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08年南京市中考数学试题及答案

发布时间:2014-02-09 09:54:09  

南京市2008年初中毕业生学业考试

数 学

注意事项:

1.本试卷共6页.全卷满分120分.考试时间为120分钟.考生答题全部答在答题卡上,答在本试卷上无效.

2.请认真核对监考教师在答题卡上所粘贴条形码的姓名、考试证号是否与本人相符合,再将自己的姓名、准考证号用0.5毫米黑色墨水签字笔填写在答题卡及本试卷上.

3.答选择题必须用2B铅笔将答题卡上对应的答案标号涂黑.如需改动,请用橡皮擦干净后,再选涂其他答案.答非选择题必须用0.5毫米黑色墨水签字笔写在答题卡上的指定位置,在其他位置答题一律无效.

4.作图必须用2B铅笔作答,并请加黑加粗,描写清楚.

一、选择题(本大题共10小题,每小题2分,共计20分.在每小题所给出的四个选项中,恰有一项是符....合题目要求的,请将正确选项的序号填涂在答题卡上)

1.?3的绝对值是( )

A.?3 B.3 C.?1 3D.1 3

2.2008年5月27日,北京2008年奥运会火炬接力传递活动在南京境内举行,火炬传递路线全程约12 900m,将12 900m用科学记数法表示应为( )

A.0.129?10

235B.1.29?10 4 C.12.9?10 3 D.129?10 23.计算(ab)的结果是( )

A.ab 5B.ab 6C.ab 35D.ab 36

4.2的平方根是( )

A.4 B

C

. D

.5.已知反比例函数的图象经过点P(?2,1),则这个函数的图象位于( )

A.第一、三象限 B.第二、三象限

C.第二、四象限 D.第三、四象限

6.如图,将一张等腰梯形纸片沿中位线剪开,拼成一个新的图形,

这个新的图形可以是下列图形中的( ) (第6题)

A.三角形 B.平行四边形 C.矩形 D.正方形

7.小刚身高1.7m,测得他站立在阳光下的影子长为0.85m,紧接着他把手臂竖直举起,测得影子长为1.1m,

那么小刚举起的手臂超出头顶( )

A.0.5m B.0.55m C.0.6m D.2.2m

8.如图,?O是等边三角形ABC的外接圆,?O的半径为2,

则等边三角形ABC的边长为( )

A

B

C

.D

.(第8题)

9.超市为了制定某个时间段收银台开放方案,统计了这个时间段本超市顾客在收银台排队付款的等待时间,并绘制成如下的频数分布直方图(图中等待时间6分钟到7分钟表示大于或等于6分钟而小于7

分钟,其

它类同).这个时间段内顾客等待时间不少于6分钟的人数为( )

A.5 B.7 C.16 D.33

(第10题)

/min (第9题)

10.如图,已知?O的半径为1,AB与?O相切于点A,OB与?O交于点C,OD?OA,垂足为D,则cos?AOB的值等于( )

A.OD B.OA C.CD D.AB

二、填空题(本大题共6小题,每小题3分,共计18分.不需写出解答过程,请把答案直接填写在答题卡...相应位置上) ....

11

12.函数y?1?x中,自变量x的取值范围是 x

13.已知?O1和?O2的半径分别为3cm和5cm,且它们内切,则圆心距O1O2等于cm.

14.若等腰三角形的一个外角为70,则它的底角为 度.

15.口袋内装有一些除颜色外完全相同的红球、白球和黑球,从中摸出一

球,摸出红球的概率是0.2,摸出白球的概率是0.5,那么摸出黑球的概率

是 .

16.如图,有一圆形展厅,在其圆形边缘上的点A处安装了一台监视器,

它的监控角度是65.为了监控整个展厅,最少需在圆形边缘上共安装 ...??(第16题) 这样的监视器 台.

三、解答题(本大题共12小题,共计82分.请在答题卡指定区域内作答,解答时应写出文字说明、证明.......

过程或演算步骤)

17.(6分)先化简,再求值:(2a?1)?2(2a?1)?

3,其中a?

18.(6分)解方程

22x?2?0. x?1x?1

?2?x?0,?19.(6分)解不等式组?5x?12x?1并把解集在数轴上表示出来. ?1≥,?3?2

(第19题)

20.(6分)我国从2008年6月1日起执行“限塑令”.“限塑令”执行前,某校为了了解本校学生所在家庭使用塑料袋的数量情况,随机调查了10名学生所在家庭月使用塑料袋的数量,结果如下(单位:只) 65,70,85,75,85,79,74,91,81,95.

(1)计算这10名学生所在家庭平均月使用塑料袋多少只?

(2)“限塑令”执行后,家庭月使用塑料袋数量预计将减少50%.根据上面的计算结果,估计该校1 000名学生所在家庭月使用塑料袋可减少多少只?

21.(6分)如图,在ABCD中,E,F为BC上两点,且BE?CF,AF?DE.

求证:(1)△ABF≌△DCE;

(2)四边形ABCD是矩形. A D

B C E

F

(第21题)

22.(6分)如图,菱形ABCD(图1)与菱形EFGH(图2)的形状、大小完全相同.

(1)请从下列序号中选择正确选项的序号填写;

①点E,F,G,H;②点G,F,E,H;③点E,H,G,F;④点G,H,E,F.

D

A C

E ?

B 图1 (第22题) F 图2

如果图1经过一次平移后得到图2,那么点A,B,C,D对应点分别是 ;

如果图1经过一次轴对称后得到图2,那么点A,B,C,D对应点分别是 ;

如果图1经过一次旋转后得到图2,那么点A,B,C,D对应点分别是 ;

(2)①图1,图2关于点O成中心对称,请画出对称中心(保留画图痕迹,不写画法);

②写出两个图形成中心对称的一条性质: .(可以结合所画图形叙述) ..

23.(6分)如图,山顶建有一座铁塔,塔高CD?30m,某人在点A处测得塔底C的仰角为20,塔顶D的仰角为23,求此人距CD的水平距离AB.

(参考数据:sin20≈0.342,cos20≈0.940,tan20≈0.364,sin23≈0.391,cos23≈0.921,

?

?

?

?

?

?

?

tan23?≈0.424)

?

?23 B (第23题) 24.(7分)小明和小颖做掷骰子的游戏,规则如下: ①游戏前,每人选一个数字; ②每次同时掷两枚均匀骰子;

③如果同时掷得的两枚骰子点数之和,与谁所选数字相同,那么谁就获胜. (1)在下表中列出同时掷两枚均匀骰子所有可能出现的结果:

D C

(2)小明选的数字是5,小颖选的数字是6.如果你也加入游戏,你会选什么数字,使自己获胜的概率比他们大?请说明理由. 25.(7分)某村计划建造如图所示的矩形蔬菜温室,要求长与宽的比为2:1.在温室内,沿前侧内墙保留3m宽的空地,其它三侧内墙各保留1m宽的通道.当矩形温室的长与宽各为多少时,蔬菜种植区域的面积是288m?

2

2

(第25题)

26.(8分)已知二次函数y?x?bx?c中,函数y与自变量x的部分对应值如下表:

(1)求该二次函数的关系式;

(2)当x为何值时,y有最小值,最小值是多少?

,y2)两点都在该函数的图象上,试比较y1与y2的大小. (3)若A(m,y1),B(m?1

27.(8分)如图,已知?O的半径为6cm,射线PM经过点O,OP?10cm,射线PN与?O相切于点点A以5cm/s的速度沿射线PM方向运动,点B以4cm/s的速度沿射线PNQ.A,B两点同时从点P出发,方向运动.设运动时间为ts. (1)求PQ的长;

(2)当t为何值时,直线AB与?O相切?

(第27题)

28.(10分)一列快车从甲地驶往乙地,一列慢车从乙地驶往甲地,两车同时出发,设慢车行驶的时间为x(h),两车之间的距离为y(km),图中的折线表示y与x

.......y根据图象进行以下探究:

信息读取 (1)甲、乙两地之间的距离为 km;

(2)请解释图中点B的实际意义;

(第28题)

图象理解

(3)求慢车和快车的速度;

(4)求线段BC所表示的y与x之间的函数关系式,并写出自变量x的取值范围;

问题解决

(5)若第二列快车也从甲地出发驶往乙地,速度与第一列快车相同.在第一列快车与慢车相遇30分钟后,第二列快车与慢车相遇.求第二列快车比第一列快车晚出发多少小时?

南京市2008年初中毕业生学业考试 数学试题参考答案及评分标准

说明:本评分标准每题给出了一种或几种解法供参考,如果考生的解法与本解答不同,参照本评分标准的精神给分.

一、选择题(每小题2分,共计20分)

二、填空题(每小题3分,共计18分) 1112.x?0

13.2

14.35

15.0.3

16.3

三、解答题(本大题共12小题,共计82分) 17.(本题6分)

解:原式?4a?4a?1?4a?2?3 ····················································································· 3分 ············

·········

······································································································ 4分 ?4a2?2. ·

当a?时,4a?2?4??2?10. ···································································· 6分 18.(本题6分)

解:方程两边同乘(x?1)(x?1),得

·················································································································· 3分 2(x?1)?x?0. ·

解这个方程,得

·································································································································· 5分 x?2. ·

检验:当x?2时,(x?1)(x?1)?0.

所以x?2是原方程的解. ····································································································· 6分

19.(本题6分)

解:解不等式①,得x?2. ································································································· 2分 解不等式②,得x≥?1. ····································································································· 4分 所以,不等式组的解集是?1≤x?2. ··············································································· 5分 不等式组的解集在数轴上表示如下:

···········································································

···································· 6分 20.(本题6分) 解:(1)

2

2

2

1

(65?70?85?75?85?79?74?91?81?95)?80. 10

答:这10名学生所在家庭平均月使用塑料袋80只. ························································· 3分

(2)80?1000?50%?40000.

答:执行“限塑令”后,估计1 000名学生所在家庭月使用塑料袋可减少40 000只. ···· 6分

21.(本题6分)

解:(1)?BE?CF,

BF?BE?EF,CE?CF?EF,

······················································································································· 1分 ?BF?CE. ·

?四边形ABCD是平行四边形,

······················································································································· 2分 ?AB?DC. ·

在△ABF和△DCE中,

?AB?DC,BF?CE,AF?DE,

········································································································· 3分 ?△ABF≌△DCE. ·

(2)解法一:?△ABF≌△DCE,

······················································································································ 4分 ??B??C. ·

?四边形ABCD是平行四边形,

?AB∥CD.

??B??C?180?.

············································································································· 5分 ??B??C?90?. ·

···································································································· 6分 ?四边形ABCD是矩形. ·

解法二:连接AC,DB.

?△ABF≌△DCE,

??AFB??DEC.

············································································································ 4分 ??AFC??DEB.·

在△AFC和△DEB中,

?AF?DE,?AFC??DEB,CF?BE,

?△AFC≌△DEB.

······················································································································· 5分 ?AC?DB. ·

?四边形ABCD是平行四边形,

···································································································· 6分 ?四边形ABCD是矩形. ·

22.(本题6分)

解:(1)①;②;④; ··········································································································· 3分

(2)①画图正确; ················································································································· 5分 ②答案不惟一,例如:对应线段相等,

························································································································ 6分 OC?OE等.

23.(本题6分)

解:在Rt△ABC中,?CAB?20, ?

?BC?AB?tan?CAB?AB?tan20?. ·············································································· 2分 在Rt△ABD中,?DAB?23, ?

············································································· 4分 ?BD?AB?tan?DAB?AB?tan23?. ·

?CD?BD?BC?AB?tan23??AB?tan20??AB(tan23??tan20?).

?AB?CD30≈?500(m). ??tan23?tan200.424?0.364

答:此人距CD的水平距离AB约为500m. ······································································· 6分

24.(本题7分)

解:(1)填表正确; ··············································································································· 3分

(2)由上表可以看出,同时投掷两枚骰子,可能出现的结果有36种,它们出现的可能性相同.

所有的结果中,满足两枚骰子点数和为5(记为事件A)的结果有4种,即(1,4),(2,3),(3,2)(4,

41···································································· 4分 ?; ·369

满足两枚骰子点数和为6(记为事件B)的结果有5种,即(1,5),(2,4),(3,3)(4,2),(5,1),所5以小颖获胜的概率为P(B)?; ······················································································· 5分 361),所以小明获胜的概率为P(A)?

要想使自己获胜的概率比他们大,必须满足两枚骰子点数和出现的结果多于5种,由所列表格可知,只有两枚骰子点数和为7(记为事件C)的结果多于5种,有6种,即(1,6),(2,5),(3,4)(4,3),(5,

2),(6,1),所以P(C)?61?.因此,要想使自己获胜的概率比他们大,所选数字应为7. 7分 366

.(本题7分)

解法一:设矩形温室的宽为xm,则长为2xm.根据题意,得

······································································································ 4分 (x?2)?(2x?4)?288. ·

解这个方程,得

x1??10(不合题意,舍去),x2?14. ············································································ 6分

所以x?14,2x?2?14?28.

答:当矩形温室的长为28m,宽为14m时,蔬菜种植区域的面积是288m. ················· 7分

解法二:设矩形温室的长为xm,则宽为21xm.根据题意,得 2

?1?x?2····································································································· 4分 ???(x?4)?288. 2??

解这个方程,得

x1??20(不合题意,舍去),x2?28. ··········································································· 6分

所以x?28,11x??28?14. 22

2答:当矩形温室的长为28m,宽为14m时,蔬菜种植区域的面积是288m. ················· 7分

26.(本题8分)

解:(1)根据题意,当x?0时,y?5;当x?1时,y?2.

所以??5?c, ?2?1?b?c.

解得??b??4, c?5.?

2所以,该二次函数关系式为y?x?4x?5. ····································································· 2分

(2)因为y?x?4x?5?(x?2)?1,

所以当x?2时,y有最小值,最小值是1. ······································································· 4分 22

,y2)两点都在函数y?x?4x?5的图象上, (3)因为A(m,y1),B(m?1

所以,y1?m?4m?5,y2?(m?1)?4(m?1)?5?m?2m?2. 2222

y2?y1?(m2?2m?2)?(m2?4m?5)?2m?3. ·························································· 5分 所以,当2m?3?0,即m?

当2m?3?0,即m?3时,y1?y2; 23时,y1?y2; 2

3当2m?3?0,即m?时,y1?y2. ··············································································· 8分 2

27.(本题8分)

(1)连接OQ.

?PN与?O相切于点Q,

······················································································· 2分 ?OQ?PN,即?OQP?90?. ·

?OP?10,OQ?6,

?PQ??8(cm). ······························································································ 3分

(2)过点O作OC?AB,垂足为C.

?点A的运动速度为5cm/s,点B的运动速度为4cm/s,运动时间为ts,

?PA?5t,PB?4t.

?PO?10,PQ?8,

?PAPB?. POPQ

??P??P,

?△PAB∽△POQ.

·································································································· 4分 ??PBA??PQO?90?. ·

??BQO??CBQ??OCB?90?,

?四边形OCBQ为矩形.

?BQ?OC.

??O的半径为6,

?BQ?OC?6时,直线AB与?O相切.

①当AB运动到如图1所示的位置.

图1

BQ?PQ?PB?8?4t.

由BQ?6,得8?4t?6.

解得t?0.5(s). ···················································································································· 6分 ②当AB运动到如图2所示的位置.

图2

BQ?PB?PQ?4t?8.

由BQ?6,得4t?8?6.

解得t?3.5(s).

所以,当t为0.5s或3.5s时直线AB与?O相切. ····························································· 8分

28.(本题10分)

解:(1)900; ························································································································ 1分

(2)图中点B的实际意义是:当慢车行驶4h时,慢车和快车相遇. ······························ 2分

(3)由图象可知,慢车12h行驶的路程为900km, 所以慢车的速度为900··············································································· 3分 ?75(km/h); ·12

当慢车行驶4h时,慢车和快车相遇,两车行驶的路程之和为900km,所以慢车和快车行驶的速度之和为900·························································· 4分 ?225(km/h),所以快车的速度为150km/h. ·4

900(4)根据题意,快车行驶900km到达乙地,所以快车行驶?6(h)到达乙地,此时两车之间的距离为150

6?75?450(km),所以点C的坐标为(6,450).

设线段BC所表示的y与x之间的函数关系式为y?kx?b,把(4,0),(6,450)代入得

?0?4k?b, ??450?6k?b.

解得??k?225, b??900.?

所以,线段BC所表示的y与x之间的函数关系式为y?225x?900. ··························· 6分

自变量x的取值范围是4≤x≤6. ···················································································· 7分

(5)慢车与第一列快车相遇30分钟后与第二列快车相遇,此时,慢车的行驶时间是4.5h.

把x?4.5代入y?225x?900,得y?112.5.

此时,慢车与第一列快车之间的距离等于两列快车之间的距离是112.5km,所以两列快车出发的间隔时间是

······························ 10分 112.5?150?0.75(h),即第二列快车比第一列快车晚出发0.75h. ·

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