山西省中考数学试卷版含答案
山西省中考数学试卷版
含答案
TTA standardization office【TTA 5AB- TTAK 08- TTA 2C】
2009年山西省初中毕业学业考试试卷
数 学
一、填空题(每小题2分,共20分)
1.比较大小:2- 3-(填“>”、“=”或“<“).
2.山西有着丰富的旅游资源,如五台山、平遥古城、乔家大院等着名景点,吸引了众多的海内外游客,2008
年全省旅游总收入亿元,这个数据用科学记数法可表示
为 .
3.请你写出一个有一根为1的一元二次方程: . 4= .
5.如图所示,A 、B 、C 、D 是圆上的点,17040A ∠=∠=°,°, 则C ∠= 度.
6.李师傅随机抽查了本单位今年四月份里6天的日用水量(单位:吨)结果如下:7,8,8,7,6,6,根据这些数据,估计四月份本单位用水总量为 吨. 7.如图,ABC △与A B C '''△是位似图形,且顶点都在格点上,则位似中心的坐标是 .
相交于点O ,点E 是CD 的中点,ABD △的周.
1x <-时,y 的取值范围是 . .下列图案是晋商大院窗格的一部分,其中“○”代表窗纸上所贴的剪纸,则第n 个图中所贴剪纸“○”的个数为
.
A B C
D 1
(第5
A C B
E O (第8B C 3 4 5 6 7 8 9 10 (第7……
二、选择题(在下列各小题中,均给出四个备选答案,其中只有一个正确答案,请将正确答案的字母号填入下表相应的空格内,每小题3分,共24分) 11.下列计算正确的是( )
A .623a a a ÷=
B .()1
22--=
C .()236326x x x -=-·
D .()0
π31-= 12.反比例函数k
y x
=的图象经过点()23-,,那么k 的值是( )
A .32-
B .2
3
- C .6- D .6
13.不等式组21
318x x --??->
≥的解集在数轴上可表示为( )
A .
C
14.解分式方程11
222x x x
-+
=
--,可知方程( ) A .解为2x = B .解为4x = C .解为3x = D .无解
15.如图是由几个相同的小正方体搭成的几何体的三视图,则搭成这个几何体的小正方体的个数是( ) A .5
B .6
C .7
D .8
16.如图,AB 是AD 是O ⊙的切线,点C 在O ⊙上,BC OD ∥,
23AB OD ==,,则BC 的长为(
) A .23 B .3
2
C D .2
17.如图(1),把一个长为m 、宽为n 的长方形(m n >)沿虚线剪开,拼接成图(2),成为在一角去掉一个小正方形后的一个大正方形,则去掉的小正方形的边长为( )
A .2m n -
B .m n -
C .2m
D .2n 18.如图,在Rt ABC △中,90ACB ∠=°,3BC =,4AC =,AB 的垂
直平分线DE 交BC 的延长线于点E ,则CE 的长为( )
A .32
B .76
C .256
D .2
三、解答题(本题共76分)
主视左视俯视(第15
A B C
(第16
m
n n n (2
(1(第17 A D
B E
C (第18
19.(每小题4分,共12分) (1)计算:()()()2
312x x x +---
(2)化简:2222
42
x x x x +--- (3)解方程:2230x x --=
20.(本题6分)已知每个网格中小正方形的边长都是1,图1 中的阴影图案是由三段以格点为圆心,半径分别为1和2
(1)填空:图1中阴影部分的面积是 (结果保留π(2)请你在图2中以图1为基本图案,借助轴对称、平移或旋转设计 一个完整的花边图案(要求至少含有两种图形变换).
21.(本题8分)根据山西省统计信息网公布的数据,绘制了山西省2004~2008固定
电话和移动电话年末用户条形统计图如下:
(1
(222.(本题8里放有4个相同的小球,球上分别标有“0元”、“10元”、“20元”和“30元”的字样.规定:顾客在本商场同一日内,每消费满200元,就可以在箱子里先后摸出两个球(第一次摸出后不放回).商场根据两小球所标金额的和返还相应价格的购物券,可以重新在本商场消费.某顾客刚好消费200元.
(1)该顾客至少可得到 元购物券,至多可得到 元购物券;
(第20题 图(第20题 图
年份
(第21
(2)请你用画树状图或列表的方法,求出该顾客所获得购物券的金额不低于30元的概率.
23.(本题8分)有一水库大坝的横截面是梯形ABCD ,AD BC EF ∥,为水库的水面,点E 在DC 上,某课题小组在老师的带领下想测量水的深度,他们测得背水坡AB 的长为12米,迎水坡上DE 的长为2米,
,求水深.(精确到0.1 1.73==)
24.(本题8分)某批发市场批发甲、乙两种水果,根据以往经验和市场行情,预计夏季某一段时间内,甲种水果的销售利润y 甲(万元)与进货量x (吨)近似满足函数关系0.3y x =甲;乙种水果的销售利润y 乙(万元)与进货量x (吨)近似满足函数关系2y ax bx =+乙(其中0a a b ≠,,为常数),且进货量x 为1吨时,销售利润
y 乙为万元;进货量x 为2吨时,销售利润y 乙为万元.
(1)求y 乙(万元)与x (吨)之间的函数关系式.
(2)如果市场准备进甲、乙两种水果共10吨,设乙种水果的进货量为t 吨,请你写出这两种水果所获得的销售利润之和W (万元)与t (吨)之间的函数关系式.并求出这两种水果各进多少吨时获得的销售利润之和最大,最大利润是多少?
25.(本题12分)在ABC △中,2120AB BC ABC ==∠=,°,
将ABC △绕点B 顺时针旋转角α(0<°α90)<°得A BC A B 111△,交AC 于点E ,11A C 分别交AC BC 、于
D F 、两点.
(1)如图1,观察并猜想,在旋转过程中,线段1EA 与FC 有怎样的数量关系?并证明你的结论;
(2)如图2,当α30=°时,试判断四边形1BC DA 的形状,并说明理由; (第23A D
E
C
F A D
B E
C
F
(3)在(2)的情况下,求ED 的长. 26.(本题14分)如图,已知直线128
:33
l y x =
+与直线2:216l y x =-+相交于点C l l 12,、分别交x 轴于A B 、两点.矩形DEFG 的顶点D E 、分别在直线12l l 、上,
顶点F G 、都在x 轴上,且点G 与点B 重合. (1)求ABC △的面积;
(2)求矩形DEFG 的边DE 与EF 的长;
(3)若矩形DEFG 从原点出发,沿x 轴的反方向以每秒1个单位长度的速度平移,设移动时间为(012)t t ≤≤秒,矩形DEFG 与ABC △重叠部分的面积为S ,
求S 关于t 的函数关系式,并写出相应的t 的取值范围.
2009数 学
一、填空题(每小题2分,共20分)
1.> 2.107.39310? 3.答案不唯一,如21x = 4 5.30 6.210 7.(9,0) 8.8 9.30y -<< 10.32n +
二、选择题(在下列各小题中,均给出四个备选答案,其中只有一个正确答案,请将正确答案的字母号填入下表相应的空格内,每小题3分,共24分)
三、解答题(本题共76分)
19.(1)解:原式=()226932x x x x ++--+ ·············································· (2
分)
=226932x x x x ++-+- ····························································· (3
分)
=97x +. ··············································································· (
4
(第26
分)
(2)解:原式=()()()22
222
x x x x x +-
+-- ······················································ (2
分) =2
22
x x x -
-- ·········································································· (3
分)
=1. ······················································································ (4
分)
(3)解:移项,得223x x -=,配方,得()2
14x -=, ·································· (2
分)
∴12x -=±,∴1213x x =-=,. ························································ (4
分)
(注:此题还可用公式法,分解因式法求解,请参照给分)
20.解:(1)π2-; ··········································································· (
2
分)
(2)答案不唯一,以下提供三种图案.
(
满四个“田”字格的,每缺1个扣1分.) 21.(1),; ····················································································· (
4
分)
(2)解:①2004~2008移动电话年末用户逐年递增.
②2008年末固定电话用户达万户.·················································· (
8
分)
(第20题 图2) ····························· (6
(注:答案不唯一,只要符合数据特征即可得分)
22.解:(1)10,50; ········································································· (
2
分)
(2)解:解法一(树状图):
··············
·······································
··················
···························· (6分)
从上图可以看出,共有12种可能结果,其中大于或等于30元共有8种可能结果,因此P (不低于30元)=82
123
=.
·························································· (8
分)
解法二(列表法):
····························································································· (6
分)
(以下过程同“解法一”) ························································ (8
分)
23.解:分别过A D 、作AM BC ⊥于M DG BC ⊥,于G .
过E 作EH DG ⊥于H ,则四边形AMGD 为矩形. ∴456030B DCG GDC ∠=∠=∠=°,°,°.
在Rt ABM △中,sin 12AM AB
B ===· 0 1
20 3120 3120 3130 40 1320 2
3520 3
150 34第一第二
和
(第23
∴DG = ·················································································· (3
分)
在Rt DHE △
中,cos 22
DH DE
EDH =∠=?=· ······························· (6
分)
∴ 1.41 1.73HG DG DH =-=?-6≈6.7. ································· (7
分)
答:水深约为米. ··········································································· (8分)
(其它解法可参照给分)
24.解:(1)由题意,得: 1.442 2.6a b a b +=??+=?,.解得0.11.5a b =-??=?,.
····························· (
2
分)
∴20.1 1.5y x x =-+乙. ····································································· (
3
分)
(2)()()20.3100.1 1.5W y y t t t =+=-+-+乙甲.
∴20.1 1.23W t t =-++. ······························································· (
5
分)
()2
0.16 6.6W t =--+.∴6t =时,W 有最大值为. ·························· (
7
分)
∴1064-=(吨).
答:甲、乙两种水果的进货量分别为4吨和6吨时,获得的销售利润之和最大,最大利润是万元. ························································ (8
分)
D
C
25.解:(1)1EA FC =. ········································································ (
1
分)
证明:(证法一)
AB BC A C =∴∠=∠,.
由旋转可知,111AB BC A C ABE C BF =∠=∠∠=∠,,, ∴ABE C BF 1△≌△. ········································ (
3
分)
∴BE BF =,又1BA BC =,
∴1BA BE BC BF -=-.即1EA FC =. ···················· (
4分)
(证法二)
AB BC A C =∴∠=∠,.
由旋转可知,11A C A B CB ∠=∠,=,而1EBC FBA ∠=∠, ∴1A BF CBE △≌△. ········································ (
3
分)
∴BE BF =,∴1BA BE BC BF -=-,
即1EA FC =. ·················································· (
4
分)
(2)四边形1BC DA 是菱形. ······························································ (5
分)
证明:111130A ABA AC AB ∠=∠=∴°,∥,
同理AC BC 1∥. ∴四边形1BC DA 是平行四边形. ···································· (7
分) 又
1AB BC =,∴四边形1BC DA 是菱形. ··························· (
8
分)
(3)(解法一)过点E 作EG AB ⊥于点G ,则1AG BG ==.
在Rt AEG △中,
1cos cos30AG AE A =
==°……(10分) 由(2)知四边形1BC DA 是菱形, ∴2AD AB ==,
∴2ED AD AE =-= ····································· (12分)
(解法二)12030ABC ABE ∠=∠=°,°,∴90EBC ∠=°.
在Rt EBC △
中,tan 2tan 30BE BC C ==?=
·°
112EA BA BE ∴=-= ····································· (10分)
∴12ED EA == ············································ (12分)
(其它解法可参照给分)
26.(1)解:由28
033
x +=,得4x A =-∴.点坐标为()40-,.
由2160x -+=,得8x B =∴.点坐标为()80,.
∴()8412AB =--=. ···························································· (
2
分)
由2833216y x y x ?=+???=-+?,.
解得56x y =??
=?,.∴C 点的坐标为()56,. ····················· (3
分)
∴11
1263622
ABC C S AB y =
=??=△·.
········································· (4
分)
(2)解:∵点D 在1l 上且28
88833
D B D x x y ==∴=?+=,.
∴D 点坐标为()88,. ······································································ (5
分)
又∵点E 在2l 上且821684E D E E y y x x ==∴-+=∴=,..
∴E 点坐标为()48,.····························································· (6
分)
∴8448OE EF =-==,. ······················································ (7
分)
(3)解法一:①当03t <≤时,如图1,矩形DEFG 与ABC △重叠部分为五边
形CHFGR (0t =时,为四边形CHFG ).过C 作CM AB ⊥于M ,则Rt Rt RGB CMB △∽△.
)t -. ········ (10分)
(图3)
(图1) (图2)