第一学期期末八年级初二质量调研数学试卷

2015学年度第一学期期末普陀区初二质量调研数学试卷（2016．1）（时间90分钟，满分100分）一、填空题（本大题共有14题，每题2分，满分28分）18xx01．化简：2．．2xx0的根是2．方程23．函数yx2的定义域是．4．某件商品原价为100元，经过两次促销降价后的价格为64元，如果连续两次降价的百分率相同，那么这件商品降价的百分率是．2x3x1=5．在实数范围内分解因式：2．2fxf3=x1，那么6．如果函数．kxx10有两个不相等的实数根，那么k的取值范围7．已知关于x的一元二次方程2是．(21)yax的图像经过第二、四象限，那么8．正比例函数a的取值范围是．kB(x,y)在反比例函数0xx，可得y的图像上，如果当22x1A(x,y)和点9．已知点112yy，那么______0．（填“>”、“=”、“<”）k1210．经过定点A且半径为2cm的圆的圆心的轨迹是．11．请写出“等腰三角形的两个底角相等”的逆命题：．12．如图1，在△ABC中，C90，∠CAB的平分线AD交BC于点D，BC=8，BD=5，那么点D到AB的距图1离等于．3，1），点13．如果点A的坐标为（B的坐标为（1，4），那么线段AB的长等于____________．1C90，将这个三角形折叠，使点B与点A重合，折痕交AB于点14．在Rt△ABC中，BN2AC，那么BM，交BC于点N，如果度．二、选择题（本大题共有4题，每题3分，满分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．已知等腰三角形的别是…周长等于20，那么底边长y与腰长x的函数解析式和定义域分()（A）y202x(0x20)；（C）y202x(5x10)；（B）y202x(0x10)；20xy(5x10)．（D）217．下列问题中，两个变量成正比例的是…………………()（A）圆的面积S与它的半径r；（B）正方形的周长C与它的边长a；（C）三角形（D）路程18．如图2，在△ABC中，AB=AC，∠A=120°，如果那么AE︰BE的值等于…………………(面积一定时，它的底边a和底边上的高h；所需要的时间t与运动的D是BC的中点，不变时，匀速通过全程速度v．DE⊥AB，垂足是E，)1113（A）；（B）；（C）；（D）．3453三、（本大题共有7题，满分60分）19．（本题满分7分）图211计算：(0.56)(75)．38220．（本题满分7分）3x6x10．用配方法解方程：221．（本题满分7分）yyyyy2xy1当时，；1已知，并且与x成正比例，与x－2成反比例．121x3当时，．求关于的函数解析式．y5yx22．（本题满分8分）△ABC中，ACB453，在，AD是边BC上的高，是GE、F分别是AB、CG的中点，且DEDF．已知：如图AD上一点，联结CG，点求证：△≌△．ABDCGD图323．（本题满分4，在D，且∠B=2∠D．AB+AC=CD．8分）已知：如图△ABC中，∠ACB=90°，AD为△ABC的外角平分线，交BC的延长线于点求证：ABCD图4324．（本题满分5，在平面直角坐标系xOy中，已知直线A，且点1，点B是x轴正半轴上一点，且AB⊥OA．（1）求反比例函数的解析式；（2）求点（3）先在AOB的内部求作点不写作法，保留作图痕迹，在图上标注清楚点P的坐标．（P）11分）ky(k0)的图xy3x与反比例函数如图像交于点A的横坐标为B的坐标；P，使点P到AOB的两边OA、OB的距离相等，且PAPB；再写出点图525．（本题满分6，在△ABC中，DF⊥DE交边BC于点F（点12分）如图D是AB的中点，E是边AC上一动点，联结DE，过点D作F与点B、C不重合），延长FD到点G，使DGDF，联结EF、AG，已知AB10，BC6，AC8．（1）求证：ACAG；（2）设AEx，，求与x的函数CFyy解析式，并写出定义域；（3）当△BDF是以BF为腰的求AE的长．等腰三角形时，GAEDCBF4图62015学年度第一学期期末普陀区初二质量调研数学试卷参考答案一、填空题（本大题共14题，每题2分，满分28分）1．32x；x0,x12；2．3．≥2；x4．20%；122(x317)(x317)311k4k；05．；6．；7．且4418．a<；9．>；10．以点A为圆心，2cm为半径的圆；211．有两个角相等的三角形是等腰三角形（写两个“底角”相等不给分）；12．3；13．5；14．15二、选择题（本大题共4题，每题3分，满分12分）17．B；15．B；16．C；18．A．三、简答题（本大题共5题，每题7分，满分35分）19．解：原式=(223)(253)·················································（4分）24223253·······················································（1分）==24233．····································································（2分）43x6x1．·································································（1分）20．解：移项，得2x2x1．················································（1分）二次项系数化为1，得23x2x111，配方，得234(x1)2.·······························································（2分）323x13利用开平方法，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与x成正比例，可设(1111kx20).·································（1分）2由y与x－2成反比例，可设(ky222kykxx2.···············································（1分）1∵yyy，∴212把x1，y1和x3，y5分别代入上式，1,······································································（1分）3kk5.kk得12121,k解得k2.···········································································（2分）122yx析式是所以y关于x的函数解x2.··································（1分）22．证明：∵AD⊥BC，E是AB的中点，∴DE1AB（直角三角形斜边上的中线等于斜边的一半）.···········（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．证明：过点D作DE⊥AB，垂足为点E．················································（1分）又∵∠ACB=90°(已知)∴DE=DC（在角的平分线上的点到这个角的两边的距离相等）．········（2分）在Rt△ACD和Rt△AED中DE=DC（已证）AD=AD（公共边）∴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）由题意，设点A的坐标为（1，m），A在正比例函数y3x的图像上，∵点m3.∴点A的坐标为.········································（1分）(1,3)∴kA在反比例函数y的图像上，∵点xk∴，解得k3.······················································（1分）313y∴反比例函数的解析式为.·············································（1分）x（2）过点A作AC⊥，垂足为点，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）∴点B的坐标是(4,0).···························································（1分）【说明】其他方法相应给分．（3）作图略.···············································································（2分）点P的坐标是（3,3）.·····························································（2分）25．（1）证明：∵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的中点，∴ADBD．在△ADG和△BDF中，ADBD,ADGBDF,DGDF.∴△ADG≌△BDF（S.A.S）．∴GABB．·································································（1分）∵ACB90，∴CABB90（直角三角形的两个锐角互余）．·················（1分）∴CABGAB90．∴EAG90．····························（1分）即：ACAG．8（2）联结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．································