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2012湖北省咸宁市数学中考试题 2

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2012年湖北省咸宁市中考数学试题及答案

考生注意:1.本试卷分试题卷(共4页)和答题卷;全卷24小题,满分120分;考试时间

120分钟.

2.考生答题前,请将自己的学校、姓名、准考证号填写在试题卷和答题卷指定的位置,同时认真阅读答题卷上的注意事项.考生答题时,请按题号顺序在答题卷上各题目的答题区域内作答,写在试题卷上无效.

一、精心选一选(本大题共8小题,每小题3分,满分24分.每小题给出的4个选项中只

有一个符合题意,请在答题卷上将正确答案的代号涂黑) 1.8的相反数是( ).

A.8

B.8

C.1 8D.

1 82.南海是我国固有领海,它的面积超过东海、黄海、渤海面积的总和,约为360万平方千米,360万用科学记数法表示为( ). A.3.6×102 B.360×104 C.3.6×104 D.3.6×106 3.某班团支部统计了该班甲、乙、丙、丁四名同

甲 乙 丙 丁

学在5月份“书香校园”活动中的课外阅读时

x1111

间,他们平均每天课外阅读时间x与方差s2如

.2 .5 .5 .2

右表所示,你认为表现最好的是( ).

s0000

A.甲 B.乙

C.丙 D.丁

,x1≥04.不等式组的解集在数轴上表示为( ).

42x0.> 

0 0 1 2 1

A B

5.下列运算正确的是( ).

A.a3a2a6

C.(ab)2a2b2

2

0

1

2

0 1

2 y F E B C C D

B.(ab3)2a2b6 D.5a3a2

6.如图,正方形OABC与正方形ODEF是位似图形,O为位似中心, 相似比为1∶2,点A的坐标为(1,0),则E点的坐标为( ).

A.(2,0)

B.(

O A D (第6题) E O D x 33,) 22C.(2,2) D.(2,2)

7.如图,⊙O的外切正六边形ABCDEF的边长为2,则图中阴影部分 F

的面积为( ).

π2ππ2π

A.3 B.3 C.23 D.23 2323

C

8.电视台有一个非常受欢迎的娱乐节目:墙来了!选手需按墙上的空洞造型摆出相同

姿势,才能穿墙而过,否则会被墙推入水池.类似地,有一个几何体恰好无缝隙地以三个不同形状的“姿势”穿过“墙”上的三个空洞,则该几何体为( ).

A B (第7题)

墙 A B D

二、细心填一填(本大题共8小题,每小题3分,满分24分.请将答案填写在答题卷相应

15% 题号的位置) 9.因式分解:a22a .

球类

110.在函数y中,自变量x的取值范围是 .

x345% 田径 跳绳 其它

11.某校为了解学生喜爱的体育活动项目,随机抽查了100名学生, 10%

让每人选一项自已喜欢的项目,并制成如图所示的扇形统计图. (第11题)

B 如果该校有1200名学生,则喜爱跳绳的学生约有 人. 12.如图,某公园入口处原有三级台阶,每级台阶高

30 为18cm,深为30cm,为方便残疾人士,拟将台 18 阶改为斜坡,设台阶的起点为A,斜坡的起始点 C A (第12题) 为C,现设计斜坡BC的坡度i1:5,则AC的

长度是 cm.

13.某宾馆有单人间和双人间两种房间,入住3个单人间和6个双人间共需1020元,入住

1个单人间和5个双人间共需700元,则入住单人间和双人间各5个共需 元. 14.如图,量角器的直径与直角三角板ABC的斜边AB重合,其中量 角器0刻度线的端点N与点A重合,射线CP从CA处出发沿顺 B O 时针方向以每秒2度的速度旋转,CP与量角器的半圆弧交于点 P E E,第35秒时,点E在量角器上对应的读数是 度.

C A (N) 15.如图,在梯形ABCD中,AD∥BC,C90,BE平分∠ABC (第14题)

且交CD于E,E为CD的中点,EF∥BC交AB于F,EG∥AB

交BC于G,当AD2,BC12时,四边形BGEF的周长为 . 16.对于二次函数yx22mx3,有下列说法:

①它的图象与x轴有两个公共点; ②如果当x≤1时y随x的增大而减小,则m1;

F A D E

③如果将它的图象向左平移3个单位后过原点,则m1;

C G

④如果当x4时的函数值与x2008时的函数值相等,

(第15题)

则当x2012时的函数值为3.

其中正确的说法是 .(把你认为正确说法的序号都填上)

三、专心解一解(本大题共8小题,满分72分.请认真读题,冷静思考.解答题应写出文

字说明、证明过程或演算步骤,请将答案写在答题卷相应题号的位置) 17.(本题满分6分)

计算:|223|()218.

18.(本题满分8分)

12解方程:

x812. x2x4y A B O x

19.(本题满分8分)

如图,一次函数y1kxb的图象与反比例函数y2的图象交于A(1,6),B(a,2)两点. (1)求一次函数与反比例函数的解析式;

(2)直接写出y1≥y2时x的取值范围.

m(x0) x(第19题)

20.(本题满分9分)

某校举行以“助人为乐,乐在其中”为主题的演讲比赛,比赛设一个第一名,一个第二名,两个并列第三名.前四名中七、八年级各有一名同学,九年级有两名同学,小蒙同学认

为前两名是九年级同学的概率是

1,你赞成他的观点吗?请用列表法或画树形图法分2析说明.

21.(本题满分9分)

如图,AB是⊙O的直径,点E是AB上的一点,CD是过 E点的弦,过点B的切线交AC的延长线于点F,BF∥CD, 连接BC.

(1)已知AB18,BC6,求弦CD的长;

(2)连接BD,如果四边形BDCF为平行四边形,则点E位

于AB的什么位置?试说明理由.

A O C F

E D B

(第21题)

22.(本题满分10分)

某景区的旅游线路如图1所示,其中A为入口,B,C,D为风景点,E为三岔路的交汇点,图1中所给数据为相应两点间的路程(单位:km).甲游客以一定的速度沿线路“A→D→C→E→A”步行游览,在每个景点逗留的时间相同,当他回到A处时,共用去3h.甲步行的路程s(km)与游览时间t(h)之间的部分函数图象如图2所示. (1)求甲在每个景点逗留的时间,并补全图象;

s/(km) D 1 C 4 3 2.6 1.3 2 1.6 E B 1

(2)求C,E两点间的路程;

(3)乙游客与甲同时从

A处出发,打算游 完三个景点后回到 A处,两人相约先 到者在A处等候, 等候时间不超过10 分钟.如果乙的步 行速度为3km/h,在

每个景点逗留的时间与甲相同,他们的约定能否实现?请说明理由.

23.(本题满分10分)

如图1,矩形MNPQ中,点E,F,G,H分别在NP,PQ,QM,MN上,若1234,则称四边形EFGH为矩形MNPQ的反射四边形.图2,图3,图4中,四边形ABCD为矩形,且AB4,BC8. 理解与作图:

(1)在图2,图3中,点E,F分别在BC,CD边上,试利用正方形网格在图上作出

矩形ABCD的反射四边形EFGH.

计算与猜想:

(2)求图2,图3中反射四边形EFGH的周长,并猜想矩形ABCD的反射四边形的周

长是否为定值? 启发与证明:

(3)如图4,为了证明上述猜想,小华同学尝试延长GF交BC的延长线于M,试利用

小华同学给我们的启发证明(2)中的猜想.

M 3 H 4 E 图1

P

B

E 图2 A 3 4 B

E

图4 (第23题)

C

C B

图3

G 1 H

2

M

D F E

G 1 F 2 F

F C Q A D A D N

24.(本题满分12分)

如图,在平面直角坐标系中,点C的坐标为(0,4),动点A以每秒1个单位长的速度,从点O出发沿x轴的正方向运动,M是线段AC的中点.将线段AM以点A为中心,沿顺时针方向旋转90,得到线段AB.过点B作x轴的垂线,垂足为E,过点C作y轴的垂线,交直线BE于点D.运动时间为t秒. (1)当点B与点D重合时,求t的值; y y

(2)设△BCD的面积为S,当t为何值

C D C M O A B E x O 备用图

x 25时,S?

4

(3)连接MB,当MB∥OA时,如果抛

物线yax210ax的顶点在△ABM

(第24题)

内部(不包括边),求a的取值范围.

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湖北省咸宁市2012年初中毕业生学业考试

数学试题参及评分说明

说明:

1.如果考生的解答正确,思路与本参不同,可参照本评分说明制定相应的评分细则评分.

2.每题都要评阅完毕,不要因为考生的解答中出现错误而中断对该题的评阅.当考生的解答在某一步出现

错误,影响了后继部分时,如果该步以后的解答未改变这道题的内容和难度,则可视影响的程度决定后面部分的给分,但不得超过后面部分应给分数的一半;如果这一步以后的解答有较严重的错误,就不给分.

3.为阅卷方便,解答题的解题步骤写得较为详细,但允许考生在解答过程中,合理地省略非关键性的步骤. 4.解答右端所注分数,表示考生正确做到这一步应得的累加分数. 5.每题评分时只给整数分数.

一.精心选一选(每小题3分,本大题满分24分)

题号 答案

1 B 2 D 3 C 4 D 5 B 6 C

7 A

8 A

二.细心填一填(每小题3分,本大题满分24分) 9.a(a2) 10.x3 11.360 12.210 13.1100 14.140 15.28 16.①④(多填、少填或错填均不给分) 三.专心解一解(本大题满分72分) 17.解:原式322432 ··············································································· 4分

21. ·············································································································· 6分

(说明:第一步中写对322得1分,写对4得2分,写对32得1分,共4分)

x8118.解:原方程即:. ························································· 1分 x2(x2)(x2)方程两边同时乘以(x2)(x2),得 x(x2)(x2)(x2)8. ··············································································· 4分

化简,得 2x48. 解得 x2. ········································································································ 7分 检验:x2时(x2)(x2)0,x2不是原分式方程的解,原分式方程无解.

·········································································· 8分

19.解:(1)∵点A(1,6),B(a,2)在y2∴

m的图象上, xm······························································································· 1分 6,m6. ·

1mm·························································································· 2分 2,a3. ·

a2∵点A(1,6),B(3,2)在函数y1kxb的图象上,

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kb6,∴ ······································································································· 4分 3kb2.k2,解这个方程组,得

b8.∴一次函数的解析式为y12x8,反比例函数的解析式为y26. ··········· 6分 x(2)1≤x≤3. ···································································································· 8分 20.解:不赞成小蒙同学的观点. ··············································································· 1分

记七、八年级两名同学为A,B,九年级两名同学为C,D. 画树形图分析如下:

D 第一名: C B A

第二名: B C D A C D A B D A B C 第三名: CD BD BC CD AD AC BD AD AB BC AC AB

·········································································· 5分

由上图可知所有的结果有12种,它们出现的可能性相等,满足前两名是九年级同学的结果有2种,所以前两名是九年级同学的概率为21·························································· 9分 . ·12621.(1)解:∵BF与⊙O相切, A ∴BFAB. ·································································· 1分

而BF∥CD,∴CDAB. 又∵AB是直径,∴CEED. ······································ 2分 O 连接CO,设OEx,则BE9x. E C D F B (第21题) 由勾股定理可知:CO2OE2BC2BE2CE2,

即92x262(9x)2,x7. ································· 4分 因此CD2CO2OE22927282. ············· 5分

(2)∵四边形BDCF为平行四边形, ∴BFCD. 而CEED11··························································· 7分 CD, ∴CEBF.·22∵BF∥CD, ∴△AEC∽△ABF. ···································································· 8分

AEEC1···················································· 9分 . ∴点E是AB的中点. ·

ABBF21.622.(1)解法一:由图2可知甲步行的速度为···························· 1分 2(km/h) ·

0.8∴

因此甲在每个景点逗留的时间为

1.80.82.61.6············································································· 3分 0.5(h) ·

2解法二:甲沿A→D步行时s与t的函数关系式为s2t. ································ 1分 设甲沿D→C步行时s与t的函数关系式为s2tb. 则21.8b2.6. ∴b1.

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∴s2t1. ········································································································· 2分 当s1.6时,2t11.6,t1.3.

因此甲在每个景点逗留的时间为1.30.80.5(h). ······································· 3分 补全图象如下: ······································································································ 5分

s/(km) 4 3 2.6 2 1.6 1 O 0.8 1.8 2.3 (第22题)

(2)解法一:甲步行的总时间为30.522(h). ∴甲的总行程为224(km). ·············· 7分 ∴C,E两点间的路程为41.610.80.6(km).

·············································· 8分

解法二:设甲沿C→E→A步行时 s与t的函数关系式为s2tm.

3 t/(h) 则22.3m2.6.

∴m2.

∴s2t2. ········································································································· 6分 当t3时,s2324. ··············································································· 7分 ∴C,E两点间的路程为41.610.80.6(km).······································· 8分 (3)他们的约定能实现. 乙游览的最短线路为:A→D→C→E→B→E→A(或A→E→B→E→C→D→A),总行程为1.610.60.420.84.8(km). ······························································ 9分

∴乙游完三个景点后回到A处的总时间为4.80.533.1(h). 3∴乙比甲晚6分钟到A处. ················································································ 10分

(说明:图象的第四段由第二段平移得到,第五段与第一、三段平行,且右端点的横坐标为3,如果学生补全的图象可看出这些,但未标出2.3也可得2分.第3问学生只说能实现约定,但未说理由不给分.) 23.(1)作图如下: ···································································································· 2分

A H B G D F F E 图2

C B

图3

E

C A H G D (2)解:在图2中,EFFGGHHE22422025,

∴四边形EFGH的周长为85. ··········································································· 3分 在图3中,EFGH22125,FGHE32624535. ∴四边形EFGH的周长为2523585. ············································ 4分 猜想:矩形ABCD的反射四边形的周长为定值. ················································ 5分 (3)证法一:延长GH交CB的延长线于点N.

∵12,15, G D A ∴25. 1 F 3 2 5 H 21世纪教育网4 -- 中国最大型、最专业的中小学教育资源门户网站。 版权所有@21世纪教育网 M K C B N E

图4

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而FCFC,

∴Rt△FCE≌Rt△FCM. ∴EFMF,ECMC.

······························· 6分

同理:NHEH,NBEB.

∴MN2BC16. ······························································································ 7分 ∵M905901,N903, ∴MN. ∴GMGN. ····································································· 8分 过点G作GK⊥BC于K,则KM1················································· 9分 MN8. ·

2∴GMGK2KM2428245.

∴四边形EFGH的周长为2GM85. ··························································· 10分 证法二:∵12,15, ∴25. 而FCFC, ∴Rt△FCE≌Rt△FCM. ∴EFMF,ECMC. ··················································································· 6分 ∵M905901,HEB904,

而14, ∴MHEB. ∴HE∥GF. 同理:GH∥EF. ∴四边形EFGH是平行四边形. ·········································································· 7分 ∴FGHE. 而14, ∴Rt△FDG≌Rt△HBE. ∴DGBE. ····················································· 8分 过点G作GK⊥BC于K,则KMKCCMGDCMBEEC8. ······ 9分 ∴GMGK2KM2428245. ∴四边形EFGH的周长为2GM85. ··························································· 10分 24.解:(1)∵CAOBAE90,ABEBAE90, ∴CAOABE. ∴Rt△CAO∽Rt△ABE. ······················································································· 2分

CAAO. ABBE2ABt∴························································································ 3分 .∴t8. ·AB41(2)由Rt△CAO∽Rt△ABE可知:BEt,AE2. ································· 4分

211t25当0<t<8时,SCDBD(2t)(4).

2224∴

∴t1t23. ········································································································· 6分 当t>8时,SCDBD121t25. (2t)(4)224∴t1352,t2352(为负数,舍去). 当t3或352时,S25. ··········································································· 8分 421世纪教育网 -- 中国最大型、最专业的中小学教育资源门户网站。 版权所有@21世纪教育网

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(3)过M作MN⊥x轴于N,则MNCO2.

y C x=5 D B 12当MB∥OA时,BEMN2,OA2BE4. ········· 9分 抛物线yax210ax的顶点坐标为(5,25a). ·· 10分 它的顶点在直线x5上移动.

直线x5交MB于点(5,2),交AB于点(5,1).11分 ∴1<25a<2. ∴ O A (第24题) E x 21<a<. ··················································· 12分 2525

试题平稳 稳中求新

点评人:温中数学老师 石 娟

总体来说,今年中考数学试卷试题整体坡度平缓,依标靠本,基础性强,大部分题目都立足于考查初中数学的核心基础知识、基本技能及隐含于其中的基本数学思想方法,同时注意结合现实背景,体现对数学本质理解的考查。

与去年相比,题型、题量、题目的赋分比较平稳,题型没有变化,填空题、选择题各有一道创新题,呈现出稳中求新的特点。其中,选择题第8题和填空题14题是本套试题最大的亮点,有新意。第8题以学生非常喜欢的一档电视娱乐节目为题干,考查学生视图的基本能力。 14题考查圆的相关知识,包含的知识点丰富,有圆周角、圆心角等,考生需要认真读懂题意,理清头绪,才能准确作答。试题入手容易,细做难,且难度有所分解,“三基”考查到位,基本的活动经验有体现,对优等生来说,做起来顺手,中等生要将试题完整地解答出来,有一定的难度。

此外,试卷的信度和区分度较高,试题的综合能力强,考查学生的数学解读能力,知识掌握能力和知识迁移能力,需要学生将课内知识与课外知识有效结合,发现规律后懂得拓展运用。

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