2009年广西南宁市中考数学试题含答案

2009年南宁市中等学校招生考试

数 学

本试卷分第Ⅰ卷和第Ⅱ卷，满分120分，考试时间120分钟.

注意：答案一律填写在答题卷上，在试题卷上作答无效．考试结束，将本试卷和答题．．．．．．．．．

卷一并交回．

第Ⅰ卷（选择题 共36分）

一、选择题：（本大题共12小题，每小题3分，共36分）每小题都给出代号为A、B、C、D的四个结论，其中只有一个是正确的．使用机改卷的考生，请用2B铅笔在答题卷上将选．．．．．．．．定的答案标号涂黑；使用非机改卷的六县考生，请用黑（蓝黑）墨水笔将每小题选定的答．．．．．．．．．．．案的序号填写在答题卷相应的表格内．

1的相反数是（ ） 31A．3 B． C．?3

31．

D．?1 32．图1是一个五边形木架，它的内角和是（ ）

图1 A．720° B．540° C．360° D．180°

3．今年6月，南宁市举行了第五届泛珠三角区域经贸合作洽谈会.据估算，本届大会合同投资总额达2260亿元.将2260用科学记数法表示为（结果保留2个有效数字）（ ） A．2.3?10

3B．2.2?10

3C．2.26?10

3

D．0.23?10

44．与左边三视图所对应的直观图是（ ）

A．

B． C． D．

?1?x≤15．不等式组?2的解集在数轴上表示为（ ）

??2?x?3 -1

0 1 A．

2 -1 0 1 B．

2 -1 0 1 C．

2 -1 0 1 D．

2 6．要使式子

x?1有意义，x的取值范围是（ ） xA．x?1 B．x?0 C．x??1且x?0 D．x≥-1且x?0

7．如图2，将一个长为10cm，宽为8cm的矩形纸片对折两次后，沿所得矩形两邻边中点的连线（虚线）剪下，再打开，得到的菱形的面积为（ ）

A．10cm

22B．20cm

2C．40cm

2D D．80cm

2A C 图2

B 8．把多项式2x?8x?8分解因式，结果正确的是（ ） A．?2x?4?

2

B．2?x?4?

2

C．2?x?2?

2

D．2?x?2?

29．在反比例函数y?是（ ） A．?1 B．0

1?k的图象的每一条曲线上，y都随x的增大而增大，则k的值可以x

C．1

D．2

10．如图3，AB是⊙O的直径，弦CD?AB于点E，?CDB?30° ，⊙O的半径为3cm，则弦CD的长为（ ） A．

3cm 2

B．3cm C．23cm D．9cm

y C A

E B

3 O 1 x O

图3

D

图4

211．已知二次函数y?ax?bx?c（a?0）的图象如图4所示，有下列四个结论：

①b?0②c?0③b2?4ac?0④a?b?c?0，其中正确的个数有（ ）

A．1个

B．2个

C．3个

D．4个

12．从2、3、4、5这四个数中，任取两个数p和q?p?q?，构成函数y?px?2和y?x?q，并使这两个函数图象的交点在直线x?2的右侧，则这样的有序数对?p，q?共有（ ） A．12对

B．6对

C．5对

D．3对

第Ⅱ卷（非选择题，共84分）

二、填空题：（本大题共6小题，每小题2分，共12分）

，则?2? °． 13．如图5，直线a、b被c所截，且a∥b，?1?120°14．计算：ab

??22?a ．

北

A 45°

A′

P 30°

B 图7

C

东

c

a

1 2 图5

b

O 灯

A 三角尺 图6

投影

15．三角尺在灯泡O的照射下在墙上形成影子（如图6所示）.现测得

OA?20cm，O?A?是 ．

50cm，这个三角尺的周长与它在墙上形成的影子的周长的比

16．有五张分别印有圆、等腰三角形、矩形、菱形、正方形图案的卡片（卡片中除图案不同外，其余均相同），现将有图案的一面朝下任意摆放，从中任意抽取一张，抽到有中心对称图案的卡片的概率是 ．

17．如图7，一艘海轮位于灯塔P的东北方向，距离灯塔402海里的A处，它沿正南方向

航行一段时间后，到达位于灯塔P的南偏东30°方向上的B处，则海轮行驶的路程AB为 _____________海里（结果保留根号）．

18．正整数按图8的规律排列．请写出第20行，第21列的数字 ．

第一列 第二列

第一行 第二行 第三行 第四行 第五行 ??

1 4 9 16 25

2 3 8 15 24 第三列 第四列 第五列 5 6 7 14 23 图8

10 11 12 13 22

17 18 19 20 21

? ? ? ? ?

考生注意：第三至第八大题为解答题，要求在答题卷上写出解答过程． ．．．三、（本大题共2小题，每小题满分6分，共12分） 19．计算：??1?20093?1???????sin60°

2?2??1

20．先化简，再求值：

1?1?1????x?2?，其中x?2 ??2?x?1?x?1

四、（本大题共2小题，每小题满分10分，共20分）

21．为迎接国庆60周年，某校举行以“祖国成长我成长”为主题的图片制作比赛，赛后整理参赛同学的成绩，并制作成图表如下： 分数段 60≤x＜70 70≤x＜80 80≤x＜90 90≤x＜100 频数 30 m 60 20 频率 0.15 0.45 n 0.1 频数 120 90 60 30 0 60 70 80 90 100 图9 请根据以上图表提供的信息，解答下列问题：

分数（分）

（1）表中m和n所表示的数分别为：m?__________，n?__________； （2）请在图9中，补全频数分布直方图； （3）比赛成绩的中位数落在哪个分数段？

（4）如果比赛成绩80分以上（含80分）可以获得奖励，那么获奖率是多少？

22．已知△ABC在平面直角坐标系中的位置如图10所示． （1）分别写出图中点A和点C的坐标；

（2）画出△ABC绕点C按顺时针方向旋转90°后的△A?B?C?；

（3）求点A旋转到点A?所经过的路线长（结果保留π）．

y

8

7 6 5 A 4 B 3

2 1 C x 0 1 2 3 4 5 6 7 8

图10

五、（本大题满分10分）

23．如图11，PA、PB是半径为1的⊙O的两条切线，点A、B分别为切点，

?APB?60°，OP与弦AB交于点C，与⊙O交于点D．

（1）在不添加任何辅助线的情况下，写出图中所有的全等三角形；

A （2）求阴影部分的面积（结果保留π）．

C D O P

B 图11

六、（本大题满分10分）

24．南宁市狮山公园计划在健身区铺设广场砖．现有甲、y元 乙两个工程队参加竞标，甲工程队铺设广场砖的造价y甲2（元）与铺设面积xm的函数关系如图12所示；乙工

48000 28000 0 500 1000 ??2程队铺设广场砖的造价y乙（元）与铺设面积xm满足

??图12

函数关系式：y乙?kx．

x?m2?

2（1）根据图12写出甲工程队铺设广场砖的造价y甲（元）与铺设面积xm的函数关系式；

??（2）如果狮山公园铺设广场砖的面积为1600m，那么公园应选择哪个工程队施工更合算？

2

七、（本大题满分10分）

25．如图13-1，在边长为5的正方形ABCD中，点E、F分别是BC、DC边上的点，且AE?EF，BE?2. （1）求EC∶CF的值；

（2）延长EF交正方形外角平分线CP于点P（如图13-2），试判断AE与EP的大小关系，并说明理由；

（3）在图13-2的AB边上是否存在一点M，使得四边形DMEP是平行四边形？若存在，请给予证明；若不存在，请说明理由． A D

F

B E C

图13-1

A D F P B E C 图13-2

八、（本大题满分10分）

26．如图14，要设计一个等腰梯形的花坛，花坛上底长120米，下底长180米，上下底相距80米，在两腰中点连线（虚线）处有一条横向甬道，上下底之间有两条纵向甬道，各甬道的宽度相等．设甬道的宽为x米．

（1）用含x的式子表示横向甬道的面积；

（2）当三条甬道的面积是梯形面积的八分之一时，求甬道的宽；

（3）根据设计的要求，甬道的宽不能超过6米.如果修建甬道的总费用（万元）与甬道的宽度成正比例关系，比例系数是5.7，花坛其余部分的绿化费用为每平方米0.02万元，那么当甬道的宽度为多少米时，所建花坛的总费用最少？最少费用是多少万元？ 图14

2009年南宁市中等学校招生考试 数学试题参考答案与评分标准

一、选择题（本大题共12小题，每小题3分，共36分） 题号 答案 1 D 2 B 3 A 4 A 5 C 6 D 7 A 8 C 9 D 10 B 11 C 12 B 二、填空题（本大题共6小题，每小题2分，共12分） 13．60 14．ab 15．

3224 16． 17．403?40 18．420 55??三、（本大题共2小题，每小题满分6分，共 12分） 19．解：??1?20093?1???????sin60°

2?2??1=??1??33 ············································································································ 4分 ?2?22=?1?2 ··································································································································· 5分

??3 ········································································································································ 6分 20．解：?1???1?1???x?2? ?2x?1?x?1=

x?x?1??x?1?·?x?2 ································································································· 3分 x?11?x2?2 ·································································································································· 4分

当x?2时，原式??2?2·························································································· 5分 ?2 ·

?4 ········································································································ 6分

四、（本大题共2小题，每小题满分10分，共20分）

21．解：（1）m?90························································································· 4分 ，n?0.3； ·（2）图略． ···························································································································· 6分 （3）比赛成绩的中位数落在：70分~80分． ······································································· 8分 （4）获奖率为：

60?20?100%=40%（或0.3+0.1=0.4） ··············································· 10分 20022．解：（1）A?0，······················································································· 2分 4?、C?31，?； ·（2）图略． ···························································································································· 6分 （3）AC?32 ···················································································································· 7分

90?32?π? ·························································································································· 9分 AA??180

?32··········································································································································10分 π ·2五、（本大题满分10分）

△APC≌△BPC，△PAO≌△PBO ·23．解：（1）△ACO≌△BCO，························· 3分

（写出一个全等式子得1分）

（2）?PA、PB为⊙O的切线

A ?PO平分?APB，PA?PB，?PAO?90° ······················ 5分

?PO?AB············································································· 6分

C D O P ····························· 7分 ?由圆的对称性可知：S阴影?S扇形AOD ·

11?在Rt△PAO中，?APO??APB??60°?30?

22B ??AOP?90°??APO?90°?30??60? ········································································ 8分

?S阴影?S扇形AOD60?π?12? ····························································································· 9分

360?π········································································································· 10分 6六、（本大题满分10分）

24．解：（1）当0≤x≤500时，设y甲?k1x，把?500，28000?代入上式得：

28000?500k1，?k1?28000?56 500···························································································································· 2分 ?y甲?56x ·

当x≥500时，设y甲?k2x?b，把?500，28000?、?1000，48000?代入上式得：

?500k2?b?28000 ············································································································ 3分 ??1000k2?b?48000?k2?40解得：? ··················································································································· 4分

b?8000??y甲?40x?8000

??56x?0≤x?500??y甲?? ···························································································· 5分 ??40x?8000?x≥500?（2）当x?1600时，y甲?40?1600?8000?72000 ······················································ 6分

···················································································· 7分 y乙?1600k ·

①当y甲?y乙时，即：72000?1600k

得：k?45 ····························································································································· 8分

②当y甲?y乙时，即：72000?1600k

得：0?k?45 ······················································································································· 9分

③当y甲?y乙时，即72000?1600k，?k?45

答：当k?45时，选择甲工程队更合算，当0?k?45时，选择乙工程队更合算，当k?45时，选择两个工程队的花费一样． ······················································································ 10分 七、（本大题满分10分） 25．解：（1）?AE?EF

A D ??2??3?90°

?四边形ABCD为正方形

??B??C?90°

??1??3?90° ?1??2 ·········································································· 1分

B 1 3 E

2 F C

??DAM??ABE?90°，DA?AB ?△DAM≌△ABE ?DM?AE ·························································································································· 9分 ?AE?EP ?DM?PE

······················································································ 10分 ?四边形DMEP是平行四边形． ·

（备注：作平行四边形DMEP，并计算出AM或BM的长度，但没有证明点M在AB边

上的扣1分）

解法②：在AB边上存在一点M，使四边形DMEP是平行四边形 ································ 8分 证明：在AB边上取一点M，使AM?BE，连接ME、MD、DP． AD?BA，?DAM??ABE?90°

A ?Rt△DAM≌Rt△ABE

5 ?DM?AE，?1??4 ········································· 9分

1 ??1??5?90° M ??4??5?90°

D 4 F

P

?AE?DM

B E C ?AE?EP

?DM?EP

·························································································· 10分 ?四边形DMEP为平行四边形 ·

（备注：此小题若有其他的证明方法，只要证出判定平行四边形的一个条件，即可得1分）

八、（本大题满分10分）

120?180x?150x?m2? ············································· 2分 21120?1802?80 ·（2）依题意：2?80x?150x?2x??················································· 4分

8226．解：（1）横向甬道的面积为：整理得：x?155x?750?0

2

············································································· 6分 x1?5，x2?150（不符合题意，舍去） ·

?甬道的宽为5米．

（3）设建设花坛的总费用为y万元．

?120?180?··············································· 7分 y?0.02???80??160x?150x?2x2???5.7x ·

2???0.04x2?0.5x?240

当x??b0.5??6.25时，y的值最小． ································································ 8分 2a2?0.04因为根据设计的要求，甬道的宽不能超过6米，

?当x?6米时，总费用最少． ····························································································· 9分

最少费用为：0.04?6?0.5?6?240?238.44万元 ······················································· 10分

2

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