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2013年初三中考复习题 (14)

2013年初三中考复习题 (14)
2013年初三中考复习题 (14)

2013年中考数学模拟卷(14)

姓名分数

一.选择题(本大题共6题,每题4分,满分24分)

一、选择题:(本大题共6题,每题4分,满分24分)

1.如果延长线段AB到C,使得,那么AC∶AB等于

(A)2∶1;(B)2∶3;(C)3∶1;(D)3∶2.

2.已知在Rt△ABC中,∠C=90°,∠A=,AB= 2,那么BC的长等于

(A);(B);(C);(D).

3.如果将抛物线向左平移2个单位,那么所得抛物线的表达式为

(A);(B);

(C);(D).

4.如果抛物线经过点(-1,0)和(3,0),那么它的对称轴是直线

(A)x= 0;(B)x= 1;(C)x= 2;(D)x=3.5.如果乙船在甲船的北偏东40°方向上,丙船在甲船的南偏西40°方向上,那么丙船在乙船的方向是

(A)北偏东40°;(B)北偏西40°;(C)南偏东40°;

(D)南偏西40°.

6.如图,已知在△ABC中,边BC = 6,高AD = 3,正方形

EFGH的顶点F、G在边BC上,顶点E、H分别在边AB

和AC上,那么这个正方形的边长等于

(A)3;(B)2.5;

(C)2;(D)1.5.

二、填空题:(本大题共12题,每题4分,满分48分)

7.已知线段b是线段a、c的比例中项,且a= 1,b= 2,那么c= ▲ .

8.计算:= ▲ .

9.如果抛物线的开口方向向下,那么a的取值

范围是▲ .

10.二次函数图像的最低点坐标是▲ .

11.在边长为6的正方形中间挖去一个边长为x()

的小正方形,如果设剩余部分的面积为y,那么y关于x

的函数解析式为▲ .

12.已知为锐角,,那么= ▲ 度.

13.已知从地面进入地下车库的斜坡的坡度为1︰2.4,地下车库的地坪与地面的垂直距离等于5米,那么此斜坡的长度等于▲ 米.

14.小明用自制的直角三角形纸板DEF测量树AB

的高度.测量时,使直角边DF保持水平状态,

其延长线交AB于点G;使斜边DE与点A在

同一条直线上.测得边DF离地面的高度等于

1.4m,点D到AB的距离等于6m(如图所示).

已知DF = 30cm,EF = 20cm,那么树AB的高

度等于▲ m.

15.如图,将△ABC沿射线BC方向平移得到△DEF,边DE与AC相交于点G,如果BC=3cm,△ABC的面积等于9cm2,△GEC的面积等于4cm2,那么BE = ▲cm.

16.相邻两边长的比值是黄金分割数的矩形,叫做黄金矩形,

从外形上看,它最具美感.现在想要制作一张“黄金矩形”的贺年卡,如果较长的一条边长等于20厘米,那么相邻一条边长等于▲厘米.

17.九年级数学课本上,用“描点法”画二次函数的图像时,列出了如下的表格:

x …0 1 2 3 4 …

… 3 0 -1 0 3 …

那么该二次函数在=5时,y= ▲ .

18.已知在Rt△ABC中,∠A=90°,,BC = a,点D在边BC上,将这个三角形沿直线AD折叠,点C恰好落在边AB上,那么BD = ▲ .(用a的代数式表示)

三、解答题:(本大题共7题,满分78分)

19.(本题满分10分,其中第(1)小题6分,第(2)小题4分)

已知:抛物线经过B(3,0)、C(0,3)两点,顶点为A.

求:(1)抛物线的表达式;

(2)顶点A的坐标.

20.(本题满分10分,其中第(1)小题6分,第(2)小题4分)

如图,已知在平行四边形ABCD中,M、N分别是边AD、DC的中点,设,.

(1)求向量、(用向量、表示);

(2)求作向量在、方向上的分向量.

(不要求写作法,但要指出所作图中表示结论的向量)

21.(本题满分10分)

某条道路上通行车辆限速为60千米/时,在离道路50米的点P处建一个监测点,道路的AB段为监测区(如图).在△AB P中,已知∠PAB = 32o,∠PBA = 45o,那么车辆通过AB段的时间在多少秒以内时,可认定为超速(精确到0.1秒)?

(参考数据:,,,)

22.(本题满分10分)

如图,在平行四边形ABCD中,点E在边BC上,联结AE并延长,交对角线BD于点F、DC的延长线于点G,如果.

求的值.

23.(本题满分12分,每小题各6分)

已知:如图,在梯形ABCD中,AD//BC,AB⊥BC,点M在边BC上,且∠MDB=∠ADB,

(1)求证:BM=CM;

(2)作BE⊥DM,垂足为点E,并交CD于点F.求证:

24.(本题满分12分,其中第(1)小题3分,第(2)小题4分,第(3)小题5分)

如图,在直角坐标系xOy中,二次函数的图像与x轴、y轴的公共点分别为A(5,0)、B,点C在这个二次函数的图像上,且横坐标为3.

(1)求这个二次函数的解析式;

(2)求∠BAC的正切值;

(3)如果点D在这个二次函数的图像上,且

∠DAC = 45°,求点D的坐标.

25.(本题满分14分,其中第(1)小题4分,第(2)、(3)小题各5分)

如图,已知在△ABC中,∠A =90°,,经过这个三角形重心的直线DE // BC,分别交边AB、AC于点D和点E,P是线段DE上的一个动点,过点P分别作PM⊥BC,PF⊥AB,PG⊥AC,垂足分别为点M、F、G.设BM = x,四边形AFPG的面积为y.(1)求PM的长;

(2)求y关于x的函数解析式,并写出它的定义域;

(3)联结MF、MG,当△PMF与△PMG相似时,求BM

的长.

答案

一、选择题:(本大题共6题,每题4分,满分24分)

1.D;2.A;3.C;4.B;5.D;6.C.

二、填空题:(本大题共12题,每题4分,满分48分)

7.4;8.;9.;10.(0,-3);11.;12.60;13.13;14.5.4;15.1;16.(或12.36);17.8;18..

三、解答题:(本大题共7题,满分78分)

19.解:(1)∵抛物线经过B(3,0)、C(0,3)两点,

∴…………………………………………………(2分)

解得…………………………………………………………(2分)

∴抛物线的解析式是.……………………………(2分)

(2)由,…………………………………(2分)得顶点A的坐标为(1,4).…………………………………………(2分)

20.解:(1)∵M是边AD的中点,∴.……………………(2分)∵四边形ABCD是平行四边形,∴DC // AB,DC = AB.

∴.……………………………………………………(1分)

又∵N是边DC的中点,∴.…………………………(1分)

∴.……………………………………(2分)(2)作图正确,3分;结论正确,1分.

21.解:过点P作PC⊥AB,垂足为点C.…………………………………………(1分)根据题意,可知PC = 50米.

在Rt△PBC中,∠PCB = 90o,∠B = 45o,

∴.……………………………………(2分)

在Rt△PAC中,∠PCA = 90o,∠PAB = 32o,

∴.………………………………(3分)

∴AB = AC +BC≈ 80 +50 = 130(米).…………………………………(1分)

∵(秒),…………………………………………(2分)

∴车辆通过AB段的时间在7.8秒以内时,可认定为超速.…………(1分)22.解:∵四边形ABCD是平行四边形,

∴BC // AD,AB // CD,BC = AD.………………………………………(2分)

∴,.………………………………………………(2分)

又∵,∴.……………………………………………(2分)

即得,.∴.…………………………(2分)

∴.

即得.……………………………………………………………(2分)23.证明:(1)∵AB⊥BC,∴∠ABC = 90o.

∵AD // BC,∴∠CBD =∠ADB,∠BAD +∠ABC = 180o.

即得∠BAD = 90o.

∵,∴.……………………………(1分)

又∵∠CBD =∠ADB,

∴△BCD∽△DBA.………………………………………………(1分)

∴∠BDC =∠BAD = 90o.…………………………………………(1分)

∴∠DBC +∠C = 90o.

∵∠MDB=∠ADB,∠MBD =∠ADB,

∴∠MBD =∠MDB.∴BM = MD.……………………………(1分)

又∵∠BDM +∠CDM =∠BDC = 90o,

∴∠C =∠CDM.…………………………………………………(1分)

∴CM = MD.∴BM = CM.……………………………………(1分)(2)∵BE⊥DM,

∴∠DEF =∠BDC = 90o.

∴∠FDE +∠DFE = 90o,∠DBF +∠DFE = 90o.

∴∠FDE =∠DBF.………………………………………………(1分)

又∵∠FDE =∠C,

∴∠DBF =∠C.…………………………………………………(1分)

于是,由∠FDB =∠BDC = 90o,∠DBF =∠C,

得△FDB∽△BDC.………………………………………………(1分)

∴.即.……………………………(1分)

∵BM = CM,∠BDC = 90o,∴BC = 2DM.…………………(1分)

又∵,

∴.…………………………………………(1分)24.解:(1)∵二次函数的图像经过点A(5,0),

∴.……………………………………………(1分)

解得.…………………………………………………………(1分)

∴二次函数的解析式是.………………………(1分)(2)当x = 0时,得y = 5.∴ B(0,5).……………………………(1分)

当x = 3时,得,∴C(3,6).……(1分)

联结BC.

∵,

∴.

∴.……………………………………………………(1分)

∴.……………………………………(1分)(3)设D(m,n).

过点D作DE⊥x轴,垂足为点E.则,DE = n.

∵A(5,0),B(0,5),∴OA = OB.

又∵,∴,……………………………(1分)

即得∠DAE +∠BAD = 45o.

又∵∠DAC = 45o,即∠BAD +∠BAC = 45o,

∴∠DAE =∠BAC.

又∵∠DEA =∠ACB = 90o,

∴△DAE∽△BAC.…………………………………………………(1分)

∴.……………………………………………………(1分)

∴.即得.

∵点D在二次函数的图像上,

∴.

解得,m2 = 5(不合题意,舍去).………………………(1分)

∴.

∴.……………………………………………………(1分)25.解:(1)过点A作AH⊥BC,垂足为点H,交DE于点Q.

∵∠BAC =90°,,∴BC = 6.…………………(1分)

又∵AH⊥BC,

∴,Q是△ABC的重心.

∴.…………………………………………………(2分)

∵DE // BC,PM⊥BC,AH⊥BC,

∴PM = QH = 1.……………………………………………………(1分)(2)延长FP,交BC于点N.

∵∠BAC =90°,AB = AC,∴∠B = 45°.

于是,由FN⊥AB,得∠PNM = 45°.

又由PM⊥BC,得MN = PM = 1,.

∴BN = BM +MN = x +1,.…………………(1分)

∴,

.…………………(1分)∵PF⊥AB,PG⊥AC,∠BAC =90°,

∴∠BAC =∠PFA =∠PGA = 90°.

∴四边形AFPG是矩形.

∴.……………………………(1分)

∴所求函数解析式为.…………………………(1分)

定义域为.……………………………………………………(1分)

(3)∵四边形AFPG是矩形,∴.…………(1分)由∠FPM =∠GPM = 135°,可知,当△PMF与△PMG相似时,有两种

情况:∠PFM =∠PGM或∠PFM =∠PMG.

(ⅰ)如果∠PFM =∠PGM,那么.

即得PF = PG.

∴.………………………………………(1分)

解得x = 3.

即得BM = 3.………………………………………………………(1分)

(ⅱ)如果∠PFM =∠PMG,那么.

即得.

∴.………………………………………(1分)

解得,.

即得或.………………………………(1分)

∴当△PMF与△PMG相似时,BM的长等于或3或.

1.下列二次根式中与是同类二次根式的是

A.;B.;C.;D..

2.将抛物线向下平移2个单位后,所得抛物线解析式为

A.;B.;C.;D..3.如果关于的一元二次方程有两个不相等的实数根,那么的取值

范围是

A.>;B.<;C.>且;D.<且.4.下列一组数据:、、、、的平均数和方差分别是

A.和;B.和;C.和;D.和.

5.下列正方形的性质中,菱形(非正方形)不具有的性质是

A.四边相等;B.对角线相等;

C.对角线平分一组对角;D.对角线互相平分且垂直.

6.在中,,,那么半径长为的⊙和直线的位置

关系是

A.相离;B.相切;C.相交;D.无法确定.

二.填空题(本大题共12题,每题4分,满分48分)

7.化简:▲ .

8.计算:_______▲_________.

9.方程的解是▲ .

10.已知函数,那么▲ .

11.如图1,点在反比例函数的图像上,那么该反比例函数的解析式是▲.12.如图2,在中,中线和相交于点,如果,=,那么向量▲.

13.如图3,∥,平分,如果,那么▲.

14.在形状、大小、颜色都一样的卡片上,分别画有线段、直角三角形、等腰三角形、等边三角形、平行四边形、菱形、等腰梯形、正五边形、正六边形、圆等10个图形,小杰随机抽取一张卡片,抽得图形既是轴对称图形,又是中心对称图形的概率是___▲_____.15.为了解某校初三年级学生一次数学测试成绩的情况,从近450名九年级学生中,随机抽取50名学生这次数学测试的成绩,通过数据整理,绘制如下统计表(给出部分数据,除[90,100]组外每组数据含最低值,不含最高值):

分数段[ 0, 60] [60, 70] [70, 80] [80, 90] [90,100]

频数 5 20

频率0.12 0.1 根据上表的信息,估计该校初三年级本次数学测试的优良率(80分及80分以上)约为▲ (填百分数).

16.如图4,⊙半径为,的顶点在⊙上,,,垂足是,

,那么的长为▲.

17.一个二元一次方程和一个二元二次方程组成的二元二次方程组的解是或

,试写出一个符合要求的方程组__________▲_____________(只需写一个).

18.在

中,,,将绕点旋转后,点落在射线上,点落到点处,那么的值等于▲.

三.(本大题共7题,第19—22题每题10分;第23、24题每题12分;第25题14分;

满分78分)

19.(本题满分10分)

计算:.

20.(本题满分10分)

解不等式组:;并将解集在数轴上表示出来.

21.(本题满分10分,每小题5分)

销售某种商品,根据经验,销售单价不少于30元∕件,但不超过50元∕件时,销售数量(件)与商品单价(元∕件)的函数关系的图像如图5所示中的线段.

(1)求关于的函数关系式;

(2)如果计划每天的销售额为2400元时,那么该商品的单价应该定多少元?

22.(本题满分10分,每小题5分)

如图6,梯形中,∥,和相交于点,,,

,.

求:(1)的值;

(2)的面积.

23.(本题满分12分)

如图7,四边形是平行四边形,在边的延长线上截取,点在

的延长线上,和交于点,和交于点.

(1)求证:四边形是平行四边形;(4分)

(2)如果,求证:. (8分)

24.(本题满分12分)

抛物线()经过点,对称轴是直线,顶点是,与轴正半轴的交点为点.

(1)求抛物线()的解析式和顶点的坐标;(6分)

(2)过点作轴的垂线交轴于点,点在射线上,当以为直径的⊙和以为半径的⊙相切时,求点的坐标.(6分)

25.(本题满分14分)

如图8,在中,,,,点是边上任意

一点,过点作交于点,截取,联结,线段交

于点,设,.

(1)求关于的函数解析式及定义域;(4分)

(2)如图9,联结,当和相似时,求的值;(5分)

(3)当以点为圆心,为半径的⊙和以点为圆心,为半径的⊙相交的另一个交点在边上时,求的长.(5分)

答案和评分标准

一、选择题:(本大题共6题,每题4分,满分24分)

1.C;2.D;3.B;4.A;5.B;6.B.

二.填空题:(本大题共12题,满分48分)

7.;8.;9.或;10.;11.;12.;

13.;14.;15.﹪;16.;17.不唯一,如等;18.或.三、(本大题共7题,第19、20、21、22题每题10分,第23、24题每题12分,第25题

14分,满分78分)

19. 解:原式…………………………………………………(8分)

……………………………………………………………………(2分)20.解:由不等式(1)解得<………………………………………………………(3分)

由不等式(2)解得≥…………………………………………………………(3分)

∴原不等式组的解集是≤<……………………………………………(2分)图正确.……………………………………………………………………………(2分)21.解:(1)设关于的函数关系式为.…………………………(1分)

由题意,得……………………………………………(2分)

解得,……………………………………………………………(1分)

∴关于的函数关系式为.…………………………(1分)(2)设该商品的单价应该定元.………………………………………………(1分)由题意,得…………………………………………(1分)

化简整理,得.………………………………………(1分)

解得,,.………………………………………………(1分)

经检验,不合题意,舍去;………………………………………(1分)

答:计划每天的销售额为2400元时,该商品的单价应该定元.

22.解:(1)∵∥,∴.……………………………………(2分)

∵,∴.………………………………………(1分)

在中,,

∴.…………………………………………………(2分)(2)∵…………………………………………(2分)

∴.…………………………………(3分)23.证明:(1)∵四边形是平行四边形,

∴∥,;…………………………………………(2分)

∵,∴;…………………………………………(1分)

又∥,

∴四边形是平行四边形.………………………………………(1分)

(2)∵,∴,………………………………(1分)又,∴∽,∴;……(1分)

∵∥,∴;………………………………(1分)

∵四边形是平行四边形,∴∥,∴;(1分)

∵四边形是平行四边形,∴∥,∴;(1分)

∴;…………………………………………………(1分)

又,∴∽,∴,…(1分)

∵,∴,

∴.………………………………………………(1分)24.解:(1)由题意,得,…………………………………………………(2分)解得……………………………………………………………(2分)

∴………………………………………………………(1分)

∴顶点.…………………………………………………………(1分)(2)设⊙的半径为.

由题意,可得,,∴⊙的半径为;;……(2分)

当⊙和⊙相切时,分下列两种情况:

当⊙和⊙外切时,此时点在线段上,

可得.

解得,∴.……………………………………………(2分)

当⊙和⊙外切时,此时点在线段的延长线上,

可得.

解得,∴.…………………………………………(2分)综合,当⊙和⊙相切时,或.

25.解:(1)过点作,垂足为.

由题意,可知是等腰直角三角形,∴;……………(1分)

易得∽,∴;

设,,∴,∴,

∴……………………………………………………………(1分)

∴.………………………………………………………(1分)

定义域是:≤≤.………………………………………………(1分)(注:其它解法参照评分.)

(2)∵,∴当和相似时,分以下两种情况:(1分)

当时,∴∥,易得四边形是正方形;

∴.…………………………………………………(2分)

当时,∴,

由上述(1)的解法,可得,

∴,∴;

∴,解得.………………………………(2分)综合,当和相似时,的值为或.

(3)如图,设⊙与⊙相交的另一个交点为,联结交于点

∴,.易得∽,∽,

∴,设,,∴;…(1分)

∴,∴;∵,∴;…(1分)

又,∴,解得;……………(2分)∴.……………………………………………(1分)

一、选择题:(本大题共6题,每题4分,满分24分)

[下列各题的四个选项中,有且只有一个选项是正确的,选择正确项的代号并填涂在答题纸的相应位置上.]

1.在下列各数中,属于无理数的是

A.;

B.;

C.;

D..

2.在下列一元二次方程中,没有实数根的是

A.;

B.;

C.;

D..

3.在平面直角坐标系中,直线经过

A.第一、二、三象限;B.第一、二、四象限;

C.第一、三、四象限;D.第二、三、四象限.

4.某小区20户家庭某月的用电量如下表所示:

用电量(度)120 140 160 180 200

户数 2 3 6 7 2

则这20户家庭该月用电量的众数和中位数分别是

A.180,160;B.160,180;C.160,160;D.180,180. 5.已知两圆内切,圆心距为5,其中一个圆的半径长为8 ,那么另一个圆的半径长是A.3;B.13;C.3或13;D.以上都不对. 6.在下列命题中,属于假命题

...的是

A.对角线相等的梯形是等腰梯形;

B.两腰相等的梯形是等腰梯形;

C.底角相等的梯形是等腰梯形;

D.等腰三角形被平行于底边的直线截成两部分,所截得的四边形是等腰梯形.

二、填空题:(本大题共12题,每题4分,满分48分)

[请将结果直接填入答题纸的相应位置]

7.计算:▲ .

8.不等式组的解集是▲ .

9.用换元法解分式方程时,如果设,那么原方程化为关于的整式方程可以是▲ .

10.方程的解是▲.

11.对于双曲线,若在每个象限内,y随x的增大而增大,则k的取值范围是▲.

12.将抛物线向左平移2个单位,所得抛物线的表达式为▲.

13.在一个不透明的盒子中装有8个白球和若干个黄球,它们除颜色不同外,其余均相同.若

从中随机摸出1个球,它恰好是白球的概率是,则该盒中黄球的个数为▲ .14.为了解某校九年级学生体能情况,随机抽查了其中的25名学生,测试了1分钟仰卧起坐的次数,并绘制成频数分布直方图(如图所示),那么仰卧起坐的次数在20~25的频率是▲.

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