第一學(xué)期期末八年級初二質(zhì)量調(diào)研數(shù)學(xué)試卷

2015學(xué)年度第一學(xué)期期末普陀區(qū)初二質(zhì)量調(diào)研數(shù)學(xué)試卷（2016．1）（時間90分鐘，滿分100分）一、填空題（本大題共有14題，每題2分，滿分28分）18xx01．化簡：2．．2xx0的根是2．方程23．函數(shù)yx2的定義域是．4．某件商品原價為100元，經(jīng)過兩次促銷降價后的價格為64元，如果連續(xù)兩次降價的百分率相同，那么這件商品降價的百分率是．2x3x1=5．在實(shí)數(shù)范圍內(nèi)分解因式：2．2fxf3=x1，那么6．如果函數(shù)．kxx10有兩個不相等的實(shí)數(shù)根，那么k的取值范圍7．已知關(guān)于x的一元二次方程2是．(21)yax的圖像經(jīng)過第二、四象限，那么8．正比例函數(shù)a的取值范圍是．kB(x,y)在反比例函數(shù)0xx，可得y的圖像上，如果當(dāng)22x1A(x,y)和點(diǎn)9．已知點(diǎn)112yy，那么______0．（填“>”、“=”、“<”）k1210．經(jīng)過定點(diǎn)A且半徑為2cm的圓的圓心的軌跡是．11．請寫出“等腰三角形的兩個底角相等”的逆命題：．12．如圖1，在△ABC中，C90，∠CAB的平分線AD交BC于點(diǎn)D，BC=8，BD=5，那么點(diǎn)D到AB的距圖1離等于．3，1），點(diǎn)13．如果點(diǎn)A的坐標(biāo)為（B的坐標(biāo)為（1，4），那么線段AB的長等于____________．1C90，將這個三角形折疊，使點(diǎn)B與點(diǎn)A重合，折痕交AB于點(diǎn)14．在Rt△ABC中，BN2AC，那么BM，交BC于點(diǎn)N，如果度．二、選擇題（本大題共有4題，每題3分，滿分12分）15．下列方程中，是一元二次方程的是……………………()x(x1)5x1；（B）（A）x43y；221（C）x35x26；（D）3x10．x216．已知等腰三角形的別是…周長等于20，那么底邊長y與腰長x的函數(shù)解析式和定義域分()（A）y202x(0x20)；（C）y202x(5x10)；（B）y202x(0x10)；20xy(5x10)．（D）217．下列問題中，兩個變量成正比例的是…………………()（A）圓的面積S與它的半徑r；（B）正方形的周長C與它的邊長a；（C）三角形（D）路程18．如圖2，在△ABC中，AB=AC，∠A=120°，如果那么AE︰BE的值等于…………………(面積一定時，它的底邊a和底邊上的高h(yuǎn)；所需要的時間t與運(yùn)動的D是BC的中點(diǎn)，不變時，勻速通過全程速度v．DE⊥AB，垂足是E，)1113（A）；（B）；（C）；（D）．3453三、（本大題共有7題，滿分60分）19．（本題滿分7分）圖211計(jì)算：(0.56)(75)．38220．（本題滿分7分）3x6x10．用配方法解方程：221．（本題滿分7分）yyyyy2xy1當(dāng)時，；1已知，并且與x成正比例，與x－2成反比例．121x3當(dāng)時，．求關(guān)于的函數(shù)解析式．y5yx22．（本題滿分8分）△ABC中，ACB453，在，AD是邊BC上的高，是GE、F分別是AB、CG的中點(diǎn)，且DEDF．已知：如圖AD上一點(diǎn)，聯(lián)結(jié)CG，點(diǎn)求證：△≌△．ABDCGD圖323．（本題滿分4，在D，且∠B=2∠D．AB+AC=CD．8分）已知：如圖△ABC中，∠ACB=90°，AD為△ABC的外角平分線，交BC的延長線于點(diǎn)求證：ABCD圖4324．（本題滿分5，在平面直角坐標(biāo)系xOy中，已知直線A，且點(diǎn)1，點(diǎn)B是x軸正半軸上一點(diǎn)，且AB⊥OA．（1）求反比例函數(shù)的解析式；（2）求點(diǎn)（3）先在AOB的內(nèi)部求作點(diǎn)不寫作法，保留作圖痕跡，在圖上標(biāo)注清楚點(diǎn)P的坐標(biāo)．（P）11分）ky(k0)的圖xy3x與反比例函數(shù)如圖像交于點(diǎn)A的橫坐標(biāo)為B的坐標(biāo)；P，使點(diǎn)P到AOB的兩邊OA、OB的距離相等，且PAPB；再寫出點(diǎn)圖525．（本題滿分6，在△ABC中，DF⊥DE交邊BC于點(diǎn)F（點(diǎn)12分）如圖D是AB的中點(diǎn)，E是邊AC上一動點(diǎn)，聯(lián)結(jié)DE，過點(diǎn)D作F與點(diǎn)B、C不重合），延長FD到點(diǎn)G，使DGDF，聯(lián)結(jié)EF、AG，已知AB10，BC6，AC8．（1）求證：ACAG；（2）設(shè)AEx，，求與x的函數(shù)CFyy解析式，并寫出定義域；（3）當(dāng)△BDF是以BF為腰的求AE的長．等腰三角形時，GAEDCBF4圖62015學(xué)年度第一學(xué)期期末普陀區(qū)初二質(zhì)量調(diào)研數(shù)學(xué)試卷參考答案一、填空題（本大題共14題，每題2分，滿分28分）1．32x；x0,x12；2．3．≥2；x4．20%；122(x317)(x317)311k4k；05．；6．；7．且4418．a(chǎn)<；9．>；10．以點(diǎn)A為圓心，2cm為半徑的圓；211．有兩個角相等的三角形是等腰三角形（寫兩個“底角”相等不給分）；12．3；13．5；14．15二、選擇題（本大題共4題，每題3分，滿分12分）17．B；15．B；16．C；18．A．三、簡答題（本大題共5題，每題7分，滿分35分）19．解：原式=(223)(253)·················································（4分）24223253·······················································（1分）==24233．····································································（2分）43x6x1．·································································（1分）20．解：移項(xiàng)，得2x2x1．················································（1分）二次項(xiàng)系數(shù)化為1，得23x2x111，配方，得234(x1)2.·······························································（2分）323x13利用開平方法，得.523，x123x1解得.···············································（2分）331123，x123x1所以，原方程的根是.···························（1分）3311ykxk0)···········································（1分）21．解：由y與x成正比例，可設(shè)(1111kx20).·································（1分）2由y與x－2成反比例，可設(shè)(ky222kykxx2.···············································（1分）1∵yyy，∴212把x1，y1和x3，y5分別代入上式，1,······································································（1分）3kk5.kk得12121,k解得k2.···········································································（2分）122yx析式是所以y關(guān)于x的函數(shù)解x2.··································（1分）22．證明：∵AD⊥BC，E是AB的中點(diǎn)，∴DE1AB（直角三角形斜邊上的中線等于斜邊的一半）.···········（2分）21DF同理：2CG.·······························································（1分）∵DEDF，∴ABCG.··················································（1分）∵AD⊥BC，ACB45，∴DAC45.··························（1分）∴ACDDAC.································································（1分）∴ADCD.·······································································（1分）在Rt△ABD和Rt△CGD中，ADCD,ABCG.6∴Rt△ABD≌Rt△CGD（H.L）．·············································（1分）23．證明：過點(diǎn)D作DE⊥AB，垂足為點(diǎn)E．················································（1分）又∵∠ACB=90°(已知)∴DE=DC（在角的平分線上的點(diǎn)到這個角的兩邊的距離相等）．········（2分）在Rt△ACD和Rt△AED中DE=DC（已證）AD=AD（公共邊）∴Rt△ACD≌Rt△AED（H.L）．∴AC=AE，∠CDA=∠EDA．·······················································（1分）B=2∠D(已知)，∴∠B=∠BDE．············································（1分）···················································（1分）∵∠∴BE=DE．··············································································（1分）AB+AE=BE，∴AB+AC=CD．········································································（1分）又∵24．解：（1）由題意，設(shè)點(diǎn)A的坐標(biāo)為（1，m），A在正比例函數(shù)y3x的圖像上，∵點(diǎn)m3.∴點(diǎn)A的坐標(biāo)為.········································（1分）(1,3)∴kA在反比例函數(shù)y的圖像上，∵點(diǎn)xk∴，解得k3.······················································（1分）313y∴反比例函數(shù)的解析式為.·············································（1分）x（2）過點(diǎn)A作AC⊥，垂足為點(diǎn)，OBCOC1，AC3.可得ACO90°．∵AC⊥，∴∠OB由勾股定理，得AO2．·······················································（1分）∴OC12AO．∴∠OAC30°.7∴∠AOC60°.∵AB⊥OA，∴∠OAB90°.∴∠ABO30°.································································（1分）∴OB2OA.∴OB4.··········································································（1分）∴點(diǎn)B的坐標(biāo)是(4,0).···························································（1分）【說明】其他方法相應(yīng)給分．（3）作圖略.···············································································（2分）點(diǎn)P的坐標(biāo)是（3,3）.·····························································（2分）25．（1）證明：∵BC6，AC8，BCAC23664100．∴2∵AB2100，∴BCACAB22．2∴△ABC是直角三角形，且∠ACB=90°（勾股定理的逆定理）．··（1分）∵D是AB的中點(diǎn)，∴ADBD．在△ADG和△BDF中，ADBD,ADGBDF,DGDF.∴△ADG≌△BDF（S.A.S）．∴GABB．·································································（1分）∵ACB90，∴CABB90（直角三角形的兩個銳角互余）．·················（1分）∴CABGAB90．∴EAG90．····························（1分）即：ACAG．8（2）聯(lián)結(jié)EG．∵AEx，AC8，∴EC8x．∵ACB90，EF(8x)2y2．····································（1分）得由勾股定理，2∵△ADG≌△BDF，∴AGBF．BC6∵CFy，，∴AGBF6y．∵EAG90，得EG2x2(6y)2．································