOCBCBDOCBBDOCCDBDCDOCOCDBCABACACDABOCAC相似的切线为圆为直径连结解26．提示:连结OD,OF,设BE=BD=x，则CD=CF=6-x，由AE=AF得10+x=8+(6-x),解得x=227．.33103612:33142AEEBAEEDEFBECECFCDCB，求得，再由，由勾股定理求得，求得、由提示28．1:432:)322(:323131390233)22(.22,2.:222222RRRBEAERRROEOBBERRROAODOEODOAOEODADODOEDOBDOBDEOODAACDERRROBAOABRRRADODOAADODOADRADRBCOD相似又的切线为圆则设连结解29．解:如图,OA=4cm设OB与OA切于点M,连结AM∵AM=2,OA=4∴sin∠α=AM/OA=1/2∴∠α=30°∴(1)当Oα30°时,射线OB与⊙A相交(2)当α=30°时,射线OB与⊙A相切(3)当30°α90°时,射线OB与⊙A相离30．相交与圆时当相离与圆时当相切与圆时当解得即相似公用中在则连结为交于切圆作的解析式为直线的解析式为设的坐标为的坐标为解',552,',3552',552552:132:,'''MBO'2MO'3,12BO':BMO'RtBE,MO'M,O'E,OCM,O'BM(2)33xyBC3k3-k03kxy)0,4()0,2(')1(:2OBEBOOBEbOBEbbbbEBBOOEMOMBOEBOBCAO31．(1)作OG⊥AB于G，OH⊥CD于H．∵∠EPO＝∠FPO，∴OG＝OH，∴AB＝CD．(2)成立(略)．32．2160coscos603228322289222:ABCCPABCPAQCCCPCABQAABQAQCCPABQ解33．解:(1)由题意知,轮船航向为避免触礁应至少沿⊙A的切线BD方向航行.∵AB=45°,AD=15,∠ADB=90°∴sinα=AD/AB=1/3(2)由(1)可知:AE⊥CE,在Rt△CEA中,AE=15,CA=45-15=30(海里),∴sin∠ACE=AE/AC=15/30=1/2∴∠ACE=30°∴轮船航向改变的角度至少应为东偏南30°才能避免触礁.34．)(4353,5:,321OA,,O:2222cmACOAOCcmACcmOAAOCRtcmABACCABOC中在则连结于作过解35．解:取AB的中点O即可如图36．)32,2(),0,4(),0,0(BAO37．4416,:41620)26(8)6410(2)26(42)26(2212.,..,:22222