武汉市中考五月调考数学试题及答案.docx
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武汉市中考五月调考数学试题及答案.docx
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武汉市中考五月调考数学试题及答案
2011~2012学年武汉市九年级五月供题数学试卷2012.5一、选择题(共12小题,每小题3分,共36分)1.在0,3,-1,-3这四个数中,最小的数是A.0.B.3.C.-1.D.-3.x32.式子在实数范围内有意义,则x的取值范围是A.x>3.B.x≥3.C.x<3.D.x≤3.x1>03.不等式组的解集在数轴上表示为0x-1≤01-100-1101-11-1A.B.C.D.4.下列事件是必然事件的是A.某运动员射击一次击中靶心.B.抛一枚硬币,正面朝上.C.3个人分成两组,一定有2个人分在一组.D.明天一定是晴天.25.若x,x是一元二次方程x-5x-6=0的两个根,则x·x的值是1212A.-5.B.5.C.-6.D.6.6.2012年武汉市约有71000个初中毕业生,其中71000这个数用科学计数法表示为3545A.71×10.B.7.1×10.C.7.1×10.D.0.71×10.7.如图,AD是△ABC的中线,∠ADC=60°,把△ADC沿直线AD翻折,点C落在点C的位置,如果DC=2,1A那么BC=13CA..1B.2.23C..D.4.乙图甲图CDB8.如图,甲、乙两图是分别由五个棱长为“1”的立方块组成的两个几何体,它们的三视图中完全一致的是A.主视图.B.左视图.C.俯视图.D.三视图都一致.9.课题研究小组对附着在物体表面的三个微生物(课题小组成员把他们分别标号为1,2,3)的生长情况进行观察记录.这三个微生物第一天各自一分为二,产生新的微生物(分别被标号为4,5,6,7,8,9),接下去每天12都按照这样的规律变化,即每个微生物一分为二,形成新的微生物(课题组成员用1113如图所示的图形进行形象的记录).那么标号为100的微生物会出现在105414211A.第3天.B.第4天.69C.第5天.D.第6天.15322010.B为线段OA的中点,P为以O为圆心,OB为半径的圆上的动点,当PA的中P8197点Q落在⊙O上时,如图,则cos∠OQB的值等于1618Q1711A..B..23OAB12C..D..43第1页共11页
11.今年的“六·一”儿童节是个星期五,某校学生会在初一年级进行了学生对学校作息安排的三种期望(全天休息、半天休息、全天上课)的抽样调查,并把调查结果绘成了如图1、2的统计图,已知此次被调查的男、女学生人数相同.根据图中信息,下列判断:
①在被调查的学生中,期望全天休息的人数占53%;②本次调查了200名学生;③在被调查的学生中,有30%的女生期望休息半天;④若该校现有初一学生900人,根据调查结果估计期望至少休息半天的学生超过了720人.其中正确的判断有A.4个.B.3个.C.2个.D.1个.图1图212.如图,在△ABC中,∠B、∠C的角平分线交于点F,分别过B、C作BF、CF的垂线,交CF、BF的延长线于点D、E,且BD、EC交于点G.则下列结论:
①∠D+∠E=∠A;②∠BFC-∠G=∠A;③∠BCA+∠AE=2∠ABD;④AB·BC=BD·BG.正确的有AA.①②④.B.①③④.FCC.①②③.DD.①②③④.B二、填空题共4小题,每小题3分,共12分)G13.计算:
tan30°=.14.小潘射击5次成绩分别为(单位:
环)5,9,8,8,10.这组数据的众数是,中位数是,平均数是.kyy15.如图,过A(2,-1)分别作y轴,x轴的平行线交双曲线于点B,点C,过点C作CE⊥x轴于点E,x过点B作BD⊥y轴于点D,连接ED.若五边形ABDEC的面积为34,则实数k=.DBy/kmlE1xOl2ACP4.8O2.81.6x/h第15题图第16题图第2页共11页
l、lP16.小敏从A地出发向B地行走,同时小聪从B地出发向A地行走,如图所示,相交于点的两条线段分12y(km)xh)(别表示小敏、小聪离B地的距离与已用时间之间的关系,则x=h时,小敏、小聪两人相距7km.三、解答题(共9小题,共72分)17.(本小题满分6分)解方程:
1x2.2x2x218.(本小题满分6分)直线y=kx+4经过点A(1,6),求关于x的不等式kx+4≤0的解集.19.(本小题满分6分)如图,在△ABC中,AB=CB,∠ABC=90°,F为AB延长线上一点,点E在线段BC上,且AE=CF.求证:
∠AEB=∠CFB.C52352366aaaaa032333ACDBEFBA20.(本小题满分7分)有4张形状、大小和质地都相同的卡片,正面分别写有字母A,B,C,D和一个算式,背面完全一致.将这4张卡片背面向上洗匀,从中随机抽取1张,不放回,接着再随机抽取1张.
(1)请用画树形图或列表法表示出所有的可能结果;(卡片可用A,B,C,D表示)
(2)将“第一张卡片上的算式是正确,同时第二张卡片上的算式是错误”记为事件A,求事件A的概率.第3页共11页
21.(本小题满分7分)如图,网格中每个小正方形的边长都是1个单位.折线段ABC的位置如图所示.ABC
(1)现把折线段ABC向右平移4个单位,画出相应的图形;ABCABCAA
(2)把折线段绕线段的中点D顺时针旋转90°,画出相应的图形;CCCAAA(3)在上述两次变换中,点的路径的长度比点的路径的长度大个单位.CAB22.(本小题满分8分)如图,AB为⊙O的直径,AM和BN是它的两条切线,E为⊙O的半圆弧上一动点(不与A、B重合),过点E的直线分别交射线AM、BN于D、C两点,且CB=CE.
(1)求证:
CD为⊙O的切线;2AH
(2)若tan∠BAC=,求的值.CH2ADMEHOCNB23.(本小题满分10分)某跳水运动员进行10米跳台跳水训练时,身体(看成一点)在空中的运动路线是如图所示坐标系下经过原点O的一条抛物线(图中标出的数据为已知条件).在跳某个规定动作时,正常情况下,该运动员在空中的最高处距210水面米,入水处距池边的距离为4米,运动员在距水面高度为5米以前,必须完成规定的翻腾动作,并调整3好入水姿势,否则就会出现失误.
(1)求这条抛物线的解析式;
(2)在某次试跳中,测得运动员在空中的运动路线是
(1)中的抛物线,且运动员在空中完成规定的翻腾动作并33调整好入水姿势时,距池边的水平距离为米,问此次跳水会不会失误?
并通过计算说明理由.5第4页共11页
O(本小题满分10分)24.如图,已知正方形ABCD,点P为射线BA上的一点(不和点A,B重合),过P作PE⊥CP,且CP=PE.过E作EF∥CD交射线BD于F.
(1)若CB=6,PB=2,则EF=;DF=;
(2)请探究BF,DG和CD这三条线段之间的数量关系,写出你的结论并证明;(3)如图2,点P在线段BA的延长线上,当tan∠BPC=时,四边形EFCD与四边形PEFC的面积12E之比为.35EPAFDGADFPBCBC图1图2第5页共11页
25.(本小题满分12分)2yx2x3如图1,已知抛物线与x轴交于点A和点B,与y轴相交于点C.
(1)求A、B、C三点的坐标;
(2)点D为射线CB上的一动点(点D、B不重合),过点B作x轴的垂线BE与以点D为顶点的抛物线y=2(x-t)+h相交于点E,从△ADE和△ADB中任选一个三角形,求出当其面积等于△ABE的面积时的t的值;(友情提示:
1、只选取一个三角形求解即可;2、若对两个三角形都作了解答,只按第一个解答给分.)yx(3)如图2,若点P是直线上的一个动点,点Q是抛物线上的一个动点,若以点O,C,P和Q为顶点的四边形为直角梯形,求相应的点P的坐标.yyy=xEBxOAxABODCC图1图2第6页共11页
2011-2012学年度武汉市部分学校九年级五月供题数学参考答案一、选择题(本大题共12小题,每小题3分,共36分)题号123456789101112DBBCCCBACCAD答案二、填空题(本大题共4小题,每小题3分,共12分)313.14.8;8;815.816.0.6或2.63三、解答下列各题(共9小题,共72分)17.(本小题满分6分)解:
方程两边同乘以2(x-2),去分母得,„„„„„„„„„„„„„„„„1分1+4(x-2)=2x.„„„„„„„„„„„„„„„„„„„„2分去括号得,1+4x-8=2x.„„„„„„„„„„„„„„„„„„„„3分7∴x=.„„„„„„„„„„„„„„„„„„„„„„„4分27经检验,x=是原方程的解.„„„„„„„„„„„„„„„„„5分27∴原方程的解是x=.„„„„„„„„„„„„„„„„„„„6分218.(本小题满分6分)解:
把(1,6)代入直线的函数关系式y=kx+4中,得,6=k+4,„„„„„„„„„„„„„„„„„„„„2分解得:
k=2.„„„„„„„„„„„„„„„„„„„„3分y2x4∴直线的函数关系式为.2x40≤∴.„„„„„„„„„„„„„„„„„„„„5分x≤-2∴.„„„„„„„„„„„„„„„„„„„„6分19.(本小题满分6分)证明:
在Rt△ABE和Rt△CBF中,ABCB∵„„„„„„„„„„„„„„„„„„„„3分AECF∴Rt△ABE≌Rt△CBF.„„„„„„„„„„„„„„„„„„„„4分∴∠AEB=∠CFB.„„„„„„„„„„„„„„„„„„„„6分20.(本小题满分7分)解:
(1)根据题意,可以列出如下的表格:
ABCDAABACADBCBDBBACBCDCCA第7页共11页
DDADBDC„„„„„„„„„„„„„„„„„3分由表可知,随机抽取1张,不放回,接着再随机抽取1张的所有可能的结果有12种.„4分它们出现的可能性相等;„„„„„„„„„„„„„„„„„5分
(2)由表可知,事件A的结果有3种,„„„„„„„„„„„„„„„„„6分1∴P(A)=.„„„„„„„„„„„„„„„„„7分421.(本小题满分7分)
(1)、
(2)问画图如图:
BB′DAA′CC′A″C″B″„„„„„„„„„„„„„„„„„5分(3)(5-1)π.„„„„„„„„„„„„„„„„„7分22.(本小题满分8分)
(1)证明:
连接OE.„„„„„„„„„„„„„„„„„1分∵OB=OE,∴∠OBE=∠OEB.∵BC=EC,∴∠CBE=∠CEB.„„„„„„„„„„„„„„„„„2分∴∠OBC=∠OEC.∵BC为⊙O的切线,∴∠OEC=∠OBC=90°,„„„„„„„„„„„„„„„„„3分∵OE为半径,∴CD为⊙O的切线.„„„„„„„„„„„„„„„„„4分
(2)延长BE交AM于点G,连接AE,过点D作DT⊥BC于点T.因为DA、DC、CB为⊙O的切线,∴DA=DE,CB=CE.2在Rt△ABC中,因为tan∠BAC=,令AB=2x,则BC=2x.2∴CE=BC=2x.„„„„„„„„„„„„„„„„„5分令AD=DE=a,则在Rt△DTC中,CT=CB-AD=2x-a,DC=CE+DE=2x+a,DT=AB=2x,222∵DT=DC-CT,第8页共11页
222∴(2x)=(2x+a)-(2x-a).„„„„„„„„„„„„„„„„„6分解之得,x=2a.„„„„„„„„„„„„„„„„„7分∵AB为直径,∴∠AEG=90°.∵AD=ED,∴AD=ED=DG=a.∴AG=2a.„„„„„„„„„„„„„„„„„8分因为AD、BC为⊙O的切线,AB为直径,∴AG∥BC.所以△AHG∽△CHB.AHAG2a∴==.„„„„„„„„„„„„„„„„„9分CHCB2xAH∴=1.„„„„„„„„„„„„„„„„„10分CH23.(本小题满分10分)
(1)解:
如图所示,在给定的平面直角坐标系中,设最高点为A,入水点为B.210米,跳台支柱10米,∵A点距水面32∴A点的纵坐标为,由题意可得O(0,0),B(2,-10).„„„1分32yaxbxca0,a,b,c设该抛物线的关系式为,(为常数)2过点O(0,0),B(2,-10),且函数的最大值为,„„„„„„2分3=0,c4a+2b+c=﹣10,则有:
„„„„„„„„„„„„„„„„„„5分24ac-b2=.4a325a610b解得:
„„„„„„„„„„„„„„„„„„„„„6分3c025102xxy.„„„„„„„„„„7分∴所求抛物线的关系式为63
(2)解:
试跳会出现失误.168332时,y=.„„„„„„„„„„„„„„„8分∵当x=3551614此时,运动员距水面的高为10=<5,„„„„„„„„„„9分33∴试跳会出现失误.„„„„„„„„„„„„„„„„„„„„„10分(本小题满分10分)24.42==
(1)EF6;DF.„„„„„„„„„„„„„„„„„„„2分第9页共11页
2
(2)BF+2DG=CD.理由如下:
如图⑴,连接AE,AC.∵△EPC为等腰Rt△;四边形ABCD为正方形,CECA2∴.CPCB∠ECP=∠ACB=45°,∴∠ECA=∠PCB.∴△EAC∽△PCB.„„„„„„„„„„„„„„„„„„„„4分∴∠EAC=∠PBC=90°.∵∠BAC=∠ABD=45°,∴∠EAB+∠ABF=180°.∴EA∥BF.又AB∥EF,∴四边形EABF为平行四边形.„„„„„„„„„„„„„„„„5分∴EF=AB=CD.又∵AB∥CD,∴EF∥CD.∴△EFG∽△CDG.EFGF1∴.„„„„„„„„„„„„„„„„„„„„„6分CDDG∴DF=2GF=2DG.„„„„„„„„„„„„„„„„„„„„7分2∴BF+2DG=BD=CD.„„„„„„„„„„„„„„„„„8分23(3)tan∠BPC=或.„„„„„„„„„„„„„„„„„„„10分57EADGPFCB25.(本小题满分12分)2解:
(1)当y=0时,x-2x-3=0,解之得x=﹣1,x=3,12所以A、B两点的坐标分别为(﹣1,0),(3,0).„„„„„„„„„„„„„„„„„2分当x=0时,y=﹣3,∴C点的坐标为(0,﹣3).„„„„„„„„„„„„„„„„„3分2
(2)由题意可知,抛物线y=(x-t)+h沿射线CB作平移变换,其顶点D(t,h)在射线CB上运动,易知直线CB的函数关系式为y=x-3,∴h=t-3.„„„„„„„„„4分①选取△ADE.△ADE与△ABE共边AE,当它们的面积相等时,点D和点B到AE的距离相等,此时直线AE∥BC,∴直线AE的函数关系式为y=x+1,∴点E的坐标为(3,4).„„„„„„5分22因为点E在抛物线上,∴4=(3-t)+h,∴4=(3-t)+(t-3),„„„„„„6分第10页共11页
5+175-17解之得,t=,t=.„„„„„„„„„„„„„7分1222②选取△ADB.△ADB与△ABE共边AB,当它们的面积相等时,点D和点E到x轴的距离相等,2∵点D到x轴的距离为|t-3|,点E到x轴的距离为|(3-t)+(t-3)|,2∴|t-3|=|(3-t)+(t-3)|.„„„„„„„„„5分22t-3=(3-t)+(t-3),或3-t=(3-t)+(t-3),„„„„„„„„„6分解之得t=3或t=1,其中t=3时,点D、B重合,舍去,∴t=1.„„„„7分3333(3)(-3,-3),(-1,-1),(2,2),(,),(-,-).2222„„„„„„„„本小问5分,写对一个坐标给一分.第11页共11页
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