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本溪09年中考数学

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2009年本溪市初中毕业生学业考试

数学试卷

考试时间120分钟试卷满分150分

一、选择题（下列各题的备选答案中，只有一个是正确的，请将正确答案的选项填在下表中相应题号下的空格内，每小题3分，共24分）

1．2009年6月，全国参加高等院校统一招生考试的学生约10 200 000人，其中10 200 000用科学记数法表示应为（） A ．6

10.210?

B ．8

1.0210?

C ．8

0.10210?

D ．9

1.0210?

2．如果a 与1互为相反数，则|2|a +等于（） A ．2

B ．2-

C ．1

D ．1-

3．反比例函数(0)k

y k x

≠的图象经过点(23)-，，则该反比例函数图象在（） A ．第一、三象限 B ．第二、四象限 C ．第二、三象限 D ．第一、二象限

4．有一个铁制零件（正方体中间挖去一个圆柱形孔）如图放置，它的左视图是（）

5．小新抛一枚质地均匀的硬币，连续抛三次，硬币落地均正面朝上，如果他第四次抛硬币，那么硬币正面朝上的概率为（） A ．

B ．

C ．1

D ．

6．下列图案中，既是轴对称图形又是中心对称图形的是（）

7．某男子排球队20名队员的身高如下表：

身

高（cm ）

180

186

188 1

92 2

人

A ．

B ．

C ．

D ． A ． B ． C ． D ．

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数（个）

则此男子排球队20名队员的身高的众数和中位数分别是（） A ．186cm ，186cm B ．186cm ，187cm C ．208cm ，188cm D ．188cm ，187cm 8．估算171+的值在（）

A ．2和3之间

B ．3和4之间

C ．4和5之间

D ．5和6之间二、填空题（每小题3分，共24分） 9．函数1

y x =

-中自变量x 的取值范围是． 10．分解因式：2

9xy x -= ．

11．由于甲型H1N1流感（起初叫猪流感）的影响，在一个月内猪肉价格两次大幅下降．由原来每斤16元下调到每斤9元，求平均每次下调的百分率是多少？设平均每次下调的百分率为x ，则根据题意可列方程为．

12．如图所示，在ABCD

中，对角线AC BD 、相交于点O ，过点O 的直线分别交AD BC 、于点M N 、，若CON △的面积为2，DOM △的面积为4，则AOB △的面积为．

13．如图所示，抛物线2

y ax bx c =++（0a ≠）与x 轴的两个交点分别为(1

0)A -，和(20)B ，，当0y <时，x 的取值范围是．

14．如图所示，菱形ABCD 中，对角线AC BD 、相交于点O ，H 为AD 边中点，菱形ABCD 的周长为24，则OH 的长等于． 15．圆锥的高为4cm ，底面圆直径长6cm ，则该圆锥的侧面积等于 2

cm （结果保留π）．

16．如图所示，已知：点(00)A ，，(30)B ，，(01)C ，

在ABC △内

依次作等边三角形，使一边在x 轴上，另一个顶点在BC 边上，

作出的等边三角形分别是第1个11AA B △，第2个122B A B △，

第3个233B A B △，…，则第n 个等边三角形的边长等

于．

M D

N B A 12题图 1 1- 2 y x B A

O 13题图

A H C C O 14题图 O y

(A )

A 1

1 1

2 B A 2

A 3

B 3 B 2 B 1 16题图

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三、解答题（每题8分，共16分） 17．先化简，再求值：2113y x

x y x

??--÷

???，其中23x y ==，．

18．如图所示，正方形网格中，ABC △为格点三角形（即三角形的顶点都在格点上）．（1）把ABC △沿BA 方向平移后，点A 移到点1A ，在网格中画出平移后得到的11A B C 1△；

（2）把11A B C 1△绕点1A 按逆时针方向旋转

90°，在网格中画出旋转后的22A B C 1△；（3）如果网格中小正方形的边长为1，求点B 经过（1）、（2）变换的路径总长．

四、解答题（每题10分，共20分）

19．“五·一”期间，九年一班同学从学校出发，去距学校6千米的本溪水洞游玩，同学们分为步行和骑自行车两组，在去水洞的全过程中，骑自行车的同学比步行的同学少用40分钟，已知骑自行车的速度是步行速度的3倍．

（1）求步行同学每分钟．．．

走多少千米？（2）右图是两组同学前往水洞时的路程y （千米）与时间x （分钟）的函数图象．完成下列填空：

①表示骑车同学的函数图象是线段； ②已知A 点坐标(300)，，则B 点的坐标为（）．

20．甲、乙两同学只有一张乒乓球比赛的门票，谁都想去，最后商定通过转盘游戏决定．游戏规则是：转动下面

B C A 1 M N A

B x y （千米）（分钟） 6

4 3

1 10 20 30 O

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平均分成三个扇形且标有不同颜色的转盘，转盘连续转动两次，若指针前后所指颜色相同，则甲去；否则乙去．（如果指针恰好停在分割线上，那么重转一次，直到指针指向一种颜色为止）

（1）转盘连续转动两次，指针所指颜色共有几种情况？通过画树状图或列表法加以说明；（2）你认为这个游戏公平吗？请说明理由．

五、解答题（每题10分，共20分） 21．初中生对待学习的态度一直是教育工作者关注的问题之一．为此某市教育局对该市部分学校的八年级学生对待学习的态度进行了一次抽样调查（把学习态度分为三个层级，A 级：对学习很感兴趣；B 级：对学习较感兴趣；C 级：对学习不感兴趣），并将调查结果绘制成图①和图②的统计图（不完整）．请根据图中提供的信息，解答下列问题：

（1）此次抽样调查中，共调查了名学生；（2）将图①补充完整；

（3）求出图②中C 级所占的圆心角的度数；

（4）根据抽样调查结果，请你估计该市近20000名初中生中大约有多少名学生学习态度达标（达标包括A 级和B 级）？

22．如图所示，AB 是O ⊙直径，OD ⊥弦BC 于点F ，且交O ⊙于点E ，若AEC ODB ∠=∠．（1）判断直线BD 和O ⊙的位置关系，并给出证明；

（2）当108AB BC ==，时，求BD 的长．

六、解答题（每题10分，共20分）

23．为奖励在演讲比赛中获奖的同学，班主任派学习委员小明为获奖同学买奖品，要求每人一件．小明到文具店看了商品后，决定奖品在钢笔和笔记本中选择．如果买4个笔记本和2支钢笔，则需86元；如果买3个笔记本

红

黄蓝人数

120 100

50 50 120

Ａ级 B 级 C 级学习态度层级

图①

图② 25% Ａ级 B 级 C 级 60% D B O

A C

E F

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和1支钢笔，则需57元．

（1）求购买每个笔记本和钢笔分别为多少元？

（2）售货员提示，买钢笔有优惠，具体方法是：如果买钢笔超过10支，那么超出部分可以享受8折优惠，若买

(0)x x >支钢笔需要花y 元，请你求出y 与x 的函数关系式；

（3）在（2）的条件下，小明决定买同一种奖品，数量超过10个，请帮小明判断买哪种奖品省钱．

24．如图所示，山坡上有一棵与水平面垂直的大树，一场台风过后，大树被刮倾斜后折断倒在山坡上，树的顶部恰好接触到坡面．已知山坡的坡角23AEF ∠=°，量得树干倾斜角38BAC ∠=°，大树被折断部分和坡面所成

的角604m ADC AD ∠==°

，．（1）求CAE ∠的度数；

（2）求这棵大树折断前的高度？

（结果精确到个位，参考数据：2 1.4=，3 1.7=，6 2.4=）．

七、解答题（本题12分）

25．在ABC △中，AB AC =，点D 是直线BC 上一点（不与B C 、重合），以AD 为一边在AD 的右侧．．

作ADE △，使AD AE DAE BAC =∠=∠，，连接CE ．

（1）如图1，当点D 在线段BC 上，如果90BAC ∠=°，则BCE ∠= 度；

（2）设BAC α∠=，BCE β∠=．

①如图2，当点D 在线段BC 上移动，则αβ，之间有怎样的数量关系？请说明理由； ②当点D 在直线BC 上移动，则αβ，之间有怎样的数量关系？请直接写出你的结论．

C 60° 38°

B D

E 23° A

F A

C C

D D B

图1 图2 A

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八、解答题（本题14分）

26．如图所示，在平面直角坐标系中，抛物线2

y ax bx c =++（0a ≠）经过(1

0)A -，，(30)B ，，(03)C ，三点，其顶点为D ，连接BD ，点P 是线段BD 上一个动点（不与B D 、重合），过点P 作y 轴的垂线，垂足为E ，连接BE ．

（1）求抛物线的解析式，并写出顶点D 的坐标；

（2）如果P 点的坐标为()x y ，，PBE △的面积为s ，求s 与x 的函数关系式，写出自变量x 的取值范围，并求出s 的最大值；

（3）在（2）的条件下，当s 取得最大值时，过点P 作x 的垂线，垂足为F ，连接EF ，把PEF △沿直线EF 折叠，点P 的对应点为P '，请直接写出P '点坐标，并判断点P '是否在该抛物线上．

2- 3- 1 2 3 3 1 D

C B A

2 E

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2009年本溪市初中毕业生学业考试数学试题参考答案及评分标准

注：本参考答案只给出一种或两种解法（证法），若用其它方法解答正确，可参考此评分标准相应步骤赋分．

一、选择题（每小题3分，共24分）

号

458

案

C A D

二、填空题（每小题3分，共24分）

9．1x > 10．(3)(3)x y y +- 11．2

16(1)9x -= 12．6

13．1x <-或2x > 14．3 15．15π 16．32n

三、解答题（每题8分，共16分） 17．解：2113y x x y x

??--÷

??? 23y x y x

x y x

--=

÷ ·········································································································· 2分 23y x x

x y y x

-=-

············································································································ 4分

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3x y

··························································································································· 6分当23x y ==，时，原式32

=． ··············································································· 8分 18．（1）画图正确． ············································ 2分（2）画图正确． ···················································· 5分（3）2212222BB =

+= ······························· 6分

弧12B B 的长90π22π1802

． ·························· 7分点B 所走的路径总长2π

222

． ················· 8分四、（每题10分，共20分）

19．（1）解：设步行同学每分钟走x 千米，则骑自行车同学每分钟走3x 千米． ··········· 1分根据题意，得：

403x x

=+ ································································································ 3分 110

x =

···································································································································· 4分经检验，1

x =是原方程的解． ·························································································· 5分

答：步行同学每分钟走1

千米． ························································································· 6分

（2）①AM ·························································································································· 8分 ②(500)，． ···························································································································· 10分 20．（1）解法一：（列表法）

红黄蓝红

红

、红

红

、黄红

、蓝黄

黄

、红黄、黄

黄

、蓝蓝

蓝

A B

A 1

B 1

C 1 B 2 C 2

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、红、黄、蓝

由上表可知，总共有9种情况． ··························································································· 5分解法二：（树状图）

由上图可知，总共有9种情况． ··························································································· 5分（2）不公平． ······················································································································· 6分

理由：由（1）可知，总共有9种不同的情况，它们出现的可能性相同，其中颜色相同的有3种，所以P （甲去）

13=

，P （乙去）23=． ······································································································ 8分 1233

≠ ， ······························································································································ 9分 ∴这个游戏不公平． ············································································································· 10分五、（每题10分，共20分）

21．（1）200； ····················································································································· 2分（2）2001205030--=（人）． ····················································································· 3分画图正确． ····························································································································· 4分

（3）C 所占圆心角度数360(125%60%)54=?--=°°． ·············································· 7分（4）20000(25%60%)17000?+=．·············································································· 9分 ∴估计该市初中生中大约有17000名学生学习态度达标． ················································ 10分 22．（1）直线BD 和O ⊙相切． ························································································· 1分证明：

∵AEC ODB ∠=∠，AEC ABC ∠=∠， ∴ABC ODB ∠=∠． ······················································· 2分 ∵OD ⊥BC ，

∴90DBC ODB ∠+∠=°． ············································ 3分 ∴90DBC ABC ∠+∠=°．

即90DBO ∠=°． ····························································· 4分

∴直线BD 和O ⊙相切． ·················································· 5分（2）连接AC ．

红黄蓝

红红黄蓝

黄红黄蓝

蓝人数 120 100 50

120

A 级

B 级学习态度层级

C 级

D B

C E F

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∵AB 是直径，

∴90ACB ∠=°． ····························································· 6分在Rt ABC △中，108AB BC ==，， ∴226AC AB BC =

-=．

∵直径10AB =， ∴5OB =． ··························································································································· 7分由（1），BD 和O ⊙相切， ∴90OBD ∠=°． ················································································································· 8分 ∴90ACB OBD ∠=∠=°．由（1）得ABC ODB ∠=∠， ∴ABC ODB △∽△． ········································································································ 9分

∴

AC BC

OB BD =． ∴685BD =，解得203

BD =． ···························································································· 10分六、（每题10分，共20分）

23．（1）解：设每个笔记本x 元，每支钢笔y 元． ·························································· 1分

4286357.x y x y +=??

+=?

，

······················································································································ 2分解得1415.

x y =??

=?，

答：每个笔记本14元，每支钢笔15元． ············································································ 5分

（2）15(010)1230(10)x

x y x x ?

≤

（3）当141230x x <+时，15x <；

当141230x x =+时，15x =；当141230x x >+时，15x >． ··························································································· 8分综上，当买超过10件但少于15件商品时，买笔记本省钱；当买15件奖品时，买笔记本和钢笔一样；当买奖品超过15件时，买钢笔省钱． ················································································ 10分 24．解：（1）延长BA 交EF 于点G ．在Rt AGE △中，23E ∠=°， ∴67GAE ∠=°． ····························································· 2分

又∵38BAC ∠=°，

∴180673875CAE ∠=--=°°°°． ····························· 3分（2）过点A 作AH CD ⊥，垂足为H ． ······················· 4分

在ADH △中，604ADC AD ∠==°

，， ············································································································· 3分 ············································································································· 4分

····································································· 6分

····································································· 7分

38°

B H

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cos DH

ADC AD ∠=

，∴2DH =． ·································· 5分 sin AH

ADC AD

∠=，∴23AH =． ······························ 6分

在Rt ACH △中，180756045C ∠=--=°°°°， ······· 7分 ∴26AC =，23CH AH ==． ······························ 8分

∴2623210AB AC CD =+=++≈（米）． ··························································· 9分答：这棵大树折断前高约10米． ························································································ 10分七、（12分） 25．（1）90°． ···················································································································· 2分（2）①180αβ+=°． ········································································································ 3分理由：∵BAC DAE ∠=∠，

∴BAC DAC DAE DAC ∠-∠=∠-∠．即BAD CAE ∠=∠．

又AB AC AD AE ==，， ∴ABD ACE △≌△． ········································································································ 6分 ∴B ACE ∠=∠．

∴B ACB ACE ACB ∠+∠=∠+∠．

∴B ACB β∠+∠=． ·········································································································· 7分 ∵180B ACB α+∠+∠=°，

∴180αβ+=°． ················································································································· 8分 ②当点D 在射线BC 上时，180αβ+=°． ······································································ 10分当点D 在射线BC 的反向延长线上时，αβ=． ······························································ 12分八、（14分）

26．解：（1）设(1)(3)y a x x =+-， ················································································ 1分

把(03)C ，代入，得1a =-， ································································································ 2分

∴抛物线的解析式为：2

23y x x =-++． ········································································· 4分

顶点D 的坐标为(14)，． ········································································································ 5分

（2）设直线BD 解析式为：y kx b =+（0k ≠），把B D 、两点坐标代入，

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得304.k b k b +=??+=?

， ······················································································································· 6分

解得26k b =-=，．

∴直线AD 解析式为26y x =-+． ····················································································· 7分

2111(26)3222

s PE OE xy x x x x =

==-+=-+ ， ··························································· 8分 ∴23(13)s x x x =-+<< ····································································································· 9分

29939

34424

s x x x ???

?=--++

=--+ ? ?

?． ·································································· 10分 ∴当32x =

时，s 取得最大值，最大值为94

．···································································· 11分（3）当s 取得最大值，32x =

，3y =，∴332P ??

???

，． ····················································· 5分 ∴四边形PEOF 是矩形．

作点P 关于直线EF 的对称点P '，连接P E P F ''、．法一：过P '作P H y '⊥轴于H ，P F '交y 轴于点M ．设MC m =，则332

MF m P M m P E ''==-=，，．在Rt P MC '△中，由勾股定理，

3(3)2m m ??+-= ?

．解得15

m =．

∵CM P H P M P E '''= ，

∴9

P H '=．

由EHP EP M ''△∽△，可得EH EP EP EM '='，6

EH =． ∴69

355

OH =-

=． ∴P '坐标99105??

???

，． ·

········································································································ 13分法二：连接PP '，交CF 于点H ，分别过点H P '、作PC 的垂线，垂足为M N 、．

易证CMH HMP △∽△．

(E ) 1-

1- 2- 3- 1 2 3

C B

2 x

F P '

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∴

CM MH MH PM ==．设CM k =，则24MH k PM k ==，．

∴352PC k ==，3

k =．

由三角形中位线定理，

1268455

PN k P N k '==

==，． ∴12395210CN PN PC =-=

-=，即910x =-． 69

355y PF P N '=-=-=

∴P '坐标99105??

???，． ·

········································································································ 13分把P '坐标99105??

- ???

，代入抛物线解析式，不成立，所以P '不在抛物线上． ·

··················· 14分

(E ) 1-

1- 2- 3- 1 2 3

3 1 D

C B A P 2 x

O F P '

H N