2009年山东省滨州市中考数学试题及答案

滨州市二○○九年初级中学学业水平考试

数 学 试 题

温馨提示：

1． 本试题共8页，满分120分，考试时间为120分钟．

b4ac b2

2． 抛物线y ax bx c(a 0)的顶点坐标是 ．

4a 2a

2

1．截止目前，滨州市总人口数约373万，此人口数用科学记数法可表示为（ ） A．3.73 10

4

B．3.73 10

5

C．3.73 10

6

D．3.73 10

7

2．对于式子 ( 8)，下列理解：（1）可表示 8的相反数；（2）可表示 1与 8的乘积；（3）可表示 8的绝对值；（4）运算结果等于8．其中理解错误的个数是（ ） A．0 B．1 C．2 D．3

3．从编号为1到10的10张卡片中任取1张，所得编号是3的倍数的概率为（ ） A．

1 10

B．

2 10

C．

3 10

D．

1 5

4．从上面看如右图所示的几何体，得到的图形是（ ）

A． Ｂ． C． D． （第4题图）

5．顺次连接对角线互相垂直的四边形的各边中点，所得图形一定是（ ） A．矩形 B．直角梯形 C．菱形 D．正方形 6．已知两圆半径分别为2和3，圆心距为d，若两圆没有公共点，则下列结论正确的是（ ） A．0 d 1 B．d 5 C．0 d 1或d 5 D．0≤d 1或d 5 7．小明外出散步，从家走了20分钟后到达了一个离家900米的报亭，看了10分钟的报纸然后用了15分钟返回到家．则下列图象能表示小明离家距离与时间关系的是（ ）

/

A． B． C． D．

8．已知y关于x的函数图象如图所示，则当y 0时，自变量x的取值范围是（ ） A．x 0 B． 1 x 1或x 2

C．x 1 D．x 1或1 x 2 9．如图所示，给出下列条件：

① B ACD； ② ADC ACB；

ACAB2

； ④AC AD AB．

CDBC

其中单独能够判定△ABC∽△ACD的个数为（ ）

③

A

A．1 B．2 C．3 D．4

10．已知△ABC中，AB 17，AC 10，BC边上的高

C

AD 8， 则边BC的长为（ ）

（第9题图）

A．21 B．15 C．6 D．以上答案都不对

二、填空题：本大题共8小题，每小题填对得4分，满分32分．只要求填写最后结果．

m2 4mn 4n2

11．化简： ．

m2 4n2

12．数据1、5、6、5、6、5、6、6的众数是 ，中位数是 ，方差是 ． 13．已知点A是反比例函数y 的面积 ．

3

图象上的一点．若AB垂直于y轴，垂足为B，则△AOBx

2x3x2 3x

14．解方程2，则方程可化为 ． 2时，若设y 2

x 1xx 1

15．大家知道|5| |5 0|，它在数轴上的意义是表示5的点与原点（即表示0的点）之间的距离．又如式子|6 3|，它在数轴上的意义是表示6的点与表示3的点之间的距离．类似地，式子|a 5|在数轴上的意义是 ．

B

16．某楼梯的侧面视图如图所示，其中AB 4米，

BAC 30°， C 90°，因某种活动要求铺设红色地毯，

C A 则在AB段楼梯所铺地毯的长度应为 ．

（第16题图）

17．已知等腰△ABC的周长为10，若设腰长为x，则x的取值范围是 ． 18．在平面直角坐标系中，△ABC顶点A的坐标为(2，3)，若以原点O为位似中心，画

△AEC的位似图形△A B C ，使△ABC与△A B C 的相似比等于

1

，则点A 的坐标2

为 ．

三、解答题：本大题共7小题，满分58分．解答时请写出必要的文字说明与推演过程． 19．（本题满分5分）

1 0

计算： 1 |2| 5 (2009 π)．

2

2

1

20．（本题满分6分）

为推进阳光体育活动的开展，某校九年级三班同学组建了足球、篮球、乒乓球、跳绳四个体育活动小组．经调查，全班同学全员参与，各活动小组人数分布情况的扇形图和条形图

如下：

篮球足球 25%

90

（1）求该班学生人数；

（2）请你补上条形图的空缺部分；

（3）求跳绳人数所占扇形圆心角的大小．

21．（本题满分7分）

如图，PA为⊙O的切线，A为切点．直线PO与⊙O交于B、C两点， P 30°，连接AO、AB、AC．求证：△ACB≌△APO．

C P O B

（第21题图）

22．（本题满分8分）

观察下列方程及其解的特征：

1

2的解为x1 x2 1； x151

（2）x 的解为x1 2，x2 ；

x221101

（3）x 的解为x1 3，x2 ；

x33

（1）x

…… ……

解答下列问题：

126

的解为 ； x5

11

（2）请猜想：关于x的方程x 的解为x1 a，x2 (a 0)；

xa

126

（3）下面以解方程x 为例，验证（1）中猜想结论的正确性．

x5

（1）请猜想：方程x

解：原方程可化为5x 26x 5．

（下面请大家用配方法写出解此方程的详细过程）

23．（本题满分10分） 根据题意，解答下列问题：

2

（1）如图①，已知直线y 2x 4与x轴、y轴分别交于A、B两点，求线段AB的长； （2）如图②，类比（1）的求解过程，请你通过构造直角三角形的方法，求出两点M(3，4)，

N( 2， 1)之间的距离；

（3）如图③，P2(x1，y2)是平面直角坐标系内的两点．

1(x1，y1)，P求证：PP12

P(x

11 （第23题图①）

（第23题图②）

（第23题图③）

24．（本题满分10分）

某商品的进价为每件40元．当售价为每件60元时，每星期可卖出300件，现需降价处理，且经市场调查：每降价1元，每星期可多卖出20件．在确保盈利的前提下，解答下列问题： （1）若设每件降价x元、每星期售出商品的利润为y元，请写出y与x的函数关系式，并求出自变量x的取值范围；

（2）当降价多少元时，每星期的利润最大？最大利润是多少？ （3）请画出上述函数的大致图象．

25．（本题满分12分）

如图①，某产品标志的截面图形由一个等腰梯形和抛物线的一部分组成，在等腰梯形ABCD中，AB∥DC，AB 20cm，DC 30cm， ADC 45°．对于抛物线部分，其顶点为CD的中点O，且过A、B两点，开口终端的连线MN平行且等于DC． （1）如图①所示，在以点O为原点，直线OC为x轴的坐标系内，点C的坐标为(15， 0)，试求A、B两点的坐标；

（2）求标志的高度（即标志的最高点到梯形下底所在直线的距离）；

（3）现根据实际情况，需在标志截面图形的梯形部分的外围均匀镀上一层厚度为3cm的保护膜，如图②，请在图中补充完整镀膜部分的示意图，并求出镀膜的外围周长． A B

D C （第25题图①） （第25题图②）

滨州市二○○九年初级中学学业水平考试

数学试题（A）解答参考及评分标准

评卷说明：

1．选择题的每小题和填空题中的每个空，只有满分和零分两个评分档，不给中间分． 2．解答题每小题的解答中所对应的分数，是指考生正确解答到该步骤所应得的累计分数．本答案对每小题只给出一种解法，对考生的其他解法，请参照评分标准进行评分．

3．如果考生在解答的中间过程出现计算错误，但并没有改变试题的实质和难度，其后续部分酌情给分，但后续部分最多不超过正确解答分数的一半；若出现严重的逻辑错误，后续部分就不再给分．

二、填空题（本大题共8小题，每小题4分，满分32分） 11．

m 2n5

12．6，5.5，（分值分配：1分、1分、2分）

m 2n233

14．2y 2

y2

13．

15．表示数a的点与表示 5的点之间的距离 16．(2米（或5.464米） 17．

5

x 5 2

18．(4，6)或( 4， 6)

三、解答题（本大题共7小题，满分58分） 19．（本题满分5分）

解：原式 1 22 5 ································· 4分（四个考查点，做对1个就得1分）

2 ··························································································································· 5分

20．（本题满分6分）

解：（1）由扇形图可知，乒乓球小组人数占全班人数的

1

． 4

由条形图可知，乒乓球小组人数为12． ··············································································· 1分 故全班人数为12

1

································································································· 2分 48． ·

4

（注：只有最后一步做对也得满分，但只有结果不得分．） （2）由扇形图可知，篮球小组人数为48 25% 12． 由条形图可知，足球小组人数为16．

故跳绳小组人数为48 (16 12 12) 8． ········································································· 3分 所以各小组人数分布情况的条形图为

············································································ 4分（注：本小题只画对图也得满分2分．）

81

································································ 5分 ， ·

4861

所以，它所占扇形圆心角的大小为360° 60°． ·························································· 6分

6

（3）因为跳绳小组人数占全班人数的

21．（本题满分7分）

证明： PA为 O的切线， PAO 90°． ·································································· 1分 又 P 30°， AOP 60°， ······················································································ 2分

1

··································································································· 3分 C AOP 30°， ·

2

······················································································································ 4分 C P， ························································································································ 5分 AC AP． ·

又BC为 O直径， CAB PAO 90°， ································································ 6分

·························································································· 7分 △ACB≌△APO（ASA）． ·

（注：其它方法按步骤得分．）

22．（本题满分8分） 解：（1）x1 5，x2

1； ································································································· 1分 5

a2 11（2）（或a ）； ·································································································· 3分

aa

（3）二次项系数化为1，得x

2

2

2

26

···································································· 4分 x 1． ·

5

2

26 13 13

x 1 ， ·配方，得x ································································· 5分 5 5 5 13 144

x ． ················································································································ 6分

525

2

1312

····································································································· 7分 ． ·

55

1

解得x1 5，x2 ．··········································································································· 8分

5

1

经检验，x1 5，x2 都是原方程的解（此环节有无暂不得分与扣分）

5

开方，得x

23．（本题满分10分）

解：（1）由y 0，得x 2，所以点A的坐标为( 2，····················· 1分 0)，故OA 2． ·同理可得OB 4． ················································································································ 2分 所以在Rt△AOB中，AB

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

（2）作MP x轴，NP y轴，MP交NP于点P． ···················································· 4分 则MP NP，P点坐标为(3，·················································································· 5分 1)． ·故PM 4 ( 1) 5，PN 3 ( 2) 5． ······································································ 6分 所以在Rt△MPN中，MN ． ······························································· 7分 （注：若直接运用了（3）的结论不得分．）

y轴，P2P交PP（3）作P2P x轴，PP11于点P．

则P，点P的坐标为(x2，y1)． ············································································ 8分 2P PP1

x2 x1（不加绝对值符号此处不扣分）． 故P2P y2 y1，PP··································· 9分 1

所以在Rt△P2PP12 1中，PP24．（本题满分10分）

解：（1）y (60 x)(300 20x) 40(300 20x)， ······················································ 3分 即y 20x 100x 6000． ····························································································· 4分 因为降价要确保盈利，所以40 60 x≤60（或40 60 x 60也可）． 解得0≤x 20（或0 x 20）． ···················································································· 6分 （注：若出现了x 20扣1分；若直接写对结果，不扣分即得满足2分．） （2）当x

2

················································ 10分

100

······················································································ 7分 2.5时， ·

2 ( 20)

4 ( 20) 6000 1002

6125， y有最大值

4 ( 20)

即当降价2.5元时，利润最大且为6125元． ········································································ 8分 （3）函数的大致图象为（注：右侧终点应为圆圈，若画成实点扣1分；左侧终点两种情况均可．） ································································································································ 10分

25．（本题满分12分）

解：（1）作AE DC，BF DC，垂足分别为E，F．

···················· 1分 AB∥DC， 四边形AEFB为矩形， AE BF，AB EF 20． ·

又 AD BC，

1

··························· 2分 Rt△ADE≌Rt△BCF（HL）， DE FC (30 20) 5． ·

2

又 ADE BCF 45°，

···························································································· 3分 AE BF DE FC 5． ·

又OD OC 15， OE OF 10．

····································································· 4分 5)． ·5)，(10， 点A，B的坐标分别为( 10，

（2）设抛物线的函数解析式为y ax． ············································································ 5分 由点B(10，5)在其图象上得5 100a，解得a

2

1

． 20

抛物线的函数解析式为y

12

················································································· 6分 x． ·

20

∥DC， 点M，N关于y轴对称， 又 MN

点N的横坐标为15，代入y

故标志的高度为

1245

． x得y

204

45

cm． ········································································································ 8分 4

（3）镀膜示意图如下：

20cm

45°

··············································································································································· 10分 由示意图可知，镀膜外围周长l由四条线段长和四条半径为3cm的弧长构成，

故l 2 20 30

135 π 345 π 3

2 2 50 6π．

180180

所以镀膜的外围周长为50 6π)cm． ·································································· 12分