2009 年广西梧州市中考数学真题及答案
说明:1.本试卷共 8 页(试题卷 4 页,答题卷 4 页),满分 120 分,考试时间 120 分钟.
2.答卷前,将准考证号、姓名写在答题卷密封线内,答案请写在答题卷相应的
区域内,在试题卷上答题无效
..........
一、填空题(本大题共 10 小题,每小题 3 分,共 30 分.)
1.6 的相反数是 ★ .
A
2.比较大小:-3 ★ -4.(用“>”“=”或“<”表示)
3.一组数据为 1,2,3,4,5,6,则这组数据的中位数是 ★ .
4.因式分解:
2 2 x
18
= ★ .
D
B
C
图(1)
5.如图(1),△ABC中,∠A=60°,∠C=40°,延长 CB到 D ,则∠ABD= ★ 度.
6.将点 A(1,-3)向右平移 2 个单位,再向下平移 2 个单位后得到点 B(a,b),则 ab= ★ .
7.某蔬菜基地的圆弧形蔬菜大棚的剖面如图(2)所示,已知
AB=16m,半径 OA=10m,则中间柱 CD的高度为 ★ m.
8.在△ABC中,∠C=90°, BC=6 cm,
sin A
3
5
,
A
则 AB的长是 ★ cm.
C
D
B
O
图(2)
9 . 一 个 扇 形 所 在 圆 的 半 径 为 3 c m , 扇 形 的 圆 心 角 为 1 2 0 ° , 则 扇 形 的 面 积
是 ★ cm2. (结果保留 π )
10.图(3)是用火柴棍摆成的边长分别是 1,2,3 根火柴棍时的正方形.当边长为 n根火柴棍时,设摆出
的正方形所用的火柴棍的根数为 s ,则 s = ★ . (用 n的代数式表示 s )
……
n=1
n=2
n=3
图(3)
二、选择题(本大题共 8 小题,每小题 3 分,共 24 分.在每小题给出的四个选项中,
只有一项是正确的,每小题选对得 3 分,选错、不选或多选均得零分.)
11.在函数
y
1
x
2
中,自变量 x的取值范围是(
)
A.
2x
B. 2x
C. x ≤ 2
D. x ≥ 2
12.下列运算正确的是(
)
A.
C.
2
a
(
a
3
a
32)
6
a
6
a
4
B.
D.
a
a
2
2
a
3
a
a
a
13.一个布袋中有 4 个除颜色外其余都相同的小球,其中 3 个白球,1 个红球.从袋中任意摸出 1 个球是白
球的概率是(
)
A.
3
4
14.不等式组
B.
1
4
C.
2
3
D.
1
3
2
≥
2 0
1
x
x
的解集在数轴上表示为(
)
-2
-1 0
1 2
3
-2
-1 0
1 2
3
-2
-1 0
1 2
3
-2
-1 0
1 2
3
A.
B.
C.
D.
15.在下列对称图形中,对称轴的条数最少的图形是(
)
A.圆
B.等边三角形
C.正方形
(D)正六边形
16.在一个仓库里堆放有若干个相同的正方体货箱,仓库管理员将这堆货箱的三视图画出来,如图(4),
则这堆货箱共有(
)
图(4)
A.6 个
主视图
B.5 个
左视图
俯视图
C.4 个
D.3 个
x
1 7 . 已 知 点 A( 1
y, ) 、 B( 2
x
1
y, ) 是 反 比 例 函 数
2
y (
k
x
0k
) 图 象 上 的 两 点 ,
若
A.
x
1
y
1
0
0
x
2
y
2
,则有(
B.
y
2
)
0
y
1
C.
y
1
y
2
0
18.如图(5),正方形ABCD中,E为AB的中点,AF⊥DE于点O, 则
A.
C.
52
3
2
3
B.
1
3
D.
1
2
三、解答题(本大题共 8 小题,满分 66 分.)
19.(本题满分 6 分)计算:
12
11
2
2sin 60
20.(本题满分 6 分)解方程:
(
x
2
)3
(2
xx
)3
0
y
2
y
1
0
等于(
)
D.
AO
DO
D
A
C
F
O
E
B
图(5)
21.(本题满分 6 分)
为了解全市太阳能热水器的销售情况,某调查公司对人口为 100 万人的某县进行调查,对调查所得的
数据整理后绘制成如图(6)所示的统计图.请据图解答下列问题:
中档
·
高档占 10%
低档占 30%
图(6)-1
2008 年该县销售高、中、低
档太阳能热水器的数量统计图
图(6)
台
1000
700
600
600 700
1000
图(6)-2
2005 2006 2007 2008
2005-2008 年该县销售太
阳能热水器的数量统计图
年
(1)2008 年该县销售中档..太阳能热水器 ★ 台.
(2)若 2007 年销售太阳能热水器的台数是 2005 年的 1.5 倍,请补全图(6)-2 的条形图.
(3)若该县所在市的总人口约为 500 万人,估计 2008 年全市销售多少台高档太阳能热水器.
22.(本题满分 8 分)
某工厂要招聘甲、乙两种工种的工人 150 人,甲、乙两种工种的工人的月工资分别为 600 元和 1000 元.
(1)设招聘甲种工种工人 x人,工厂付给甲、乙两种工种的工人工资共 y元,写出 y(元)与 x(人)的
函数关系式;
(2)现要求招聘的乙种工种的人数不少于甲种工种人数的 2 倍,问甲、乙两种工种
各招聘多少人时,可使得每月所付的工资最少?
23.(本题满分 8 分)
如图(7),△ABC中,AC的垂直平分线 MN交 AB于
点 D,交 AC于点 O,CE∥AB交 MN于 E,连结 AE、CD.
D
M
(1)求证:AD=CE;
(2)填空:四边形 ADCE的形状是 ★ .
A
O
E
N
B
C
图(7)
24.(本题满分 10 分)
由甲、乙两个工程队承包某校校园绿化工程,甲、乙两队单独完成这项工程所需时间比是 3︰2,两队
合做 6 天可以完成.
(1)求两队单独完成此项工程各需多少天?
(2)此项工程由甲、乙两队合做 6 天完成任务后,学校付给他们 20000 元报酬,若
按各自完成的工程量分配这笔钱,问甲、乙两队各得到多少元?
25.(本题满分 10 分)
如图(8)所示,△ABC内接于⊙O,AB是⊙O的直径,点 D在⊙O 上,过点 C的切线交 AD的延长线于
点 E,且 AE⊥CE,连接 CD.
(1)求证:DC=BC;
(2)若 AB=5,AC=4,求 tan∠DCE的值.
E
D
A
C
B
·
O
图(8)
26.(本题满分 12 分)
如图(9)-1,抛物线
y
ax
交于另一点 B.
(1)求此抛物线的解析式;
2 3
ax b
经过 A( 1 ,0),C(3, 2 )两点,与 y 轴交于点 D,与 x 轴
(2)若直线
y
kx
(1
k
)0
将四边形 ABCD面积二等分,求 k 的值;
(3)如图(9)-2,过点 E(1,1)作 EF⊥ x 轴于点 F,将△AEF绕平面内某点旋转 180°得△MNQ(点 M、N、
Q分别与点 A、E、F对应),使点 M、N在抛物线上,作 MG⊥ x 轴于点 G,若线段 MG︰AG=1︰2,求点 M,N
的坐标.
y
A
O
D
图(9)-1
B
x
A
O
C
y=kx+1
y
E
F
Q
B
x
G
M
N
图(9)-2
一、填空题(本大题共 10 小题,每小题 3 分,共 30 分.)
2009 年梧州市初中毕业升学考试
数学参考答案及评分标准
题号
答案
题号
答案
1
6
题号
6
答案
15
2
>
7
4
12
C
11
B
3
3.5
13
8
A
10
三、解答题(本大题共 8 小题,满分 66 分.)
4
2(x+3)(x-3)
15
14
16
5
100
17
D
9
B
3 π
C
10
2 (
n n
A
1)
18
二、选择题(本大
题 共 8
小 题 ,
每 小 题
3 分,共
D
24 分.)
2
3
=
32
22
20.解:
19.解:原式=
······································································ 3 分
3
2
32
··········································································4 分
3 ···················································································· 6 分
2
=
)23
(
)(3
x
x
x
········································································ 2 分
0
)3
(
3)(3
x
x
··············································································3 分
03 x
03
3
x
········································································· 4 分
或
1
3
2 x
··············································································· 6 分
即
·················································································· 2 分
(2)在右图上补全条形图如图. ······································································4 分
21.解:(1) 600
1 x
或
0
台
700
600
1000
700
600
1000
900
2005
2006
2007
2008
年
图(6)-2
)
x
x
y
22.解:(1)
y
600
400
(2)依题意得,150
·····································································6 分
(3)500÷100×1000×10%=500
1000
150(
x
···························································· 2 分
150000
··································································3 分
2
x
≥ ································································ 5 分
··············································································6 分
因为-400<0,由一次函数的性质知,当 x=50 时,y有最小值··················· 7 分
所以 150-50=100
x ≤
50
x
答: 甲工种招聘 50 人,乙工种招聘 100 人时可使得每月所付的工资最少. (8 分)
A
23.(1)证明:∵MN是 AC的垂直平分线························· 1 分
D
M
O
E
N
∴OA=OC ∠AOD=∠EOC=90°··················· 3 分
∵CE∥AB
∴∠DAO=∠ECO ··································· 4 分
∴△ADO≌△CEO ····································· 5 分
∴AD=CE ··········································· 6 分
(2)四边形 ADCE是菱形. ·····································8 分
(填写平行四边形给 1 分)
24.解:(1)设甲队单独完成此项工程需 x天,由题意得······································ 1 分
1
······················································································ 3 分
6
x
6
2
x
3
解之得 15x
经检验, 15x
所以甲队单独完成此项工程需 15 天,
························································································ 4 分
是原方程的解. ······························································5 分
乙队单独完成此项工程需 15×
2
3
=10(天)················································· 6 分
(2)甲队所得报酬:
20000
1
15
6
8000
(元)············································8 分
乙队所得报酬:
20000
1
10
6
12000
(元)···················································· 10 分
25.(1)证明:连接 OC··················································································· 1 分
∵OA=OC
∴∠OAC=∠OCA
∵CE是⊙O的切线
∴∠OCE=90° ········································· 2 分
∵AE⊥CE
C
E
D
B
∴∠AEC=∠OCE=90°
∴OC∥AE ················································3 分
∴∠OCA=∠CAD
∴∠CAD=∠BAC ········································4 分
∴ DC BC
∴DC=BC ····························································································· 5 分
(2)∵AB是⊙O的直径 ∴∠ACB=90°
图(8)
·
O
A
2
2
2
5
2
4
∴
AB
BC
AC
···················································6 分
3
∵∠CAE=∠BAC ∠AEC=∠ACB=90°
∴△ACE∽△ABC·················································································· 7 分
EC
BC
AC
AB
∴
∴
EC
3
4
5
∵DC=BC=3
12EC
5
········································································ 8 分
∴
ED
2
DC
CE
2
2
3
12(
5
2
)
9
5
·················································· 9 分
∴
tan
DCE
ED
EC
9
5
12
5
3
4
······························································ 10 分
26.(1)解:把 A( 1 ,0),C(3, 2 )代入抛物线
y
ax
2 3
ax b
得
)1(
2
a
9
a
)1(3
9
ba
ba
2
0
······································································ 1 分
整理得
0
4
b
ba
2
……………… 2 分
解得
a
b
1
2
2
………………3 分
1 2
x
2
0
3
2
x
2
·························································· 4 分
x
解得 1
1
,
x
2
4
∴AB∥CD ∴四边形 ABCD是梯形.
8
··························· 5 分
y
与 x轴的交点为 H,
3 , 2 )················6 分
k
将四边形 ABCD面积二等分
∴抛物线的解析式为
y
(2)令
1 2
x
2
x
3
2
2
∴ B点坐标为(4,0)
又∵D点坐标为(0, 2 )
1
2
kx
2)35(
)0
(1
∴S 梯形 ABCD =
y
设直线
k
与 CD的交点为 T,
则 H(
∵直线
∴S 梯形 AHTD =
S 梯形 ABCD=4
1 ,0), T(
k
)0
y
(1
k
kx
1
2
1
2)311(
2
k
4k
3
k
∴
∴
H
O
A
B
x
D
T
C
y=kx+1
图(9) -1
y
E
F
Q
A
O
B
x
G
M
N
图(9) -2
4
································ 7 分
···················································· 8 分
(3)∵MG⊥ x 轴于点 G,线段 MG︰AG=1︰2
m
1
2
∴设 M(m,
),··································· 9 分
∵点 M在抛物线上
m
∴
1
2
1
2
2
m
m
2
3
2
3
,
m
解得 1
∴M点坐标为(3, 2 )··········································································· 11 分
(舍去) ························ 10 分
m
2
1
根据中心对称图形性质知,MQ∥AF,MQ=AF,NQ=EF,
∴N点坐标为(1, 3 ) ········································································· 12 分