备战2014年数学中考————吉林省2009年初中毕业暨高中招生考试题及答案

吉林省2009年初中毕业生学业考试

数 学 试 卷

一、填空题（每小题2分，共20分）

1．数轴上A、B两点所表示的有理数的和是．

0 1 2 3

（第1题）

2．计算(3a)·a

3．为鼓励大学生创业，某市为在开发区创业的每位大学生提供货款150 000元，这个数据用科学记数法表示为 元．

4．不等式2x 3 x的解集为．

5．如图，点A关于y轴的对称点的坐标是 6．方程

2

5

3

1的解是 ． x 2

（第5题）

7．若a 5,b 2,且ab 0,则a b ．

8．将一个含有60°角的三角板，按图所示的方式摆放在半圆形纸片上，O为圆心，则 ACO

B

A

B

（第9题） （第10题） （第8题）

9．如图，△OAB的顶点B的坐标为（4，0），把△OAB沿x轴向右平移得到△CDE，如果CB 1,那么OE的长为 ．

10．将一张矩形纸片折叠成如图所示的形状，则 ABC度． 二、单项选择题（每小题3分，共18分）

xy 2y

的结果是（ ） 2

x 4x 4xxyyA． B． C． D．

x 2x 2x 2x 2

11．化简

12．下列图案既是轴对称图形，又是中心对称图形的是（ ）

A． B． C． D． 13．下列几何体中，同一个几何体的主视图与俯视图不同的是（ ）

正方体 圆锥 球 圆柱

A．

B． C． D．

14．A种饮料B种饮料单价少1元，小峰买了2瓶A种饮料和3瓶B种饮料，一共花了13元，如果设B种饮料单价为x元/瓶，那么下面所列方程正确的是（ ） A．2(x 1) 3x 13 B．2(x 1) 3x 13 C．2x 3(x 1) 13 D．2x 3(x 1) 13

15．某校七年级有13名同学参加百米竞赛，预赛成绩各不相同，要取前6名参加决赛，小梅已经知道了自己的成绩，她想知道自己能否进入决赛，还需要知道这13名同学成绩的（ ）

A．中位数 B．众数 C．平均数 D．极差

16．将宽为2cm的长方形纸条折叠成如图所示的形状，那么折痕PQ的长是（ ） A2cm

60P Q

（第16题）

B C D．2cm 三、解答题（每小题5分，共20分）

17．在三个整式x 2xy,y 2xy,x中，请你任意选出两个进行加（或减）运算，使所得整式可以因式分解，并进行因式分解．

18．在一个不透明的盒子里，装有三个分别写有数字6， 2，7的小球，它们的形状、大小、质地等完全相同，先从盒子里随机取出一个小球，记下数字后放回盒子，摇匀后再随机取出一个小球，记下数字．请你用画树形图或列表的方法，求下列事件的概率： （1） 两次取出小球上的数字相同；

（2） 两次取出小球上的数字之和大于10．

19．如图，AB AC,AD BC于点D，AD AE，AB平分 DAE交DE于点F，请你写出图中三对全等三角形，并选取其中一对加以证明． ．．

E

A

B D C

（第19题）

20．如图所示，矩形ABCD的周长为14cm，E为AB的中点，以A为圆心，AE长为半径画弧交AD于点F．以C为圆心，CB长为半径画弧交CD于点G．设AB xcm，

222

BC ycm，当DF DG时，求x,y的值．

G

（第20题）

四、解答题（每小题6分，共12分）

21．下图是根据某乡2009年第一季度“家电下乡”产品的购买情况绘制成的两幅不完整的统计图，请根据统计图提供的信息解答下列问题：

电脑 电视机

35%

冰箱 热水器 洗衣机 注意：将答案写在横线上 ．．

（第21题）

（1）第一季度购买的“家电下乡”产品的总台数为 ； （2）把两幅统计图补充完整．

22．如图，⊙O中，弦AB、CD相交于AB的中点E，连接AD并延长至点F， 使DF AD，连接BC、BF． （1）求证：△CBE∽△AFB；

BE5CB

（2）当的值． 时，求

FB8AD

F

B （第22题）

五、解答题（每小题7分，共14分）

23．小鹏学完解直角三角形知识后，给同桌小艳出了一道题：“如图所示，把一张长方形卡片ABCD放在每格宽度为12mm的横格纸中，恰好四个顶点都在横格线上，已知 =36°，求长方形卡片的周长．”请你帮小艳解答这道题．（精确到1mm）（参考数据：sin36°≈0.60，cos36°≈0.80，tan36°≈0.75）

C

（第23题）

k

的图象与直线y x m在第一象限交于点P，A、B为（6，2）x

直线上的两点，点A的横坐标为2，点B的横坐标为3．D、C为反比例函数图象上的两点，

24．如图，反比例函数y 且AD、BC平行于y轴．

（1）直接写出k，m的值； （2）求梯形ABCD的面积．

六、解答题（每小题8分，共16分）

（第24题）

25．A、B两地相距45千米，图中折线表示某骑车人离A地的距离y与时间x的函数关系．有一辆客车9点从B地出发，以45千米/时的速度匀速行驶，并往返于A、B两地之间．（乘客上、下车停留时间忽略不计）

（1）从折线图可以看出，骑车人一共休息 次，共休息 小时； （2）请在图中画出9点至15点之间客车与A地距离y随时间x变化的函数图象； （3）通过计算说明，何时骑车人与客车第二次相遇．

9 10 11 12 13 14 15

（第25题）

x/时

26．两个长为2cm，宽为1cm的长方形，摆放在直线l上（如图①），将长方形ABCDCE=2cm，

绕着点C顺时针旋转 角，将长方形EFGH绕着点E逆时针旋转相同的角度． （1）当旋转到顶点D、H重合时，连接AG（如图②），求点D到AG的距离； （2）当 45°时（如图③），求证：四边形MHND为正方形．

A D H G

l B C E F

图① G A

H

F l

C E

图②

A

F C

l C E

图③

（第26题）

七、解答题（每小题10分，共20分）

27．某数学研究所门前有一个边长为4米的正方形花坛，花坛内部要用红、黄、紫三种颜色的花草种植成如图所示的图案，图案中AE MN．准备在形如Rt△AEH的四个全等三角形内种植红色花草，在形如Rt△AEH的四个全等三角形内种植黄色花草，在正方形

MNPQ内种植紫色花草，每种花草的价格如下表：

设的长为米，正方形的面积为平方米，买花草所需的费用为元，解答下列问题：

（1）S与x之间的函数关系式为S ；

（2）求W与x之间的函数关系式，并求所需的最低费用是多少元； （3）当买花草所需的费用最低时，求EM的长．

E

F

H

G C

（第27题）

28．如图所示，菱形ABCD的边长为6厘米， B 60°．从初始时刻开始，点P、Q同时从A点出发，点P以1厘米/秒的速度沿A C B的方向运动，点Q以2厘米/秒的速度沿A B C D的方向运动，当点Q运动到D点时，P、Q两点同时停止运动，设的面积为y平方厘米（这里规定：P、Q运动的时间为x秒时，△APQ与△ABC重叠部分．．．．点和线段是面积为O的三角形），解答下列问题： （1）点P、Q从出发到相遇所用时间是 秒；

（2）点P、当△APQ是等边三角形时x的值是 秒； Q从开始运动到停止的过程中，（3）求y与x之间的函数关系式．

（第28题）

吉林省2009年初中毕业生学业考试数学试卷

参考答案及评分标准

阅卷说明：

1． 评卷采分最小单位为1分，每步标出的是累计分．

2． 考生若用本“参考答案”以外的解（证）法，可参照本“参考答案”的相应步骤给分．

一、填空题（每小题2分，共20分）

1． 1 2．9a 3．1.5×105 4．x>1 5．（5，3）

7

6．x＝5 7． 7 8．120 9．7 10．73 二、单项选择题（每小题3分，共18分）

11．D 12．D 13．C 14．A 15．A 16．B 三、解答题（每小题5分，共20分）

17．解：(x 2xy) x 2x 2xy 2x(x y); 或(y 2xy) x (x y);

或(x 2xy) (y 2xy) x y (x y)(x y); 或(y 2xy) (x 2xy) y x (y x)(y x).

说明：选择整式正确得2分，整式加（减）结果正确得1分，因式分解正确得2分，累计5分.

18． 解：

树形图

2 6 2 7 6 2 7 6 2 7

2

2

2

2

2

2

2

2

2

2

2

2

2

2

········································································································································· （2分）

1

． ·························································································· （3分） 3

4

（2）P（两数和大于10）=． ················································································· （5分）

9

19． 解：（1）△ADB≌△ADC、△ABD≌△ABE、△AFD≌△AFE、

··························· （3分） △BFD≌△BFE、△ABE≌△ACD（写出其中的三对即可）.

（2）以△ADB≌ADC为例证明．

（1）P（两数相同）=

证明： AD BC, ADB ADC 90°. 在Rt△ADB和Rt△ADC中，

AB AC,AD AD,

························································································· （5分） Rt△ADB≌Rt△ADC. ·

说明：选任何一对全等三角形，只要证明正确均得分．

20． 解：根据题意，得

2x 2y 14,

·············································································································· （3分） x ·

x y y 2

解得

x 4,

···················································································································· （5分）

y 3.

答：x为4，y为3．

说明：不写答不扣分．

四、解答题（每小题6分，共12分） 21．解：（1）500． ········································································································· （1分） （2） 电脑 电视机

35%

20 30 冰箱 热水器 洗衣机

注意：将答案写在横线上

········································································································································· （6分） 说明：第（2）问中每图补对一项得1分，条形图中不标台数不扣分． 22．（1）证明： AE EB,AD DF,

ED是△ABF的中位线，

·········································································································· （1分） ED∥BF, ·

································································································ （2分） CEB ABF, ·

又 C A, ············································································································ （3分） ······························································································ （4分） △CBE∽△AFB, ·（2）解：由（1）知， △CBE∽△AFB,

CBBE5

·········································································································· （5分） .·

AFFB8

又AF 2AD,

CB5

················································································································· （6分） ． ·

AD4

五、解答题（每小题7分，共14分）

23． 解：作BE l于点E，DF l于点F． ·························································· （1分）

l

C

DAF 180° BAD 180° 90° 90°， ADF DAF 90 ， ADF 36 .

根据题意，得BE=24mm，DF=48mm. ········································································ （2分） 在Rt△ABE中，sin

BE

， ···················································································· （3分） AB

BE24

·················································································· （4分） 40mm ·

sin36°0.60

DF

在Rt△ADF中，cos ADF ， ·········································································· （5分）

AD

DF48

··············································································· （6分） AD 60mm．

cos36°0.80

······························································ （7分） 矩形ABCD的周长=2（40+60）=200mm． ·

24． 解：（1）k=12，m= 4. ······················································································ （2分） AB

（2）把x=2代入y=

12

，得y=6. D（2，6）． x

把x=2代入y x 4，得y 2. A(2， 2).

···································································································· （4分） DA 6 ( 2) 8. ·

把x=3代入y x 4，得y= 1， B(3， 1).

······························································································· （6分）

BC 4 （ 1）=5． ·

S梯形ABCD

(5 8) 113

······················································································ （7分） . ·

22

六、解答题（每小题8分，共16分）

25． 解：（1）两．两． ································································································· （2分） （2）

9 10 11 12 13 14 15 x/时

········································································································································· （4分） （3）设直线EF所表示的函数解析式为y kx b. 把E(10,0),F(11,45)分别代入y kx b，得

10k b 0

················································································································· （5分）

11k b 45

k 45，解得

b 450.

·························································· (6分) 直线EF所表示的函数解析式为y 45x 450. ·把y 30代入y 45x 450,得

············································································································ （7分） 45x 450 30. ·2

x 10．

3答：10点40分骑车人与客车第二次相遇. ···································································· （8分） 说明：第（3）问时间表达方式可以不同，只要表达正确即可得分，不写答不扣分． 26．解：（1） CD CE DE 2cm， △CDE是等边三角形．

········································································································· （1分） CDE 60°. ·

ADG 360° 2 90° 60° 120°．

A K G

H

F B

l E

图②

A

B F

又AD DG 1cm，

························································································ （2分） DAG DGA 30°． ·

如图②作DK AG于点K．

11

DK DG cm．

22

1

························································································· (4分) 点D到AG的距离为cm． ·

2

（2） 45°

NCE NEC 45°,

·············································································································· (5分) CNE 90°. ·

DNH 90°

D H 90°,

··························································································· （6分） 四边形MHND是矩形． ·又CN NE,

············································································································· （7分） DN NH， ·

··························································································· （8分） 矩形MHND是正方形． ·

七、解答题（每小题10分，共20分）

27．解：（1）x (4 x)或2x 8x 16. ································································ （2分） （2）W 60 4S△AEB 80(S正方形EFGN-S正方形MNPQ)+120S正方形MNPQ =60 4

2

2

2

2

1

··················································· （4分） x(4 x) 80[x2 (4 x)2 x2] 120x2.

2

=80x 160x 1280. ····································································································· （5分） 配方，得

W 80(x 1)2 1200. ·································································································· （6分）

·············································································· （7分） 当x 1时，W最小值 1200元． ·（3）设EM a米，则MH (a 1)米. 在Rt△EMH中，

a2 (a 1)2 12 32,

解得a

a 0,

a

EM米． ························································································· （10分） 28． 解：（1）6．··········································································································· （1分） （2）8． ·························································································································· （3分） （3）①当0≤x 3时，

Q3

P3

Q2

B

Q

1

y S△APQ 13

112

AP·AQ·sin60 x·2x· x． ····································· （5分） 112222

②当3≤x 6时，

y S△APQ=12

1AP2 P2Q22

1AP2·CQ2·sin60 21 x·(12-2x·)22

=2

x . ·········································································································· （7分） ③当6≤x≤9时，设P3Q3与AC交于点O． (解法一)

过Q3作Q3E∥CB,则△CQ3E为等边三角形．

Q3E CE CQ3 2x 12. Q3E∥CB. △COP3∽△EOQ3

OCCP3x 61

,OEEQ32x 122

11

OC CE (2x 12),

33y S△AQP3

S△ACP3-S△COP3

11

CP3·AC·sin60° OC·CP3·sin60°

22

111 (x 6)·6 (2x 12)(x 6) ．

22232

2x x ····················································································· （10分） 62

（解法二）

如右图，过点O作OF CP3于点F，OG CQ3，于点G, 过点P3作P3H DC交DC延长线于点H．

ACB ACD, OF OG.

又CP3 x 6,CQ3, 2x 12 2(x 6),

S△CQP3 S△COP3

1

S△

COQ3 21

S△CP3Q3,3

Q3

G H 3

B

11

CQ3·P3H32

11 (2x 12)(x 6)·322

x 6)2.又S△ACP3

1CP3·AC·sin60°

2

1(x 6) 6 22

x 6).2

y S△

AOP3

S△ACP3 S△

OCP3

2

(x 6) x 6)26

2x x ························································································· （10分） 62

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