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初中数学13

2009年临沂市中考 数 学 试 题

本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.第Ⅰ卷1至4页,第Ⅱ卷5至12页,满分120分.考试时间120分钟.

第Ⅰ卷(选择题 共42分)

注意事项:

1. 答第Ⅰ卷前,考生务必将自己的姓名、准考生号、考试科目用铅笔涂写在答题卡上.

2. 每小题选出答案后,用铅笔把答题卡上对应题目的答案标号涂黑,如需改动,用橡

皮擦干净后,再选涂其他答案,不能答在试卷上. 3. 考试结束,将本试卷和答题卡一并收回.

一、选择题(本大题共14小题,每小题3分,共42分)在每小题所给的4个选项中,只有一项是符合题目要求的. 1.9-的相反数是( ) A .

19

B .19

-

C .9-

D .9

2.某种流感病毒的直径是0.00000008m ,这个数据用科学记数法表示为( ) A .6

810m -? B .5

810m -?

C .8810m -?

D .4

810m -?

3.下列各式计算正确的是( ) A .3

4

x x x += B .2510

·

x x x =

C .4

2

8

()x x =

D .2

2

4

(0)x x x x +=≠

4.下列图形中,由AB CD ∥,能得到12∠=∠的是( )

5.计算12718123

--

的结果是( )

A .1

B .1-

C .32-

D .23-

6.化简

2

2

422b

a

a b

b a

+

--的结果是( )

A .2a b --

B .2b a -

C .2a b -

D .2b a +

7.已知1O ⊙和2O ⊙相切,1O ⊙的直径为9C m ,2O ⊙的直径为4cm .则12O O 的长是( )

A

C B

D 1 2 A C B D 1 2 A . B . 1 2 A C B D C . B D

C

A D .

1

2

A .5cm 或13cm

B .2.5cm

C .6.5cm

D .2.5cm 或6.5cm

8.如图,OP 平分AOB ∠,PA OA ⊥,PB OB ⊥,

垂足分别为A ,B .下列结论中不一定成立的是( )

A .PA P

B = B .PO 平分APB ∠

C .OA OB =

D .A B 垂直平分OP

9.对于数据:80,88,85,85,83,83,84.下列说法中错误的有( ) A .这组数据的平均数是84 B .这组数据的众数是85 C .这组数据的中位数是84 D .这组数据的方差是36 A .1个 B .2个 C .3个 D .4个

10.若x y >,则下列式子错误的是( ) A .33x y ->- B .33x y ->-

C .32x y +>+

D .

33

x y >

11.如图,在等腰梯形ABCD 中,AD BC ∥,对角线AC BD ⊥于

点O ,AE BC DF BC ⊥⊥,,垂足分别为E 、F ,设AD =a ,BC =b ,则四边形AEFD 的周长是( ) A .3a b +

B .2()a b +

C .2b a +

D .4a b +

12.如图是一个包装盒的三视图,则这个包装盒的体积是( )

初中数学13

A .3

192πcm

B .31152πcm

C .32883cm

D .3

3843cm

13.从1,2,3,4这四个数字中,任意抽取两个不同数字组成一个两位数,则这个两位数能被3整除的概率是( ) A .

13

B .

14

C .

16

D .

112

14.矩形ABCD 中,8cm 6cm AD AB ==,.动点E 从点C 开始沿边CB 向点B 以2cm/s 的速度运动,动点F 从点C 同时出发沿边CD 向点D 以1cm/s 的速度运动至点D 停止.如图可得到矩形CFHE ,设运动时间为x (单位:s ),此时矩形ABCD 去掉矩形CFHE 后剩余

D

C A

B

E F

O

(第11题图)

12cm 4cm (第12题图) A D

F

C

E

H

B

(第14题图)

O

(第8题图)

B A

P

部分的面积为y(单位:2

cm),则y与x之间的函数关系用图象表示大致是下图中的()

第Ⅱ卷(非选择题共78分)

注意事项:

1.第Ⅱ卷共8页,用钢笔或圆珠笔直接答在试卷上.

2.答卷前将密封线内的项目及座号填写清楚.

二、填空题(本大题共5小题,每小题3分,共15分)把答案填在题中横线上.

15.分解因式:2

2

x xy xy

-+=_________________.

16.某制药厂两年前生产1吨某种药品的成本是100万元,随着生产技术的进步,现在生产1吨这种药品的成本为81万元,.则这种药品的成本的年平均下降率为______________.

17.若一个圆锥的底面积是侧面积的1

3

,则该圆锥侧面展开图的圆心角度数是____ _度.

18.如图,在菱形ABCD中,72

AD C

∠= ,AD的垂直平分线交对角线BD于点P,垂足为E,连接CP,则CPB

∠=________度.

19.如图,过原点的直线l与反比例函数

1

y

x

=-的图象交于M,N两点,根据图象猜想线

段MN的长的最小值是___________.

三、开动脑筋,你一定能做对!(本大题共3小题,共20分)20.(本小题满分6分)

解不等式组

3(21)2

102(1)3(1)

x

x x

---

?

?

-+-<-

?

,并把解集在数轴上表示出来.

O y (cm2)

x(s)

48 16

4 6 A.

O

y (cm2)

x(s)

48

16

4 6

B.

O

y (cm2)

x(s)

48

16

4 6

C.

O

y (cm2)

x(s)

48

16

4 6

D.

D

C

B

A

E

P

(第18题图)

O

y

x

M

N

l

(第19题图)

21.(本小题满分7分)

为了了解全校1800名学生对学校设置的体操、球类、跑步、踢毽子等课外体育活动项目的喜爱情况,在全校范围内随机抽取了若干名学生.对他们最喜爱的体育项目(每人只选一项)进行了问卷调查,将数据进行了统计并绘制成了如图所示的频数分布直方图和扇形统计图(均不完整).

(1) 在这次问卷调查中,一共抽查了多少名学生? (2) 补全频数分布直方图;

(3) 估计该校1800名学生中有多少人最喜爱球类活动?

22.(本小题满分7分)

如图,A ,B 是公路l (l 为东西走向)两旁的两个村庄,A 村到公路l 的距离AC =1km ,B 村到公路l 的距离BD =2km ,B 村在A 村的南偏东45

方向上.

(1)求出A ,B 两村之间的距离;

(2)为方便村民出行,计划在公路边新建一个公共汽车站P ,要求该站到两村的距离相等,请用尺规在图中作出点P 的位置(保留清晰的作图痕迹,并简要写明作法).

踢毽子 25% 球类 跑步 12.5%

体操

北 东 B A

C

D

(第22题图) l 体操 球类 踢毽子 跑步 其他 项目 人数 40 0

20 10 30 10 36 10

4

四、认真思考,你一定能成功!(本大题共2小题,共19分) 23.(本小题满分9分)

如图,AC 是O ⊙的直径,P A ,PB 是O ⊙的切线,A ,B 为切点,AB =6,P A =5. 求(1)O ⊙的半径; (2)sin BAC ∠的值.

24.(本小题满分10分)

在全市中学运动会800m 比赛中,甲乙两名运动员同时起跑,刚跑出200m 后,甲不慎摔倒,他又迅速地爬起来继续投入比赛,并取得了优异的成绩.图中分别表示甲、乙两名运动员所跑的路程y (m )与比赛时间x (s )之间的关系,根据图像解答下列问题: (1)甲摔倒前,________的速度快(填甲或乙); (2)甲再次投入比赛后,在距离终点多远处追上乙?

五、相信自己,加油啊!(本大题共2小题,共24分) 25.(本小题满分11分)

数学课上,张老师出示了问题:如图1,四边形ABCD 是正方形,点E 是边BC 的中点.90AEF ∠=

,且EF 交正方形外角DCG ∠的平行线CF 于点F ,求证:AE =EF . 经过思考,小明展示了一种正确的解题思路:取AB 的中点M ,连接ME ,则AM =EC ,易证AME ECF △≌△,所以AE EF =.

在此基础上,同学们作了进一步的研究:

(1)小颖提出:如图2,如果把“点E 是边BC 的中点”改为“点E 是边BC 上(除B ,

P O A B C (第23题图)

O y (m) x (s) 800 200 40 120 125 C D

A B

(第24题图) 甲 乙

P

C 外)的任意一点”,其它条件不变,那么结论“AE =EF ”仍然成立,你认为小颖的观点正确吗?如果正确,写出证明过程;如果不正确,请说明理由;

(2)小华提出:如图3,点E 是BC 的延长线上(除C 点外)的任意一点,其他条件不变,结论“AE =EF ”仍然成立.你认为小华的观点正确吗?如果正确,写出证明过程;如果不正确,请说明理由.

26.(本小题满分13分)

如图,抛物线经过(40)(10)(02)A B C -,

,,,,三点. (1)求出抛物线的解析式;

(2)P 是抛物线上一动点,过P 作PM x ⊥轴,垂足为M ,是否存在P 点,使得以A ,P ,M 为顶点的三角形与OAC △相似?若存在,请求出符合条件的点P 的坐标;若不存在,请说明理由; (3)在直线AC 上方的抛物线上有一点D ,使得DCA △的面积最大,求出点D 的坐标.

2009年临沂市中考数学试题

参考答案及评分标准

说明:第三、四、五大题给出了一种或两种解法,考生若用其它解法,应参照本评分标准给分.

一、选择题(每小题3分,共42分) 题号 1 2 3 4 5 6 7 8 9 10 11 12 13 14 答案

D

C

C

B

C

A

D

D

B

B

A

C

A

A

二、填空题(每小题3分,共15分)

15.2

(1)x y - 16.10% 17.120 18.72 19.22

O x y

A B C

4 1 2- (第26题图)

A D F C G E

B 图1 A D F

C G E B 图2 A

D F

C G E B 图3 (第25题图)

三、开动脑筋,你一定能做对!(共20分)

20.解:解不等式()3212x ---≥,得3x ≤. ······················································· (2分) 解不等式102(1)3(1)x x -+-<-,得1x >-. ························································· (4分) 所以原不等式组的解集为13x -<≤. ········································································ (5分) 把解集在数轴上表示出来为

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

21.解:(1)1012580.%÷=(人). 一共抽查了80人. ········································································································· (2分) (2)802520%?=(人), 图形补充正确. ·············································································································· (4分) (3)36180081080

?

=(人).

估计全校有810人最喜欢球类活动. ············································································ (7分)

22.解:(1)方法一:设A B 与CD 的交点为O ,根据题意可得45A B ∠=∠=°.

ACO ∴△和BDO △都是等腰直角三角形.································································ (1分) 2AO ∴=

,22BO =.

∴A B ,两村的距离为22232AB AO BO =+=+=(km )

. ························ (4分) 方法二:过点B 作直线l 的平行线交AC 的延长线于E .

易证四边形CDBE 是矩形, ·························································································· (1分) ∴2CE BD ==.

在Rt AEB △中,由45A ∠=°,可得3BE EA ==.

∴22

3332AB =

+=(km )

∴A B ,两村的距离为32km . ·················································································· (4分)

(2)作图正确,痕迹清晰. ·········································· (5分) 作法:①分别以点A B ,为圆心,以大于12

A B 的长为

半径作弧,两弧交于两点M N ,,

作直线MN ;

②直线MN 交l 于点P ,点P 即为所求. ····················· (7分) 四、认真思考,你一定能成功!(共19分) 23.解:(1)连接PO O B ,.设PO 交A B 于D . PA PB ,是O ⊙的切线.

∴90PAO PBO ∠=∠=°,

PA PB =,APO BPO ∠=∠.

1 0

2

3 1- P

B D

B

A C

D

第22题图

l

N M

O

P

∴3AD BD ==,PO AB ⊥. ·································· (2分) ∴22

534PD =

-=. ·

·············································· (3分) 在Rt PAD △和Rt POA △中,

tan AD AO APD PD

PA

==∠.

∴·35154

4

AD PA AO PD

?=

=

=

,即O ⊙的半径为154

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

(2)在Rt AOD △中,2

22

2

159344D O AO AD

??=

-=

-= ?

??

. ························ (7分) ∴9

3

4sin 155

4

O D

BAC AO ∠=

==. ·

·················································································· (9分) 24.解:(1)甲. ·········································································································· (3分) (2)设线段OD 的解析式为1y k x =.

把(125800),

代入1y k x =,得1325

k =.

∴线段OD 的解析式为325

y x =

(0125x ≤≤). ··················································· (5分)

设线段BC 的解析式为2y k x b =+.

把(40200),

,(120800),分别代入2y k x b =+. 得2220040800120k b k b =+??=+?,. 解得2152

100k b .

?

=

???=-?, ∴线段BC 的解析式为151002

y x =

-(40120x ≤≤). ········································ (7分)

解方程组325151002y x ,y x .?=????=-??得100011

640011x y .?

=????=??, ····································································· (9分)

6400240080011

11

-

=.

答:甲再次投入比赛后,在距离终点

2400m 11

处追上了乙. ···································· (10分)

五、相信自己,加油啊!(共24分)

25.解:(1)正确. ··························································· (1分) 证明:在A B 上取一点M ,使AM EC =,连接M E . (2分) BM BE ∴=.45BME ∴∠=°,135AME ∴∠=°.

A D

F M

CF 是外角平分线, 45DCF ∴∠=°, 135ECF ∴∠=°. AME ECF ∴∠=∠.

90AEB BAE ∠+∠= °,90AEB CEF ∠+∠=°,

∴BAE CEF ∠=∠.

AME BCF ∴△≌△(ASA )

. ···················································································· (5分) AE EF ∴=. ················································································································ (6分)

(2)正确. ··································································· (7分)

证明:在B A 的延长线上取一点N . 使AN CE =,连接NE . ············································ (8分) BN BE ∴=. 45N PCE ∴∠=∠=°. 四边形ABCD 是正方形, AD BE ∴∥.

D A

E BEA ∴∠=∠.

NAE CEF ∴∠=∠.

ANE ECF ∴△≌△(ASA )

. ··················································································· (10分) AE EF ∴=. ·············································································································· (11分)

26.解:(1) 该抛物线过点(02)C -,

,∴可设该抛物线的解析式为2

2y ax bx =+-. 将(40)A ,

,(10)B ,代入, 得1642020a b a b .+-=??+-=?,解得12

52a b .

?

=-????=??,

∴此抛物线的解析式为2

1522

2

y x x =-

+

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

(2)存在. ···················································································································· (4分)

如图,设P 点的横坐标为m , 则P 点的纵坐标为2

1522

2

m m -+

-,

当14m <<时, 4AM m =-,2

15222

PM m m =-

+

-.

又90COA PMA ∠=∠= °,

∴①当

21

A M A O P M

O C

=

=

时,

APM ACO △∽△,

即2

15

4222

2m m m ??

-=-

+

- ??

?

. A D F C G E B N O x

y

A B C

4

1

2-

(第26题图)

D P

M E

解得1224m m ==,(舍去),(21)P ∴,. ·································································· (6分) ②当

12

A M O C P M

O A

==时,APM CAO △∽△,即2

152(4)22

2

m m m -=-

+

-.

解得14m =,25m =(均不合题意,舍去)

∴当14m <<时,(21)P ,. ························································································ (7分)

类似地可求出当4m >时,(52)P -,. ······································································· (8分) 当1m <时,(314)P --,

. 综上所述,符合条件的点P 为(21),或(52)-,或(314)--,. ································ (9分)

(3)如图,设D 点的横坐标为(04)t t <<,则D 点的纵坐标为2

1522

2

t t -+

-.

过D 作y 轴的平行线交AC 于E . 由题意可求得直线AC 的解析式为122

y x =

-. ······················································ (10分)

E ∴点的坐标为122t t ??- ???

,.

2

215

112222

222D E t t t t t ??

∴=-+

---=-+ ???

. ··················································· (11分) 222

1

1244(2)422D AC S t t t t t ??∴=

?-+?=-+=--+ ???

△. ∴当2t =时,DAC △面积最大.

(21)D ∴,. ·················································································································· (13分)

北京一线教师家教 - 一线教师一对一辅导

北京一线教师家教 - 一线教师一对一辅导老师NO.1:本人性格开朗、乐观,做事耐心仔细,有责任心。初高中时成绩优异,理科生,由以数学、化学家教较好。中考全乡第一,高考数学140分,化学优等。

北京一线教师家教 - 一线教师一对一辅导老师NO.2:我的数学成绩非常好,高考数学成绩是143,在物理家教方面,我曾获得过全国物理竞赛河南赛区二等奖。我姐姐是教师,我曾在她办的假期补习班中当教员,并且到大学后也带过家教。同时我比较有耐心,讲题思路清晰。

北京一线教师家教- 一线教师一对一辅导老师NO.3:性格开朗,善于交流,有耐心。曾做过高中数理化家教和初中数学方面的辅导,有丰富的教学辅导经验,保证能很好的完成辅导任务。成绩优异(尤其在数理化方面),荣获三好学生、人民奖学金等,在社团中担任重要职务,有严谨的工作态度和责任心。

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