2009年广东省湛江市中考数学试题及参考答案

更新时间:2023-05-24 17:03:01 阅读量: 实用文档 文档下载

说明:文章内容仅供预览,部分内容可能不全。下载后的文档,内容与下面显示的完全一致。下载之前请确认下面内容是否您想要的,是否完整无缺。

2010年广东省各地级市中考数学试题

湛江市2009年初中毕业生学业考试

数 学 试 卷

说明:

1.本试卷满分150分,考试时间90分钟. 2.本试卷共6页,共5大题.

3.答题前,请认真阅读答题卡上的“注意事项”,然后按要求将答案写在答题卡相应的位置上. 4.请考生保持答题卡的整洁,考试结束,将试卷和答题卡一并交回. 注意:在答题卡上作图必须用黑色字迹的钢笔或签字笔.

一、选择题:本大题10个小题,其中1~5每小题3分,6~10每小题4分,共35分.在每小题给出的四个选项中,只有一项是符号题目要求的)

1.下列四个数中,在 1和2之间的数是( ) A.0 B. 2 C. 3 D.3 2.下列各式中,与(x 1)相等的是( ) A.x 1

2

2

B.x 2x 1 C.x 2x 1

22

D.x

2

3.湛江是个美丽的海滨城市,三面环海,海岸线长达1556000米,数据1556000用科学记数法表示为( )

A.1.556 10 C.15.56 10

57

B.0.1556 10 D.1.556 10

6

8

4.在右图的几何体中,它的左视图是( )

第4题图

A. B. C. D.

5.沃尔玛商场为了了解本商场的服务质量,随机调查了本商场的100名顾客,调查的结果如图所示,根据图中给出的信息,这100名顾客中对该商场的服务质量表示不满意的有( ) A.6人 B.11人 C.39人 D.44人 A:很满意 A

A B:满意 44%

D

C:说不清 C

B

D:不满意 E D 39%

C B 第5题图 第6题图

6.如图,在等边△ABC中,D、E分别是AB、AC的中点,DE 3,则△ABC的周长是( ) A.6 B.9 C.18 D.24

2010年广东省各地级市中考数学试题

7.如图,在平面直角坐标系中,菱形OACB的顶 点O在原点,点C的坐标为(4,0),点B的纵坐标 是 1,则顶点A的坐标是( )

A.(2, 1) B.(1, 2) C.(1,2) D.(2,1) 8.根据右图所示程序计算函数值, 若输入的x的值为

5

,则输出的 2

函数值为( )

32A. B.

25

第8题图

425C. D.

254

9.下列说法中:

①4

的算术平方根是±2;

3)关于原点对称的点的坐标是( 2,③点P(2, 3);

④抛物线y

1

(x 3)2 1的顶点坐标是(31),. 2

其中正确的是( ) P A.①②④ B.①③ C.②④ D.②③④

10.如图,小林从P点向西直走12米后,向左转,

转动的角度为 ,再走12米,如此重复,小林共

第10题图 走了108米回到点P,则 ( )

A.30° B.40° C.80° D.不存在

二、填空题:本大题共10个小题,其中11~15每小题3分,16~20每小题4分,共35分. 11. 2的相反数是.

1

B 12.要使分式有意义,则x的取值范围是 .A

x 3

,13.如图,已知AB∥CD, 1 55°则 2. 14.分解因式:m n .

2

2

2 C 第13题图

D 15.已知在一个样本中,40个数据分别落在4个组内,第一、二、四组

数据个数分别为5、12、8,则第三组的频数为 . 16.如图,AB是⊙O的直径,C、D、E是⊙O上的点,则 1 2 C

°.

17.一件衬衣标价是132元,若以9折降价出售,仍可获利10%,则这件衬衣的进价是 元.

B 第16题图

2010年广东省各地级市中考数学试题

18.如图,现将⊙O1向⊙O2平移,当O1O2cm时,⊙O1、⊙O2的直径分别为2cm和4cm,⊙O1与⊙O2相切. 19.已知2

2233

22 ,3 32 , 3388

44aa24 42 , ,若8 8 (a、b为正整数)则a b .

1515bb

B M

第18题图 第20题图

, A B 90°,CD 5,AB 11,20.如图,在梯形ABCD中,AB∥CD点M、N分别为

AB、CD的中点,则线段MN .

三、解答题:本大题共2小题,每小题8分,共16分.

21.如图,一只蚂蚁从点A沿数轴向右直爬2个单位到达点B,点A表示,设点B所表示的数为 m.

(1)求m的值;

(2)求m (m 6)的值.

第21题图

0)(30)(3 2),将△OAB绕点O按逆时针方向旋转90°得22.如图,点O、A、B的坐标分别为(0,、,、,

到△OA B .

(1)画出旋转后的△OA B ,并求点B 的坐标;

(2)求在旋转过程中,点A所经过的路径 AA 的长度.

2010年广东省各地级市中考数学试题

四、解答题:本大题共4小题,每小题10分,共40分.

23.某语文老师为了了解中考普通话考试的成绩情况,从所任教的九年级(1)、(2)两班各随机抽取了10名学生的得分,如图所示:

九(1)班 九(2)班

(1)利用图中的信息,补全下表:

第23题图

(2)若把16分以上(含16分)记为“优秀”,两班各有60名学生,请估计两班各有多少名学生成绩优秀.

24.如图,某军港有一雷达站P,军舰M停泊在雷达站P的南偏东60°方向36海里处,另一艘军舰N位于军舰M的正西方向,与雷达站P相距 (1)军舰N在雷达站P的什么方向? (2)两军舰M、N的距离.(结果保留根号)

2010年广东省各地级市中考数学试题

25.六张大小、质地均相同的卡片上分别标有1、2、3、4、5、6,现将标有数字的一面朝下扣在桌面上,从中随机抽取一张(放回洗匀),再随机抽取第二张.

(1)用列表法或树状图表示出前后两次抽得的卡片上所标数字的所有可能结果;

(2)记前后两次抽得的数字分别为m、n,若把m、n分别作为点A的横坐标和纵坐标,求点A(m,n)在函数y

12

的图象上的概率. x

26.如图,AB是⊙O的切线,切点为B,AO交⊙O于点C,过点C作DC OA,交AB于点D. (1)求证: CDO BDO;

,⊙O的半径为4,求阴影部分的面积.(2)若 A 30°(结果保留π)

A

B D

第26题图

五、解答题:本大题共2小题,每小题12分,共24分.

27.某公司为了开发新产品,用A、B两种原料各360千克、290千克,试制甲、乙两种 新型产品共50件,下表是试验每件

新产品所需原料的相关数据:

2010年广东省各地级市中考数学试题

(1)设生产甲种产品x件,根据题意列出不等式组,求出x的取值范围;

(2)若甲种产品每件成本为70元,乙种产品每件成本为90元,设两种产品的成本总额为y元,写出

成本总额y(元)与甲种产品件数x(件)之间的函数关系式;当甲、乙两种产品各生产多少件时,产品的成本总额最少?并求出最少的成本总额.

28.已知矩形纸片OABC的长为4,宽为3,以长OA所在的直线为x轴,O为坐标原点建 立平面直角坐标系;点P是OA边上的动点(与点O、A不重合),现将△POC沿PC翻折 得到△PEC,再在AB边上选取适当的点D,将△PAD沿PD翻折,得到△PFD,使得 直线PE、PF重合.

(1)若点E落在BC边上,如图①,求点P、C、D的坐标,并求过此三点的抛物线的函数关系式; (2)若点E落在矩形纸片OABC的内部,如图②,设OP x,AD y,当x为何值时,y取得最大值? (3)在(1)的情况下,过点P、C、D三点的抛物线上是否存在点Q,使△PDQ是以PD为直角边的直角三角形?若不存在,说明理由;若存在,求出点Q的坐标

图①

图②

第28题图

湛江市2009年初中毕业生学业考试

数学试卷参考答案与评分说明

一、选择题:本大题共10小题,其中1~5小题每题3分,6~10小题每题4分,共35分.

二、填空题:本大题共10小题,其中11~15每小题3分,16~20每小题4分,共35分. 11.2 12.x 3 13.125° 14.(m n)(m n) 15.15 16.90

2010年广东省各地级市中考数学试题

17.108 18.1或3 19.71 20.3

三、解答题:本大题共2小题,每小题

8分,共16分.

21.解:(1)由题意可得m 2 ·························································································

2分 (2)把m

的值代入得:m 1 (m 6) 21 (26) ·····························

···· 3

= (8 ·········································································································

· 4分 1 1 ·························································································································

· 7分 ··································································································································· 8分

3); 22.解:(1)如图△OA B 为所示,点B 的坐标为(2,

················································································· 4

(2)△OAB绕点O逆时针旋转90°后得△OA B 点A所经过的路径 AA 是圆心角为90°,半径为3 的扇形OAA 的弧长,所以l

13

(2π 3) π. 42

3

π. ·············· 8分2

················································································ 7分即点A所经过的路径 AA 的长度为

第22题图

四、解答题:本大题共4小题,每小题10分,共40分. 23 ······················································································································································· 6分 (2) 60

76

42(名) 36(名),60 . 1010

································ 10分 九(1)班有42名学生成绩优秀,九(2)班有36名学生成绩优秀. ·

24.解:过点P作PQ MN,交MN的延长线于点Q. ···················································· 1分

,(1)在Rt△PQM中,由 MPQ 60°

得 PMQ 30° 又PM 36

北 PQ

11

PM 36 18(海里) 22

N 第24题图

···················································································· 3分 在Rt△PQN中,cos QPN

PQ PN

2010年广东省各地级市中考数学试题

QPN 45°

即军舰N到雷达站P的东南方向(或南偏东45°) ··························································· 5分 (2)由(1)知Rt△PQN为等腰直角三角形, PQ NQ 18(海里) ···················

· 7分 在Rt△PQM中,MQ PQ

·tan QPM 18·tan60°

·····································································

····· 9分 MN MQ NQ 18(海里) ·

答:两军舰的距离为18海里. ············································································ 10分

由表可看出,前后两次抽得的卡片上所标数字的所有可能结果有36种. ······························· 5分 或画树状图:

第一次: 1 2 3 4 5 6

第二次: 1 6 1 6 1 6 1 6 1 6 1 6

从树状图可以看出,所有可能出现的结果有36种,即: ·························································· 3分 (1,1)、(1、2)、(1、3)、(1、4)、(1、5)、(1、6)、 (2,1)、(2、2)、(2、3)、(2、4)、(2、5)、(2、6) (3,1)、(3、2)、(3、3)、(3、4)、(3、5)、(3、6) (4,1)、(4、2)、(4、3)、(4、4)、(4、5)、(4、6) (5,1)、(5、2)、(5、3)、(5、4)、(5、5)、(5、6) (6,1)、(6、2)、(6、3)、(6、4)、(6、5)、(6、6) ···························································· 5分 (2)有4个点(2,6)、(3,4)、(4,3)、(6,2)在函数y

12

的图象上 ·························· 8分 x

所求概率P

41 ··············································································································· 10分 369

26.解:(1) AB切⊙O于点B ∴OB AB,即 B 90° ··········································································································· 1分

OCD 90° ·又 DC OA,································································································ 2分

2010年广东省各地级市中考数学试题

在Rt△COD与Rt△BOD中 OD OD,OB OC

······························································································· 3分 Rt△COD≌Rt△BOD(HL) ·

CDO BDO. ···················································································································· 4分 ,OB 4 (2)在Rt△ABO中, A 30°

OA 8

AC OA OC 8 4 4

····································································· 5分 在Rt△ACD中,tan A

CD

AC

,AC 4 又

A 30°

CD AC·tan30°

3

B D

第26题图

A

·································································

···· 7

1····································································· 8分 S四边形OCDB 2S△OCD 2 4

2, BOC 60°. 又 A 30°

S扇形OBC

60π·428π

. ······························································································

····· 9分

3603

S阴影 S四边形OCDB S扇形OBC

···································································· 10分 .

3

五、解答题:本大题共2小题,每小题12分,共24分. 27.解:(1)依题意列不等式组得

9x 4(50 x)≤360

··············································· 3分

3x 10(50 x)≤290

由不等式①得x≤32 ················································································································· 4分

由不等式②得x≥30 ················································································································· 5分

······························································································· 6分 x的取值范围为30≤x≤32 ·

(2)y 70x 90(50 x) ··································································································· 8分 化简得y 20x 4500

20 0, y随x的增大而减小. ····················································································· 9分

而30≤x≤32

······················· 11分 当x 32,50 x 18时,y最小值 20 32 4500 3860(元) ·

答:当甲种产品生产32件,乙种18件时,甲、乙两种产品的成本总额最少,最少的成本总额为3860

元. ······································································································································· 12分

△PAD均为等腰直角三角形, 28.解:(1)由题意知,

△POC、

2010年广东省各地级市中考数学试题

可得P(3,、······························································································· 2分 0)C(0,、3)D(41),

图①

图②

第28题图

c 3

设过此三点的抛物线为y ax2 bx c(a 0),则 9a 3b c 0

16a 4b c 1

1 a 2

5

b

2 c 3

125

x x 3 ······································ 4分 22

(2)由已知PC平分 OPE,PD平分 APF,且PE、PF重合,则 CPD 90° OPC APD 90°,又 APD ADP 90° OPC ADP.

Rt△POC∽Rt△DAP.

过P、C、D三点的抛物线的函数关系式为y

OPOCx3

,即 ······························································································ 6分 ADAPy4 x

y

11414

x(4 x) x2 x (x 2)2 (0 x 4) 33333

4 ·························································································· 8分 当x 2时,y有最大值.3

(3)假设存在,分两种情况讨论:

①当 DPQ 90°时,由题意可知 DPC 90°,且点C在抛物线上,故点C与点Q重合,所求的点

Q为(0,3)····························································································································· 9分

②当 DPQ 90°时,过点D作平行于PC的直线DQ,假设直线DQ交抛物线于另一点Q

, 点

2010年广东省各地级市中考数学试题

P(3,、0)C(0,3), 直线PC的方程为y x 3,将直线PC向上平移2个单位与直线DQ重合,

直线DQ的方程为y x 5 ································································································· 10分

y x 5

x 1 x 4

由 得或 125

y x x 3 y 6 y 1 22

又点D(41),, Q( 1,.6)

故该抛物线上存在两点Q(0,、······························································ 12分 3)( 1,6)满足条件.

说明:以上各题如有其他解(证)法,请酌情给分. .

第28题图

本文来源:https://www.bwwdw.com/article/ylv4.html

Top