2009年山西省太原市中考数学试卷及答案(纯Word版)
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2009年山西省太原市初中毕业学业考试试卷
数 学
一、选择题(本大题含10个小题,每小题3分,共30分)
在每小题给出的四个选项中,只有一项符合题目要求,请选出并在答题卡上将该项涂黑. 1.在数轴上表示?2的点离开原点的距离等于( )
A.2
236B.?2 C.?2 D.4
232.下列计算中,结果正确的是( )
A.a·a?a B.?2a·??3a??6a C.a表.这个班学生体育测试成绩的众数是( ) ???a6 D.a6?a2?a3
3.学业考试体育测试结束后,某班体育委员将本班50名学生的测试成绩制成如下的统计成绩(分) 人数(人) 20 1 21 1 222 2 23 4 24 5 225 6 26 5 27 8 28 10 29 6 30 2 A.30分 B.28分 C.25分 D.10人 4.已知一个多项式与3x?9x的和等于3x?4x?1,则这个多项式是( ) A.?5x?1 B.5x?1 C.?13x?1 D.13x?1 5.用配方法解方程x?2x?5?0时,原方程应变形为( ) A.?x?1??6 C.?x?2??9
222B.?x?1??6
A?2A
D.?x?2??9
2B D.40°
6.如图,△ACB≌△A?C?B?,?BCB?=30°,则?ACA?的度数为( ) A.20°
B.30°
C.35°
B?7.如图,在Rt△ABC中,?C=90°,AB=10,若以点C为圆心,
C
CB长为半径的圆恰好经过AB的中点D,则AC的长等于( )
A.53
B.5 C.52 D.6
A C 8.如果三角形的两边分别为3和5,那么连接这个三角形三边中点 所得的三角形的周长可能是( )
A.4 B.4.5 C.5
D.5.5
D B 9.如图,AB是半圆O的直径,点P从点O出发,沿OA??设OPAB?BO的路径运动一周.为s,运动时间为t,则下列图形能大致地刻画s与t之间关系的是( ) A
A. B. C. D. 10.在二行三列的方格棋盘上沿骰子的某条棱翻动骰子(相对面上分别标有1点和6点,2点和5点,3点和4点),在每一种翻动方式中,骰子不能后退.开始时骰子如图(1)那
O
B O t O t O t O t P s s s s http://passport.http://www.wodefanwen.com//?business&aid=6&un=pkszjb#2 第 1 页 共 12 页
样摆放,朝上的点数是2;最后翻动到如图(2)所示的位置,此时骰子朝上的点数不可..能是下列数中的( ) .
图(1) 图(2) A.5 B.4 C.3 把答案填在题中的横线上或按要求作答. 11.计算
D.1 二、选择题(本大题含10个小题,每小题2分,共20分)
?2?的结果等于 .
212.若反比例函数的图象经过点A??21,?,则它的表达式是 .
13.自2005年以来,太原市城市绿化走上了快车道.目前我市园林绿化总面积达到了7101.5万平方米.这个数据用科学记数法表示为 万平方米.
25?的解是 . x?12x15.如图是一种贝壳的俯视图,点C分线段AB近似于黄金分割.已知AB=10cm,则AC的长约为 cm.(结果精确到0.1cm)
14.方程
16.甲、乙两盏路灯底部间的距离是30米,一天晚上,当小华
走到距路灯乙底部5米处时,发现自己的身影顶部正好接触路灯乙的底部.已知小华的身高为1.5米,那么路灯甲的高为 米.
小华乙 17.某种品牌的手机经过四、五月份连续两次降价,每部售价由 甲 3200元降到了2500元.设平均每月降价的百分率为x,根据题意列出的方程是 .
C AB?AAC⊙OC18.如图、是的两条弦,=30°,过点的切 线与OB的延长线交于点D,则?D的度数为 .
B 19.有两把不同的锁和三把钥匙,其中两把钥匙分别能打开其中一把
锁,第三把钥匙不能打开这两把锁,任意取出一把钥匙去开任意的一把锁,一次打开锁的概率为 .
20.如图,在等腰梯形ABCD中,AD∥BC,BC=4AD=42,A D F B E C D ?B=45°.直角三角板含45°角的顶点E在边BC上移动,一直角边始终经过点A,斜边与CD交于点F.若△ABE为等腰三角形,则CF的长等于 .
三、解答题(本大题含9个小题,共70分)
解答应写出文字说明、证明过程或演算步骤 21.(每小题满分5分)
化简:?A 1?1?4 ???2?x?4x?2?x?2http://passport.http://www.wodefanwen.com//?business&aid=6&un=pkszjb#2 第 2 页 共 12 页
22.(本小题满分5分)
已知,二次函数的表达式为y?4x2?8x.写出这个函数图象的对称轴和顶点坐标,并
求图象与x轴的交点的坐标.
23.(本小题满分6分)
某公司计划生产甲、乙两种产品共20件,其总产值w(万元)满足:1150<w<1200,相关数据如下表.为此,公司应怎样设计这两种产品的生产方案.
产品名称 甲 乙 每件产品的产值(万元) 45 75 24.(本小题满分8分)
如图,从热气球C上测得两建筑物A、B底部的俯角分别为30°和60°.如果这时气球的高度CD为90米.且点A、D、B在同一直线上,求建筑物A、B间的距离.
E C F 30°60°B A D http://passport.http://www.wodefanwen.com//?business&aid=6&un=pkszjb#2 第 3 页 共 12 页
25.(本小题满分8分)
为了解某校学生每周购买瓶装饮料的情况,课外活动小组从全校30个班中采用科学的方法选了5个班.并随机对这5个班学生某一天购买瓶装饮料的瓶数进行了统计,结果如下图所示.
(1)求该天这5个班平均每班购买饮料的瓶数;
(2)估计该校所有班级每周(以5天计)购买饮料的瓶数;
(3)若每瓶饮料售价在1.5元至2.5元之间,估计该校所有学生一周用于购买瓶装饮料
的费用范围.
瓶数/瓶
13 12 11 10 9
8 7 6 5 4 3
2 1 0
A B C D E 26.(本小题满分9分)
如图,A是?MON边OM上一点,AE∥ON.
(1)在图中作?MON的角平分线OB,交AE于点B;(要求:尺规作图,保留作图痕迹,不写作法和证明)
(2)在(1)中,过点A画OB的垂线,垂足为点D,交ON于点C,连接CB,将图形补充完整,并证明四边形OABC是菱形.
M E A O N
班数
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27(本小题满分8分)
某中学九年级有8个班,要从中选出两个班代表学校参加社区公益活动.各班都想参加,但由于特定原因,一班必须参加,另外从二至八班中再选一个班.有人提议用如下的方法:在同一个品牌的四个乒乓球上分别标上数字1,2,3,4,并放入一个不透明的袋中,摇匀后从中随机摸出两个乒乓球,两个球上的数字和是几就选几班,你认为这种方法公平吗?请用列表或画树状图的方法说明理由.
1 2 3 4
28.(本小题满分9分)
A、B两座城市之间有一条高速公路,甲、乙两辆汽车同时分别从这条路两端的入口
处驶入,并始终在高速公路上正常行驶.甲车驶往B城,乙车驶往A城,甲车在行驶过程中速度始终不变.甲车距B城高速公路入口处的距离y(千米)与行驶时间x(时)
之间的关系如图.
(1)求y关于x的表达式;
(2)已知乙车以60千米/时的速度匀速行驶,设行驶过程中,两车相距的路程为s(千
米).请直接写出s关于x的表达式;
(3)当乙车按(2)中的状态行驶与甲车相遇后,速度随即改为a(千米/时)并保
持匀速行驶,结果比甲车晚40分钟到达终点,求乙车变化后的速度a.在下图
中画出乙车离开B城高速公路入口处的距离y(千米)与行驶时间x(时)之间的函数图象.
29.(本小题满分12分)
360 300 240 180 120 60 O 1 2 345x/时 F
D
y/千米 问题解决
如图(1),将正方形纸片ABCD折叠,使点B落在CD边上一点E(不与点C,D重合),压平后得到折痕MN.当
A M CE1?时,CD2AM求的值. BN 方法指导:
为了求得AM的值,可先求BN、AM的长,不妨设:AB=2
BN
B
N
图(1)
E
C
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类比归纳
AMAMCE1CE1?,?,则的值等于 ;若则的值等
BNBNCD3CD4CE1AM?(n为整数)于 ;若,则的值等于 .(用含n的式子表示) CDnBN在图(1)中,若
联系拓广
如图(2),将矩形纸片ABCD折叠,使点B落在CD边上一点E(不与点C,D重合),压平后得到折痕MN,设
AMAB1CE1??m?1?,?,则的值等于 .(用含
BNBCmCDnF
A
M D E
m,n的式子表示)
B
N 图(2)
C
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2009年山西省太原市初中毕业生学业考试试卷
数学试题参考答案
一、选择题 题号 答案 1 A 2 C 3 B 4 A 5 B 6 B 7 A 8 D 9 C 10 D 二、填空题
11.2; 12.y??223; 13.7.1015×10; 14.x?5(或5); 15.6.2; 16.9 x217.3200?1?x??2500(或32x?64x?7?0或32(1?x)2?25)
18.30° 19.三、解答题 21.解:原式=?15 20.,2,42?3. 32??4x?21??················································ 2分
??x?2??x?2??x?2??x?2???x?2 ·??=
x?2·?x?2? ·················································································· 4分
?x?2??x?2?=1. ·················································································································· 5分
22.解:在y?4x2?8x中,a?4,b?8,c?0.
b84ac?b24?4?0?82????1,???4.∴? 2a2?44a4?4∴这个函数图象的对称轴是x??1,顶点坐标是:??1 ··················· 2分 ,?4?.评分说明:直接写出正确结果也得2分.
令y=0,则4x?8x?0. ············································································· 3.分 解得x1?0,x2??2. ························································································· 4分 ∴函数图象与x轴的交点的坐标为?0, ········································· 5分 0?,0?.??2,23.解:设计划生产甲产品x件,则生产乙产品?20?x?件, ·········································· 1分
2http://passport.http://www.wodefanwen.com//?business&aid=6&un=pkszjb#2 第 7 页 共 12 页
根据题意,得? 解得10?x???45x?75?20?x??1150, ·························································· 3分
??45x?75?20?x??1200.35. ······························································································ 4分 3. ?x为整数,∴x?11此时,20?x?9( 件). ·············································· 5分
答:公司应安排生产甲产品11件,乙产品9件. ···················································· 6分
,?FCB?60°,CD?90,24.解:由已知,得?ECA?30°
EF∥AB,CD?AB于点D.
,?B??FCB?60°. ??A??ECA?30° ···················································· 2分
,tanA= 在Rt△ACD中,?CDA?90° ?AD?CD, ADCD903??90??903. ··························································· 4分 tanA333CD, BD,tanB= 在Rt△BCD中,?CDB?90° ?DB?CD90??303. ············································································· 6分 tanB3 ?AB?AD?BD?903?303?1203(米).
答:建筑物A、B间的距离为1203米. ································································· 8分 25.解:(1)??8?9?12?11?10??10(瓶).
答:该天这5个班平均每班购买饮料10瓶. ···························································· 3分
(2)10?5?30?1500(瓶).
答:该校所有班每周购买饮料1500瓶. ···································································· 6分
(3)1.5?1500?2250(元),2.5?1500?3750(元).
答:该校所有班级学生一周用于购买瓶装饮料的费用为2250元至3750元. ········· 8分 26.解:(1)如图,射线OB为所求作的图形. ······························································ 3分
O A D C N B E M 15??AOB??BOC. (2)方法一:?OB平分?MON,
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??ABO??BOC. ?AE∥ON,
??AOB??ABO,AO?AB. ··········································································· 5分
?BD?OD. ?AD?OB, ···················································································· 6分
??ABD??COD,?BD?OD, 在△ADB和△CDO中???ADB??CDO,??△ADB≌△CDO,AB?OC. ········································································· 7分 ?AB∥OC,∴四边形OABC是平行四边形. ···················································· 8分 ?AO?AB,∴四边形OABC是菱形. ································································ 9分 AO?AB.方法二:同方法一, ?AOB??ABO, ············································ 5分
. ?AD?OB于点D,∴OD?DB,?ADO??CDO?90° ························· 6分
??AOD??COD,?OD?OD, 在△AOD和△COD中???ADO??CDO,?∴△AOD≌△COD,AD?CD. ········································································ 7分 ∴四边形OABC是平行四边形. ··········································································· 8分 ?AO?AB(或AC?OB),∴四边形OABC是菱形. ·································· 9分
27.解:这种方法不公平.一次摸球可能出现的结果列表如下: ··································· 4分
1 2 3. 4
由上表可知,一次摸球出现的结果共有16种可能的情况,且每种情况出现的可能性相同.其中和为2的一种,和为3的两种,和为4的三种,和为5的四种,和为6的三种,和为7的两种,和为8的一种. ······················································· 6分
1 2 3. 4 (1,1) (1,2) (1,3) (1,4) (2,1) (2,2) (2,3) (2,4) (3.,1) (3.,2) (3,3) (3.,4) (4,1) (4,2) (4,3) (4,4) 211?, ,P(和为3)=P(和为7)=
16816413P(和为4)=P(和为6)=,P(和为5)=?.
164161311??.所以?························································································· 7分
416816P(和为2)=P(和为8)=
因为二班至八班各班被选中的概率不全相等,所以这种方法不公平. ············ 8分
评分说明:只要计算出二至八班中有两个班被选中的概率不相等,就可得分. 28.解:(1)方法一:由图知y是x的一次函数,设y?kx?b.···································· 1分 ?图象经过点(0,300),(2,120),∴??b?300, ······························ 2分
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解得??k??90, ······························································································· 3分
?b?300. ∴y??90x?300.即y关于x的表达式为y??90x?300. ······················· 4分
方法二:由图知,当x?0时,y?300;x?2时,y?120.
所以,这条高速公路长为300千米. 甲车2小时的行程为300-120=180(千米).
∴甲车的行驶速度为180÷2=90(千米/时).··············································· 3分 ∴y关于x的表达式为y?300?90x(y??90x?300). ························ 4分
(2)s??150x?300. ····························································································· 5分 (3)在s??150x?300中.当s?0时,x?2.
即甲乙两车经过2小时相遇. ········································································ 6分 在y??90x?300中,当y?0,x?时间为
10.所以,相遇后乙车到达终点所用的3y/千米 360 300 240 180 120 60 O 1 2 345x/时
102??2?2(小时). 33 乙车与甲车相遇后的速度
a??300?2?60??2?90(千米/时).
∴a?90(千米/时). ······································ 7分 乙车离开B城高速公路入口处的距离y(千米)与行 驶时间x(时)之间的函数图象如图所示. ··············· 9分 29.问题解决
解:方法一:如图(1-1),连接BM,EM,BE.
F M A D
B
N 图(1-1)
C E
由题设,得四边形ABNM和四边形FENM关于直线MN对称.
∴MN垂直平分BE.∴BM?EM,BN?EN.················································ 1分 ∵四边形ABCD是正方形,∴?A??D??C?90°,AB?BC?CD?DA?2. ∵
CE1NC?2?x.?,?CE?DE?1.设BN?x,则NE?x,
CD2222 在Rt△CNE中,NE?CN?CE.
22 ∴x??2?x??1.解得x?255,即BN?.····················································· 3分 44http://passport.http://www.wodefanwen.com//?business&aid=6&un=pkszjb#2 第 10 页 共 12 页
在Rt△ABM和在Rt△DEM中,
AM2?AB2?BM2, DM2?DE2?EM2,
············································································· 5分 ?AM2?AB2?DM2?DE2.2222 设AM?y,则DM?2?y,∴y?2??2?y??1.
11 解得y?,即AM?. ······················································································· 6分
44AM1?. ∴ ············································································································ 7分 BN55 方法二:同方法一,BN?. ················································································ 3分
4 如图(1-2),过点N做NG∥CD,交AD于点G,连接BE.
F G M A D
E
B C N
图(1-2)
∵AD∥BC,∴四边形GDCN是平行四边形. ∴NG?CD?BC. 同理,四边形ABNG也是平行四边形.∴AG?BN?5. 4??EBC??BNM?90°. ∵MN?BE, ??MNG??BNM?90°,??EBC??MNG. ?NG?BC,
在△BCE与△NGM中
??EBC??MNG,? ?BC?NG,∴△BCE≌△NGM,EC?MG. ································ 5分
??C??NGM?90°.?∵AM?AG?MG,AM=∴
类比归纳
51?1?. ··································································· 6分 44AM1?. ·········································································································· 7分 BN52249?n?1? ·(或);; ·················································································10分 251017n?1联系拓广
n2m2?2n?1 ·············································································································12分
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