2007,2008自贡数学中考(含答案)

07年的自贡中考题

绝密★启用前 [考试时间：2007年6月12日 下上午9∶00—11∶00]

四川省自贡市2007年初中毕业暨升学考试

数 学 试 卷

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2页，第Ⅱ卷3至12页．满分120分．考试时间120分钟．考试结束后，将试卷Ⅰ、试卷Ⅱ和答题卡一并交回．装订时将试卷Ⅱ单独装订．

第Ⅰ卷(选择题 共33分)

注意事项：

1．答第Ⅰ卷前，考生务必将自己的姓名、准考证号，考试科目涂写在答题卡上

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，用橡皮擦擦干净后，再选涂其它答案标号，不能答在试卷Ⅰ上．

一、选择题：本大题共11小题，每小题3分，共33分．在每小题给出的四个选项中，只有一个选项是符合题目要求的．

1．下列各式中，p，q互为相反数的是（ ） A．pq＝1 C．p＋q=0

相信自己一定成功！

B．pq＝－1 D．p－q＝0

2．下列计算正确的是（ ） A．C．

18xx2y

18y

18(x y) 12y

B．D．

yx

1

yz

2yxz1

x 12y

x y

y x

0

3．a是实数，且x＞y，则下列不等式中，正确的是（ ） A．ax＞ay

B. a2x≤a2y

C．a2x＞a2y

D. a2x≥a2y

4．矩形、菱形、正方形都具有的性质是（ ） A．每一条对角线平分一组对角 C．对角线互相平分

B．对角线相等 D．对角线互相垂直

5．用配方法解关于x的方程x2＋mx＋n＝0，此方程可变形为（ ）

1 页 共 10 页

07年的自贡中考题

A．(x C．

m2

)

2

2

4n m

4 m

2

2

B．(x D．(x

m2m2

)

2

m

2

4n4

2

(x

m2

)

4n2

)

2

4n m

2

6．进入夏季后，某电器商场为减少库存，对电热取暖器连续进行两次降价．若设平均每次降价的百分率是x，降价后的价格为y元，原价为a元，则y与x之间的函数关系式为（ ）

A．y＝2a(x－1)

B．y＝2a(1－x)

C．y＝a(1－x)

2

D．y＝a(1－x)

2

7．若等腰三角形一腰上的高和另一腰的夹角为25°，则该三角形的一个底角为（ ） A．32.5°

B．57.5°

C．65°或57.5° D．32.5°或57.5°

8．随机抛掷一枚均匀的硬币两次，则出现两面不一样的概率是（ ） A．

14

B．

12

C．

34

D．1

9．两圆的半径分别为7和1，圆心距为10，则其内公切线长和外公切线长分别为（ ） A．6，8

B．6，10

C．8，2

D．8，6

10．我市某风景区，在“五一“长假期间，接待游人情况如下图所示，则这七天游览该风景区的平均人数为（ ）

A．2800人

B．3000人

C．3200人

D．3500人

11．小洋用彩色纸制做了一个圆锥型的生日帽，其底面半径为6cm，母线长为12cm，不考虑接缝，这个生日帽的侧面积为（ ）

A．36πcm2

B．72πcm2

C．100πcm2

D．144πcm2

2 页 共 10 页

07年的自贡中考题

绝密★启用前 [考试时间：2007年6月12日 下上午9∶00—11∶00]

四川省自贡市2007年初中毕业暨升学考试

数 学 试 卷

第Ⅱ卷(非选择题 共87分)

注意事项：1．第Ⅱ卷共10页（3至12页），用钢笔或蓝色圆珠笔将答案直接答在试题卷上．

2．答题前请将密封线内的项目填写清楚． 3．监考人员将第Ⅱ卷密封装订．

二、填空题：本大题共5小题，每小题4分，共20分

12、一生物教师在显微镜下发现，某种植物的细胞直径约为0.00012mm，用科学记数法表示这个数为____________mm．

13．请写出一个值k＝___________，使一元二次方程x2－7x＋k＝0 有两个不相等的非0实数根．（答案不唯一）

14．有4条长度分别为1，3，5，7的线段，现从中任取三条能构成三角形的概率是__________．

15．如图是中国共产主义青年团团旗上的图案（图案本身没有字母），5个角的顶点A，B，C，D，E把外面的圆5等分，则∠A＋∠B＋∠C＋∠D＋∠E＝__________________．

16．一个叫巴尔末的中学教师成功地从光谱数据

95

你可要小心点哦！

，

1612

，

2521

，

3632

， 中得到巴尔末公式，从而打开了

光谱奥秘的大门，请你按照这种规律，写出第n（n≥1）个数据是___________．

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07年的自贡中考题

三、解答题：本大题共4个小题，每小题6分，共24分．

17．解方程组：

18．解方程：

1

19

．计算：

3

2

2x y 4 0 3x y 6 0

① ②

1x 2

1x

2

( 1)

01001

·tan30°

20．学校举行百科知识抢答赛，共有20道题，规定每答对一题记10分，答错或放弃记－4分．九年级一班代表队的得分目标为不低于88分．问这个队至少要答对多少道题才能达到目标要求？

四、解答题：本大题共3个小题，每小题7分，共21分．

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07年的自贡中考题

21．按规定尺寸作出下面图形的三视图．

22．如图所示，我市某中学数学课外活动小组的同学，利用所学知识去测量沱江流经我市某段的河宽． 小凡同学在点A处观测到对岸C点，测得∠CAD＝45°，又在距A处60米远的B处测得∠CBA＝30°，请你根据这些数据算出河宽是多少？（精确到0.01m）

23．某商店按图（Ⅰ）给出的比例，从甲、乙、丙三个厂家共购回饮水机150台，商店质检员对购进的这批饮水机进行检测，并绘制了如图所示的统计图（Ⅱ）．请根据图中提供的信息回答下列问题．

（Ⅰ） （Ⅱ）

（1）求该商店从乙厂购买的饮水机台数？ （2）求所购买的饮水机中，非优等品的台数？

（3）从优等品的角度考虑，哪个工厂的产品质量较好些？为什么？

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07年的自贡中考题

五、解答题：本大题共2个小题，每小题7分，共14分．

24．如图，AB是⊙O的直径，AE平分∠BAC交⊙O于点E，过E作⊙O的切线ME交AC于点D．试判断△AED的形状，并说明理由．

25．已知：三角形ABC中，∠A＝90°，AB＝AC，D为BC的中点， （1）如图，E，F分别是AB，AC上的点，且BE＝AF，

求证：△DEF为等腰直角三角形．

（2）若E，F分别为AB，CA延长线上的点，仍有BE＝AF，其他条件不变，那么，△DEF是否仍为等

腰直角三角形？证明你的结论．

六、解答题：本大题8分．

26．△ABC中，∠A，∠B，∠C的对边分别为a，b，c，抛物线y＝x－2ax＋b交x轴于两点M，N，交y轴于点P，其中M的坐标是(a＋c，0)．

（1）求证：△ABC是直角三角形．

（2）若S△MNP＝3S△NOP，①求cosC的值；②判断△ABC的三边长能否取一组适当的值，使三角形MND（D为抛物线的顶点）是等腰直角三角形？如能，请求出这组值；如不能，请说明理由．

6 页 共 10 页

2

2

07年的自贡中考题

四川省自贡市2007年初中毕业暨升学考试

数学参考答案及评分标准

说明：

一．如果考生的解法与下面提供的参考解法不同，只要正确一律给满分，若某一步出现错误，可参照该题的评分意见进行评分．

二．评阅试卷时，不要因解答中出现错误而中断对该题的评阅，当解答中某一步出现错误，影响了后继部分，但该步以后的解答未改变这一道题的内容和难度，后来发生第二次错误前，出现错误的那一步不给分，后面部分只给应给分数之半；明显笔误，可酌情少扣；如有严重概念性错误，则不给分；在同一解答中，对发生第二次错误起的部分不给分．

三．涉及计算过程，允许合理省略非关键性步骤．

四．在几何题中，考生若使用符号“ ”进行推理，其每一步应得分数，可参照该题的评分意见进行评分． 一．选择题：本大题共11个小题，每小题3分，共33分． 1．C 2．D 3．D 4．C 5．B 6．D 7．D 8．B 9．A 10．B 11．B

二．填空题：（每小题4分，共计20分）

12．1.2×10－4 13．10（答案不唯一） 14．

(n 2)

2

14

15．180° 16．

n(n 4)

或

(n 2)

2

2

（只填一个均可）

(n 2) 4

三．解答题：（每小题6分，共计24分） 17．解：由①＋②得 5x＝10······················································································ 2分 x＝2 ························································································· 3分 将x＝2代入①得 y＝0 ······························································································ 5分 ∴原方程组的解为

x 2 y 0

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

18．解：x＋(x＋2)＝2x(x＋2) ························································································ 2分 整理得：x2＋x－1＝0 ···································································································· 3分 ∴x＝

1

2

5

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

5

经检验x＝

1

2

均为原方程的解 ··············································································· 5发

1

2

5

∴原方程的解为x＝ ························································································· 6分

33

19．解：原式＝9＋1－1＋(23－33)·＝9＋(－3)·

33

····························································· 2.5分

····································································································· 4.5分

＝9－1 ·························································································································· 5分 ＝8 ······························································································································· 6分 20．解：设九年级一班代表队至少要答对x道题才能达到目标要求． ···························· 1分 由题意得：10x－4(20－x)≥88 ······················································································· 4分 10x－80＋4x≥88 ·················································································································· 14x≥168

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07年的自贡中考题

x≥12 ···························································································································· 5分 答：这个队至少要答对12道题才能达到目标要求． ······················································ 6分 四．解答题：（每小题7分，共计21分） 21．解：

主视图 左视图

俯视图

（三个视图各2分，位置正确给1分，共7分．) 22．解：如图，过C作CE⊥AB于E···················· 1分 则CE为河宽

设CE＝x（米），于是BE＝x＋60（米） ············· 2分 在Rt△BCE中 tan30°＝

CEEB

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

∴3x＝x＋60 ·············································································································· 4分 ∴x＝30(3＋1) ·········································································································· 5分 ≈81.96（米） ············································································································· 6分 答：河宽约为81.96米． ······························································································· 7分 23．解：（1）150×40%＝60（台） ··············································································· 2分 ∴设商店从乙厂购买的饮水机台数为60台 （2）由图（II）知优等品的台数为 50＋51＋26＝127（台）

∴非优等品的台数为150－127＝23（台）····································································· 4分 （3）由题意知： 甲厂的优等品率为乙厂的优等品率为丙厂的优等品率为又

2630

50150 40%

51150 40%

26150 20%

506051602630

················································································ 4.5人 ·················································································· 5分 ················································································ 5.5分

＞

5160

＞

5060

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

∴丙厂的产品质量较好． ······························································································ 7分 五．解答题：（每小题7分，共计14分）

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07年的自贡中考题

24．解△AED为直角三角形 ····································· 1分 理由：连结BE ·························································· 2分 ∵AB是直径 ∴∠BEA＝90°························································· 3分 ∴∠B＋∠BAE＝90°················································ 4分 又∵AE平分∠BAC

∴∠BAE＝∠EAD ·················································· 4.5分

∵ME切 O于点E ∴∠AED＝∠B·············································································································· 5分 ∴∠AED＋∠EAD＝90° ······························································································ 6分 ∴△AED是直角三角形 ································································································ 7分 25．证明：①连结AD ································································································ 0.5分 ∵AB AC ∠BAC＝90° D为BC的中点 ∴AD⊥BC BD＝AD ·············································· 1分 ∴∠B＝∠DAC＝45° ···············································1.5分 又BE＝AF

∴△BDE≌△ADF （S.A.S） ···································· 2分

∴ED＝FD ∠BDE＝∠ADF ····················································································· 2.5分 ∴∠EDF＝∠EDA＋∠ADF＝∠EDA＋∠BDE＝∠BDA＝90°

∴△DEF为等腰直角三角形 ·························································································· 3分 ②若E，F分别是AB，CA延长线上的点，如图所示．

连结AD ··················································································································· 4分 ∵AB＝AC ∠BAC＝90° D为BC的中点 ∴AD＝BD AD⊥BC ········································· 5分 ∴∠DAC＝∠ABD＝45° ∴∠DAF＝∠DBE＝135°··································5.5分 又AF＝BE

∴△DAF≌△DBE （S.A.S）······························ 6分 ∴FD＝ED ∠FDA＝∠EDB ······························ 6.5分

∴∠EDF＝∠EDB＋∠FDB＝∠FDA＋∠FDB＝∠ADB＝90°

∴△DEF仍为等腰直角三角形 ······················································································ 7分 六．解答题：（共8分） 26．解：（1）证明：∵抛物线y＝x2－2ax＋b2 经过点M(a c，0) ∴(a c)2 2a(a c) b2 0 ························································································· 1分 ∴a2 2ac c2 2a2 2ac b2 0

222b c a∴ ·············································································································· 1.5分 由勾股定理的逆定理得：

△ABC为直角三角形 ················································································· 2分 （2）解：①如图所示； ∵S△MNP 3S△NOP

∴MN 3ON 即MO 4ON ························· 2.5分 又M(a c，0) ∴N

a c

4

，0

······················· 3分

∴a c，

a c4

是方程x2－2ax＋b2＝0的两根

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07年的自贡中考题

∴(a c) ∴c

35a

a c4

2a

··································································································· 3.5分

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

由（1）知：在△ABC中，∠A＝90° 由勾股定理得b ∴cosC

ba 45

45a

··································································································· 4.5分

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

②能···························································································································· 5.5分

222

2

由（1）知 y x 2ax b x

ax

2

a

2

c( x )

2

a

c

2

∴顶点D(a，··········································································································· 6分 c)

过D作DE⊥x轴于点E 则NE＝EM DN＝DM 要使△MND为等腰直角三角形，只须ED＝

2∵M(a c， c) 0) D(a，

12

MN＝EM ·············································· 6.5分

∴DE c2 EM c

∴c2 c 又c＞0，∴c＝1·························································································· 7分 由于c＝a b＝

53

45

a ∴a＝ b＝

3

543

··································································· 7.5分

∴当a＝，b＝

3

543

，c＝1时，△MNP为等腰直角三角形·············································· 8分

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