中考数学试题(word版含答案)

初中毕业生学业考试

数 学 试 卷

※考试时间120分钟 试卷满分150分

一、选择题（下列各题的备选答案中，只有一个是正确的，请将正确答案的选项填在下表中相应题号下的空格内．每小题3分，共24分）

1．目前国内规划中的第一高楼上海中心大厦，总投入约14 800 000 000元．14 800 000 000元用科学记数法表示为（ ） A．1.481011元

B．0.148109元

C．1.481010元

D．14.8109元

2．计算(2a2)3的结果为（ ） A．2a5 B．8a6 C．8a5

D．6a6

E A 45°

3．如图所示，已知直线AB∥CD，C125°，A45°， 则E的度数为（ ） A．70° C．90° B．80° D．100° B

125°

4．一个圆柱体钢块，正被挖去了一个长方体孔，其俯视图如图所示，则此圆柱体钢块的左视图是（ ） ． C 第3题图

D

A． B． C． D． 5．数据21，21，21，25，26，27的众数、中位数分别是（ ） A．21，23 B．21，21 C．23，21 D．21，25

6．为了美化环境，某市加大对绿化的投资．2007年用于绿化投资20万元，2009年用于绿化投资25万元，求这两年绿化投资的年平均增长率．设这两年绿化投资的年平均增长率为x，根据题意所列方程为（ ） A．20x25

2俯视图 第4题图 2

B．20(1x)25

D．20(1x)20(1x)25

2C．20(1x)25

7．如图所示，反比例函数y1与正比例函数y2的图象的一个交点坐标是A(2，1)，若

y2y10，则x的取值范围在数轴上表示为（ ）

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A． 0

1

2

B． 0

1

2

y 2 1 1O 0

1

2

A(2，1) y2 x C． 0 1 2

D． 1 2 y1 第7题图

8．将一等腰直角三角形纸片对折后再对折，得到如图所示的图形，然后将阴影部分剪掉，把剩余部分展开后的平面图形是（ ） B． C． 第8题图 二、填空题（每小题3分，共24分） 9．分解因式：a34a ． 10．函数y3x3垂直 A． D． 自变量x的取值范围是 ． 11．小丽想用一张半径为5cm的扇形纸片围成一个底面半径为4cm的圆锥，接缝忽略不计，则扇形纸片的面积是 cm2．（结果用π表示） 12．如图所示，小区公园里有一块圆形地面被黑白石子铺成了面积相等的八部分，阴影部分是黑色石子，小华随意向其内部抛一个小球，则小球落在黑色石子区域内的概率是 ． POA40°，C13．如图所示，AB为⊙O的直径，P点为其半圆上一点，为另一半圆上任意一点（不含A、B），则PCB 度． C y A P 第13题图 40° 第12题图 O B 1 O x 第14题图 214．已知抛物线yaxbxc（a0）经过点(1，0)，且顶点在第一象限．有下列三个结论：①a0 ②abc0 ③横线上 ．

b2a0．把正确结论的序号填在

② ① A B C ③ 第15题图

15．如图所示，在正方形网格中，图①经过 变换（填“平移”或“旋转”或“轴对称”）可以得到图②；图③是由图②经过旋转变换得到的，其旋转中心是点 （填“A”或“B”或“C”）． 16．如图所示，把同样大小的黑色棋子摆放在正多边形的边上，按照这样的规律摆下去，则第n个图形需要黑色棋子的个数是 ．

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第1个图形 第2个图形 第3个图形 第4个图形

第16题图

三、解答题（每题8分，共16分） 17．计算：12|32|(2π)0．

18．解方程：

四、解答题（每题10分，共20分）

19．如图所示，在Rt△ABC中，C90°，A30°．

（1）尺规作图：作线段AB的垂直平分线l（保留作图痕迹，不写作法）；

xx12x11．

（2）在已作的图形中，若l分别交AB、AC及BC的延长线于点D、E、F，连接BE． 求证：EF2DE．

A

C B 第19题图

20．某市开展了干部“一帮一扶贫”活动．为了解贫困群众对帮扶情况的满意程度，有关部门在该市所管辖的两个区内，分别随机抽取了若干名贫困群众进行问卷调查．根据收集的信息进行了统计，并绘制了下面尚不完整的统计图．已知在甲区所调查的贫困群众中，非常满意的人数占甲区所调查的总人数的35%．根据统计图所提供的信息解答下列问题： （1）甲区参加问卷调查的贫困群众有 人； （2）请将统计图补充完整；

（3）小红说：“因为甲区有30人不满意，乙区有40人不满意，所以甲区的不满意率比乙区低．”你认为这种说法正确吗？为什么？ 人数

800 700 600 420 400 200 760 500 250 40 30 甲 乙 非常满意 满意 比较满意 不满意 满意程度 第20题图

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12999数学网[www.12999.com] 精品资料下载 免费下载 五、解答题（每题10分，共20分）

21．小明和小亮是一对双胞胎，他们的爸爸买了两套不同品牌的运动服送给他们，小明和小亮都想先挑选．于是小明设计了如下游戏来决定谁先挑选．游戏规则是：在一个不透明的袋子里装有除数字以外其它均相同的4个小球，上面分别标有数字1、2、3、4．一人先从袋中随机摸出一个小球，另一人再从袋中剩下的3个小球中随机摸出一个小球．若摸出的两个小球上的数字和为奇数，则小明先挑选；否则小亮先挑选． （1）用树状图或列表法求出小明先挑选的概率； （2）你认为这个游戏公平吗？请说明理由．

22．如图所示，已知AB是半圆O的直径，弦CD∥AB，AB10，CD6，E是AB延长线上一点，BE

六、解答题（每题10分，共20分）

A 103．判断直线DE与半圆O的位置关系，并证明你的结论．

D O B 第22题图

E

C 23．某旅游区有一个景观奇异的望天洞，D点是洞的入口，游人从入口进洞游览后，可经山洞到达山顶的出口凉亭A处观看旅游区风景，最后坐缆车沿索道AB返回山脚下的B处．在同一平面内，若测得斜坡BD的长为100米，坡角DBC10°，在B处测得A的仰角ABC40°，在D处测得A的仰角ADF85°，过D点作地面BE的垂线，垂足为C．

（1）求ADB的度数； （2）求索道AB的长．（结果保留根号）

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B

A

D

C

第23题图

F E

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24．为迎接国庆六十周年，某校团委组织了“歌唱祖国”有奖征文活动，并设立了一、二、三等奖．学校计划派人根据设奖情况买50件奖品，其中二等奖件数比一等奖件数的2倍还少10件，三等奖所花钱数不超过二等奖所花钱数的1.5倍．各种奖品的单价如下表所示．如果计划一等奖买x件，买50件奖品的总钱数是w元． （1）求w与x的函数关系式及自变量x的取值范围；

（2）请你计算一下，如果购买这三种奖品所花的总钱数最少？最少是多少元？ 七、解答题（本题12分） 25．△ABC是等边三角形，点D是射线BC上的一个动点（点D不与点B、C重合），△ADE是以AD为边的等边三角形，过点E作BC的平行线，分别交射线AB、AC于点F、G，连接BE．

单价(元)

一等奖 12

二等奖 10

三等奖 5

（1）如图（a）所示，当点D在线段BC上时． ①求证：△AEB≌△ADC； ②探究四边形BCGE是怎样特殊的四边形？并说明理由； （2）如图（b）所示，当点D在BC的延长线上时，直接写出（1）中的两个结论是否成立？ （3）在（2）的情况下，当点D运动到什么位置时，四边形BCGE是菱形？并说明理由．

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第25题图

E B F D 图（a） A A G C B C D

F E

图（b）

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八、解答题（本题14分）

26．如图所示，已知在直角梯形OABC中，AB∥O，C⊥BC轴于点

动点P从O点出发，沿x轴正方向以每秒1个单位长度的速度移动．过C，A(，、11)B，(．3，△OP点作PQ垂直于直线PQ．．OA，垂足为Q．设P点移动的时间为t秒（0t4）直角梯形OABC重叠部分的面积为S．

（1）求经过O、A、B三点的抛物线解析式； （2）求S与t的函数关系式；

（3）将△OPQ绕着点P顺时针旋转90°，是否存在t，使得△OPQ的顶点O或Q在抛物线上？若存在，直接写出t的值；若不存在，请说明理由． y

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2 1 Q O P 1 第26题图 A B C 3 与

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2009年铁岭市初中毕业生学业考试 数学试题参及评分标准

注：本参只给出一种或几种解法（证法），若用其他方法解答并正确，可参考此评分标准相应步骤赋分．

一、选择题（每小题3分，共24分） 题号 1 2 3 4 答案 C B B C 二、填空题（每小题3分，共24分）

9．a(a2)(a2) 10．x3 11．20π 12．125 A

6 C

7 D

8 A

13．70 14．①②③ 15．平移（2分）；A（3分） 16．n(n2)或n22n或(n1)21 三、（每题8分，共16分） 17．解：原式232 331·················································································· 6分

3 ································································································· 8分

18．解：方程两边分别乘以(x1)(x1)得 ··························································································· 3分 x(x1)2(x1)x1 ·xx2x2x1 x3 ································································································· 7分

222检验：当x3时，(x1)(x1)0（或分母不等于0） ∴x3是原方程的根．································································································ 8分 四、（每题10分，共20分） 19．（1）直线l即为所求． ········································ 1分 作图正确． ······························································· 3分 （2）证明：在Rt△ABC中， A30°，ABC60°， A l D E F C

第19题图

B

又∵l为线段AB的垂直平分线， ∴EAEB， ·························································· 5分 ∴EBAA30°，AEDBED60°， ∴EBC30°EBA，FEC60°． 又∵ED⊥AB，EC⊥BC，

∴EDEC． ············································································································· 8分

EFC30°， 在Rt△ECF中，FEC60°，∴EF2EC， ∴EF2ED． ··········································································································10分

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12999数学网[www.12999.com] 精品资料下载 免费下载 20．（1）1200················································································································ 3分 （2）图形正确（甲区满意人数有500人） ···································································· 5分 （3）不正确． ·············································································································· 6分 ∵甲区的不满意率是

3012002.5%，乙区的不满意率是

40700760500402%，

∴甲区的不满意率比乙区的不满意率高．·····································································10分 五、（每题10分，共20分） 21．解：（1）根据题意可列表或树状图如下： 第一次 1 2 第二次 1 2 3 4 第一次摸球 2 第二次摸球—— （2，1） （3，1） （4，1） （1，2） —— （3，2） （4，2） 3 （1，3） （2，3） —— （4，3） 4 （1，4） （2，4） （3，4） —— ··························································································· 5分 1 2 3 4 3 4 1 3 4 1 2 4 1 2 3 （1，2） （1，3） （1，4） （1，1） （2，3） （2，4） （3，1） （3，2） （3，4）（ 4，1） （4，2）（4，3） ·········································································································· 5分 从表或树状图可以看出所有可能结果共有12种，且每种结果发生的可能性相同，符合条件的结果有8种， ∴P（和为奇数）23 ································································································· 7分 （2）不公平． ·············································································································· 8分 ∵小明先挑选的概率是P（和为奇数）∵231323，小亮先挑选的概率是P（和为偶数）13， ，∴不公平． ·································································································10分 22．直线DE与半圆O相切．······················································································· 1分 证明：法一： 连接OD，作OF⊥CD于点F． ∵CD6，∴DF12CD3． ··································· 2分 103253C A

F D ∵OEOBBE5DF．·································· 3分

B E 3OD53，， ∴第22题图 25OD5OE53DFOD∴．············································································································ 6分 ODOE∵CD∥AB，∴CDODOE．·········································································· 7分 『12999数学网』收集整理 第8页 共12页 欢迎下载教学资料

O 12999数学网[www.12999.com] 精品资料下载 免费下载 ∴△DOF∽△OED， ······························································································· 8分 ∴ODEOFD90°， ∴OD⊥DE

∴直线DE与半圆O相切． ·························································································10分 法二：连接OD，作OF⊥CD于点F，作DG⊥OE于点G． ∵CD6，∴DF12CD3．

在Rt△ODF中，OFODDF22··············································· 3分 534 ·

22∵CD∥AB，DG⊥AB，OF⊥CD， ∴四边形OFDG是矩形，

∴DGOF4，OGDF3． ∵OEOBBE51032532，GEOEOG25323163， ···························· 5分

在Rt△DGE中，DEDGGE22016． 433220252∵5， 3322∴OD2DE2OE2 ··································································································· 8分 ∴CD⊥DE． ∴直线DE与半圆O相切． ·························································································10分 六、（每题10分，共20分） 23．（1）解：∵DC⊥CE，∴BCD90°． 又∵DBC10°， ∴BDC80°，····························································· 1分 ∵ADF85°， ∴ADB360°80°90°85°105°． ····················· 2分 （2）过点D作DG⊥AB于点G． ·································· 3分 在Rt△GDB中，GBD40°10°30°， ∴BDG90°30°60 ············································· 4分 B 又∵BD100， ∴GD12BD1001250．

A G

D C

第23题图

F E

GBBDcos30°10032503． ····································································· 6分

在Rt△ADG中，GDA105°60°45 ······························································ 7分 ∴GDGA50，······································································································ 8分 ∴ABAGGB50503（米） ········································································· 9分

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答：索道长50503米． ··························································································10分 24．解：（1）12x10(2x10)5[50x(2x10)] ·········································· 2分

17x200． ····················································································· 3分

x02x100由 ································································ 5分

[50x(2x10)]05[50x(2x10)]≤1.510(2x10)得10≤x20 ············································································································· 6分 ∴自变量的取值范围是10≤x20，且x为整数． ····················································· 7分 （2）∵k170，∴随x的增大而增大，当x10时，有最小值． ·················· 8分 最小值为1710200370． ············································································· 9分 答：一等奖买10件，二等奖买10件，三等奖买30件时，所花的钱数最少， 最少钱数是370元． ····································································································10分 七、（本题12分）

25．（1）①证明：∵△ABC和△ADE都是等边三角形， ∴AEAD，ABAC，EADBAC60°． ······· 1分 又∵EABEADBAD，DACBACBAD， ∴EABDAC，

∴△AEB≌△ADC． ··················································· 3分 ②法一：由①得△AEB≌△ADC， ∴ABEC60°． 又∵BACC60°，

∴ABEBAC， ∴EB∥GC． ································································ 5分

F B

D

图（a） 第25题图

A E G C

又∵EG∥BC，

∴四边形BCGE是平行四边形． ·················································································· 6分 法二：证出△AEG≌△ADB，

得EGABBC． ···································································································· 5分 由①得△AEB≌△ADC．

得BECG．

∴四边形BCGE是平行四边形． ·················································································· 6分 （2）①②都成立． ······································································································· 8分 （3）当CDCB（BD2CD或CD12BD或CAD30°或BAD90°或

ADC30°）时，四边形BCGE是菱形．··················· 9分 理由：法一：由①得△AEB≌△ADC，

A ∴BECD ·································································· 10分 又∵CDCB，

∴BECB． ······························································· 11分 由②得四边形BCGE是平行四边形， ∴四边形BCGE是菱形．·············································· 12分

B C E

图（b）

G

D

F 第25题图

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12999数学网[www.12999.com] 精品资料下载 免费下载 法二：由①得△AEB≌△ADC，

∴BECD．·············································································································· 9分 又∵四边形BCGE是菱形， ∴BECB ················································································································ 11分 ∴CDCB．·············································································································12分 法三：∵四边形BCGE是平行四边形， ∴BE∥CG，EG∥BC，

∴FBEBAC60°，FABC60° ··························································· 9分 ∴FFBE60°，

∴△BEF是等边三角形．···························································································10分 又∵ABBC，四边形BCGE是菱形， ∴ABBEBF，

∴AE⊥FG ··············································································································· 11分 ∴EAG30°，

∵EAD60°， ∴CAD30°．·······································································································12分 八、解答题（本题14分） 26．解：（1）法一：由图象可知：抛物线经过原点， 设抛物线解析式为yax2bx(a0)． 把A(1，··················································································· 1分 1)，B(3，1)代入上式得：·1a1ab3解得 ······················································································ 3分 19a3b1b43∴所求抛物线解析式为y法二：∵A(1，1)，B(3，1)， 13x243x ········································································ 4分

∴抛物线的对称轴是直线x2． 2设抛物线解析式为ya(x2)h（a0） ····························································· 1分

把O(0，0)，A(1，1)代入得 1a20a(02)h3 解得 ··········································································· 3分 21a(12)hh43∴所求抛物线解析式为y（2）分三种情况：

13(x2)243x． ···························································· 4分

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12999数学网[www.12999.com] 精品资料下载 免费下载 ①当0t≤2，重叠部分的面积是S△OPQ，过点A作AF⊥x轴于点F， ∵A(1，1)，在Rt△OAF中，AFOF1，AOF45°， 在Rt△OPQ中，OPt，OPQQOP45°，

22y 2 1 Q A F O P 1 第26题图1 B C 3 ∴PQOQtcos45°12∴S222t，

x 12·············································· 6分 tt． ·4②当2t≤3，设PQ交AB于点G，作GH⊥x轴于点H， OPQQOP45°，则四边形OAGP是等腰梯形， y 2 Q 1 A G B C P 3 重叠部分的面积是S梯形OAGP． ∴AGFHt2， ∴S12(AGOP)AF12(tt2)1t1． ··········· 8分 O F 1 H x 第26题图2 ③当3t4，设PQ与AB交于点M，交BC于点N，重叠部分的面积是S五边形OAMNC． 因为△PNC和△BMN都是等腰直角三角形，所以重叠部分的面积是S五边形OAMNS梯形OABCS△BMN． y 2 1 A M O F 1 第26题图3 N Q B C 3 P ∵B(3，1)，OPt， ∴PCCNt3， ∴BMBN1(t3)4t， ∴S12(23)112t4t2 S(4t) 211212x ． ················································ 10分 （3）存在 t11 ····································································································12分 t22 ···································································································14分

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