2009年福建省三明市中考数学试题

2009年三明市初中毕业暨高级中等学校招生统一考试

数 学 试 题

（满分：150分 考试时间：6月21日上午8﹕30—10﹕30）

★友情提示：

1．本试卷共4页．

2．考生将自己的姓名、准考证号及所有答案均填写在答题卡上． 3．答题要求见答题卡上的“注意事项”．

4．未注明精确度、保留有效数字等的计算问题，结果应为准确数． ．．．

?b4ac?b2?b?x??5．抛物线y?ax?bx?c?a?0?的顶点坐标为?，对称轴． ?，?2a?2a4a??2一、选择题（共10小题，每小题4分，满分40分；每小题只有一个正确的选项，请在答题卡的相应位置填涂） ．．．1．6的相反数是（ ）

11A． 6 B．?6 C． D．?

662．2008年末我市常住人口约为2630000人，将2630000用科学记数法表示为（ ） A．263?10 B．2.63?10 C．2.63?10 D．0.263?10 3．下列计算正确的是 （ ）

A． a?a?2a B． (2a)2?4a2 C．3?3??3 D．4??2 4．在下面的四个几何体中，它们各自的主视图与左视图可能不相同的是（ ）

A． B． C． D． 5．下列事件是必然事件的是（ ）

A．打开电视机，正在播电视剧 B．小明坚持体育锻炼，今后会成为奥运冠军 C．买一张电影票，座位号正好是偶数 D．13个同学中，至少有2人出生的月份相同

6．九年级（1）班10名同学在某次“1分钟仰卧起坐”的测试中，成绩如下（单位：次）：39，45，40，44，37，39，46，40，41，39，这组数据的众数、中位数分别是（ ） A．39，40 B．39，38 C．40，38 D．40，39

7．如图, △ABC是边长为2的等边三角形，将△ABC沿射线BC向右平移得到△DCE，连接AD、BD，下列结论错误的是（ ） ．．A．AD//BC B．AC⊥BD

C．四边形ABCD 面积为43 D．四边形ABED是等腰梯形 8．点P （2，1）关于直线y=x对称的点的坐标是（ ）

A．（?2，1） B．（2，?1） C．（?2，?1） D．（1，9．如图，已知圆锥的高为4，底面圆的直径为6，则此圆锥的侧面积

2240?144672）

1

是（ ）

A．12π B．15π C．24π D．30π

k10．如图，直线l和双曲线y?（k?0）交于A、B两点，P是线

x段AB上的点（不与A、B重合），过点A、B、P分别向x轴 作垂线，垂足分别为C、D、E，连接OA、OB、OP，设△AOC 的面积为S1、△BOD的面积为S2、△POE的面积为S3， 则有（ ）

A．S1?S2?S3 B．S1?S2?S3 C． S1?S2?S3 D．S1?S2?S3

二、填空题（共6小题，每小题4分，满分24分．请将答案填入答题卡的相应位置） ．．．11．化简：12?3= ．

12． 分解因式：ax?4ax?4a= ．

13． 已知一个多边形的内角和等于900，则这个多边形的边数是 ． 14． 如图，△ABC 内接于⊙O，∠C=30，AB=5，则⊙O的直径为 ．

15．袋中装有2个红球和2个白球，它们除了颜色外都相同．随机从中

摸出一球，记下颜色后放回袋中，再随机摸出一球，则两次都摸到 红球的概率是 ．

16．根据下列5个图形及相应点的个数的变化规律，试猜测第n个图中有 个点．

?? （1） （2） （3） （4） （5）

三、解答题（共7小题，满分86分．请将解答过程写在答题卡的相应位置．作图或添辅助线先用铅笔画完，再．．．用水笔描黑）

17．（每小题8分，满分16分） （1）化简：(

?2?11a?)?2； a?3a?3a?9?4x?3??x，?（2）解不等式组?x?4x?21并把解集在数轴上表示出来．

?≤，?63?2

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18．（本题满分10分）

如图，A、B两点分别位于一个池塘的两端，由于受条件限制无法直接度量A、B间的距离．小明利用学过的知识，设计了如下三种测量方法，如图①、②、③所示（图中a，b，c．．．表示长度，?，?，?．．．表示角度）．

（1）请你写出小明设计的三种测量方法中AB的长度：

图①AB= ，图②AB= ，图③AB= ；（6分）

（2）请你再设计一种不同于以上三种的测量方法,画出示意图（不要求写画法），用字母

标注需测量的边或角,并写出AB的长度． （4分）

19．（本题满分10分）

2009年4月1日《三明日报》发布了“2008年三明市国民经济和社会发展统计公报”，根据其中农林牧渔业产值的情况，绘制了如下两幅统计图，请你结合图中所给信息解答下列问题： （1）2008年全市农林牧渔业的总产值为 亿元；（2分）

（2）扇形统计图中林业所在扇形的圆心角为 度（精确到度）；（2分） （3）补全条形统计图；（2分）

（4）三明作为全国重点林区之一，市政府大力发展林业产业，计划2010年林业产值达60.5亿元，求今明两

年林业产值的年平均增长率． （4分）

（第18题备用图）

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20． （本题满分12分）

如图，在直角梯形ABCD中，AB∥CD，?B?90?，AB=AD，∠BAD的平分线交BC于E，连接DE． （1）说明点D在△ABE的外接圆上；（6分）

（2）若∠AED=∠CED，试判断直线CD与△ABE外接圆的位置关系，并说明理由．（6分）

21．（本题满分12分）

为把产品打入国际市场，某企业决定从下面两个投资方案中选择一个进行投资生产．方案一：生产甲产品，每件产品成本为a万美元（a为常数，且3＜a＜8），每件产品销售价为10万美元，每年最多可生产200件；方案二：生产乙产品，每件产品成本为8万美元，每件产品销售价为18万美元，每年最多可生产120件．另外，年销售x件乙产品时需上交0.05x万美元的特别关税．在不考虑其它因素的情况下： ．．．

（1）分别写出该企业两个投资方案的年利润y1、y2与相应生产件数x（x为正整数）之间的函数关系式，并指出自变量的取值范围；（4分）

（2）分别求出这两个投资方案的最大年利润；（4分）

（3）如果你是企业决策者，为了获得最大收益，你会选择哪个投资方案？（4分）

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2

22．（本题满分12分）

已知：矩形ABCD中AD＞AB，O是对角线的交点，过O任作一直线分别交BC、AD于点M、N（如图①）． （1）求证：BM=DN；

（2）如图②，四边形AMNE是由四边形CMND沿MN翻折得到的，连接CN，求证：四边形AMCN是菱形； （3）在（2）的条件下，若△CDN的面积与△CMN的面积比为1︰3，求

23．（本题满分14分）

如图，在平面直角坐标系xOy中，抛物线y??MN的值． DN12x?bx?c与x轴交于A（1，0）、 2B（5，0）两点．

（1）求抛物线的解析式和顶点C的坐标；（4分）

（2）设抛物线的对称轴与x轴交于点D，将∠DCB绕点C按顺时针方向旋转，角的两边CD和CB与x轴分

别交于点P、Q，设旋转角为?（0???≤90?）． ①当?等于多少度时，△CPQ是等腰三角形？（5分）

②设BP?t，AQ?s，求s与t之间的函数关系式．（5分）

5

2009年三明市初中毕业暨高级中等学校招生统一考试

数学试卷参考答案及评分标准

说明：以下各题除本卷提供的解法外，其他解法本标准不一一例举，评卷时可参考评分标准，按相应给分段评分．用

计算器计算的部分，列式后可直接得到结果．

一、选择题（每小题4分，共40分）

1． B 2． C 3． B 4．A 5．D 6．A 7．C 8．D 9．B 10．C 二、填空题（每小题4分，共24分） 11．

23 12． a(x?2) 13． 7 14． 10 15．

12 16． n?n?1 4三、解答题（共86分） 17．（1）解法一：原式=

=

=

(a?3)?(a?3)a?2 ································································ 3分

(a?3)(a?3)a?92a(a?3)(a?3) ······························································· 5分 ?(a?3)(a?3)a2a ·························································································· 7分 a6

=2 ································································································ 8分

解法二：原式=（11a2a?3?a?3）??9a ····························································· 2分 =1(a?3)(a?3)a?3?a+1(a?3)(a?3)a?3?a ·········································· 4分 =a?3a?3a?a ··················································································· 6分 =

2aa ···························································································· 7分 =2 ································································································ 8分

（2）解：解不等式①，得 x??3， ·········································································· 3分 解不等式②，得 x≤3， ············································································ 6分 不等式①、②的解集在数轴上表示如下：

············································· 7分 ∴不等式组的解集为?3?x≤3． ·································································· 8分

7

18．解：（1）①a?tan? ② 2c ③ b （每空2分）

（2）示意图正确2分，AB表示正确2分．（注：本题方法多种，下面列出3种供参考）

方法1： AB=a2?b2 AB=c2?a2

方法2： 方法3：

AB=c AB=

ac b

19．解：（1） 221 （2） 81 （每空2分） （3）补全条形统计图正确（2分）

（4）设今明两年林业产值的年平均增长率为x． 根据题意，得

50(1?x)2?60.5 ····························································································· 2分

解得：x1?0.1=10% ，x2??2.1（不合题意，舍去） ······························ 3分 答：今明两年林业产值的年平均增长率为10%． ······································· 4分

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20．（1）证法一：∵∠B=90°，∴ AE是△ABE外接圆的直径． ··· …1分 D取AE的中点O，则O为圆心，连接OB、OD． ··············· 2分

C∵AB=AD，∠BAO=∠DAO，AO=AO，

E∴△AOB≌△AOD． ··············································· 4分

O∴OD=OB． ··········································································· 5分 AB∴点D在△ABE的外接圆上． ············································ 6分 证法二：∵∠B=90°，∴AE是△ABE外接圆的直径． ······································· 1分 ∵AB=AD，∠BAE=∠DAE，AE=AE， ∴△ABE≌△ADE． ·············································································· 3分 ∴∠ADE=∠B=90°． ················································································ 4分

取AE的中点O， 则O为圆心，连接OD，则OD=12AE．

∴点D在△ABE的外接圆上． ·································································· 6分 （2）证法一：直线CD与△ABE的外接圆相切． ························································· 7分

理由：∵AB∥CD， ∠B=90°． ∴∠C=90°． ····················································· 8分 ∴∠CED+∠CDE=90°． ········································································· 9分 又∵OE=OD， ∴∠ODE=∠OED． ································································· 10分 又∠AED=∠CED， ∴∠ODE=∠DEC．

∴?ODC?∠CDE+∠ODE=∠CDE+∠CED=90°． ················································ 11分 ∴CD与△ABE的外接圆相切． ··························································· 12分 证法二： 直线CD与△ABE的外接圆相切． ························································ 7分 理由：∵AB∥CD， ∠B=90°． ∴∠C=90°． ·························································· 8分 又∵OE=OD， ∴∠ODE=∠OED． ···················································· 9分 又∠AED=∠CED，∴∠ODE=∠DEC． ······················································· 10分 ∴OD∥BC．

∴?ODC?90?． ··············································································· 11分 ∴CD与△ABE的外接圆相切． ······································································ 12分 21．解：（1）y1?(10?a)x （1≤x≤200，x为正整数） ··································· 2分

y2?10x?0.05x2 （1≤x≤120，x为正整数） ·················································· 4分

（2）①∵3＜a＜8， ∴10-a＞0，即y1随x的增大而增大 ， ·································· 5分

∴当x=200时，y1最大值=（10-a）×200=2000-200a（万美元） ··················· 6分

②y22??0.05(x?100)?500 ·································································· 7分

∵-0.05＜0， ∴x=100时， y2最大值=500（万美元） ·································· 8分

（3）由2000-200a＞500，得a＜7.5， ∴当3＜a＜7.5时，选择方案一； ······································································· 9分 由2000?200a?500，得 a?7.5，

∴当a=7.5时，选择方案一或方案二均可； ························································ 10分 由2000?200a?500，得 a?7.5， ∴当7.5＜a＜8时，选择方案二． ···································································· 12分

22．（1）证法一：连接BD，则BD过点O．

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∵AD∥BC， ∴∠OBM=∠ODN． ···················· 1分 又OB=OD， ∠BOM=∠DON， ······················ 2分 ∴△OBM≌△ODN． ····························· 3分 ∴BM=DN． ···························· 4分 证法二：∵矩形ABCD是中心对称图形，点O是对称中心． ········································

∴B、D和M、N关于O点中心对称． ····························· 3分 ∴BM=DN． ····································································· 4分 （2）证法一：∵矩形ABCD，

∴AD∥BC，AD=BC． 又BM=DN， ∴AN=CM． ························ 5分

∴四边形AMCN是平行四边形． ························· 6分

由翻折得，AM=CM， ························ 7分 ∴四边形AMCN是菱形． ························· 8分 证法二：由翻折得，AN=NC，AM=MC， ∠AMN=∠CMN． ······················································· 5分 ∵AD∥BC， ∴∠ANM=∠CMN．

∴∠AMN=∠ANM． ∴AM=AN． ·································································· 6分 ∴AM=MC=CN=NA． ··············································································· 7分 ∴四边形AMCN是菱形． ·········································································· 8分 （3）解法一：∵S1?CDN?2DN?CD，S1?CMN?2CM?CD，

又S?CDN：S?CMN=1︰3，

∴DN︰CM=1︰3 ············································· 9分 设DN=k，则CN=CM=3k． 过N作NG⊥MC于点G，

则CG=DN=k，MG=CM-CG=2k． ························ 10分 NG=CN2?CG2?9k2?k2?22k

∴MN=MG2?NG2?4k2?8k2?23k ································································ 11分 ∴

DNMN?k23k?36． ··························································· 12分 解法二：∵S12?CD，S1?CDN?DN?CMN?2CM?CD，

又S?CDN：S?CMN=1︰3， ∴DN︰CM=1︰3 ················································· 9分

连接AC，则AC过点O，且AC⊥MN． 设DN=k，则CN=AN=CM=3k，AD=4 k．

CD=NC2?DN2?9k2?k2?22k ···················································· 10分 OC=12AC?12AD2?CD2?1216k2?8k2?6k

∴MN=2ON?2CN2?OC2?29k2?6k2?23k ··················································· 11分 ∴

DNMN?k23k?36． ···························································································· 12分 10

??1?b?c?．解：（1）根据题意，得 ???20, ······························································ 1分

????252?5b?c?0.? 解得?b?3,?5 ·········································· 2分 ??c?? ·

2.∴y??12x2?3x?52 ······················································ 3分 =?12(x?3)2?2

∴顶点C的坐标为（3，2）． ············································································· 4分 （2）①∵CD=DB=AD=2，CD⊥AB， ∴∠DCB=∠CBD=45°． ············································· 5分

ⅰ）若CQ=CP，则∠PCD=

12∠PCQ=22.5°． ∴当?=22.5°时，△CPQ是等腰三角形． ·············· 6分 ⅱ）若CQ=PQ，则∠CPQ=∠PCQ=45°， 此时点Q与D重合，点P与A重合． ∴当?=45°时，

△CPQ是等腰三角形． ············································ 7分 ⅲ）若PC=PQ， ∠PCQ=∠PQC=45°，此时点Q与B重合，点P与D重合．

∴?=0°，不合题意． ··········································· 8分 ∴当?=22.5°或45°时，△CPQ是等腰三角形． ················································ 9分 ②连接AC，∵AD=CD=2，CD⊥AB，

∴∠ACD=∠CAD=45?， AC= BC=22?22?22 ················································ 10分 ⅰ）当0???≤45?时，

∵∠ACQ=∠ACP+∠PCQ=∠ACP+45°． ∠BPC=∠ACP+∠CAD=∠ACP+45°．

∴∠ACQ=∠BPC． ···························································· 11分 又∵∠CAQ=∠PBC=45°， ∴△ACQ∽△BPC． ∴AQACBC?BP． ∴AQ·BP=AC·BC=22×22=8 ·············································································· 12分

ⅱ）当45????90?时，同理可得AQ·BP=AC·BC=8 ∴s?8t． 14分

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??1?b?c?．解：（1）根据题意，得 ???20, ······························································ 1分

????252?5b?c?0.? 解得?b?3,?5 ·········································· 2分 ??c?? ·

2.∴y??12x2?3x?52 ······················································ 3分 =?12(x?3)2?2

∴顶点C的坐标为（3，2）． ············································································· 4分 （2）①∵CD=DB=AD=2，CD⊥AB， ∴∠DCB=∠CBD=45°． ············································· 5分

ⅰ）若CQ=CP，则∠PCD=

12∠PCQ=22.5°． ∴当?=22.5°时，△CPQ是等腰三角形． ·············· 6分 ⅱ）若CQ=PQ，则∠CPQ=∠PCQ=45°， 此时点Q与D重合，点P与A重合． ∴当?=45°时，

△CPQ是等腰三角形． ············································ 7分 ⅲ）若PC=PQ， ∠PCQ=∠PQC=45°，此时点Q与B重合，点P与D重合．

∴?=0°，不合题意． ··········································· 8分 ∴当?=22.5°或45°时，△CPQ是等腰三角形． ················································ 9分 ②连接AC，∵AD=CD=2，CD⊥AB，

∴∠ACD=∠CAD=45?， AC= BC=22?22?22 ················································ 10分 ⅰ）当0???≤45?时，

∵∠ACQ=∠ACP+∠PCQ=∠ACP+45°． ∠BPC=∠ACP+∠CAD=∠ACP+45°．

∴∠ACQ=∠BPC． ···························································· 11分 又∵∠CAQ=∠PBC=45°， ∴△ACQ∽△BPC． ∴AQACBC?BP． ∴AQ·BP=AC·BC=22×22=8 ·············································································· 12分

ⅱ）当45????90?时，同理可得AQ·BP=AC·BC=8 ∴s?8t． 14分

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