2009 年湖南省湘西自治州中考数学真题及答案
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4、本试卷三大题,25 小题,时量 120 分钟,满分 120 分.
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一、填空题(本大题 8 小题,每小题 3 分,共 24 分. 将正确答案填在答题卡相应横线上)
1.数 3 的绝对值是
.
2.用代数式表示“a与 b的和”,式子为
.
3.如果 x-y<0,那么 x与 y的大小关系是 x
y .(填<或>符号)
4.一个圆的半径是 4,则圆的面积是
.(答案保留π)
5.一次函数 3
y
x b
的图像过坐标原点,则 b的值为
.
6.长方形一条边长为 3cm,面积为 12cm2,则该长方形另一条边长为
cm.
7.截止到 2008 年底,湘西州在校小学生中的少数民族学生数约为 21.2 万人,约占全州小学生总数的 80%,
则全州的小学生总数大致为
万.(保留小数点后一位)
8.对于任意不相等的两个数 a,b,定义一种运算※如下:a※b=
ba
ba
,
如 3※2=
23
23
5
.那么 12※4=
.
二、选择题(本大题 8 个小题,每小题 3 分,共 24 分. 将每个小题所给四个选项中惟一正确选项的代号
在答题卡上填涂)
9. 一个角是 80°,它的余角是(
)
A.10°
B.100°
C.80°
D.120°
10.要了解一批电视机的使用寿命,从中任意抽取 40 台电视机进行试验,在这个问题中,40 是(
)
A.个体
B.总体
C.样本容量
D.总体的一个样本
11.在下列运算中,计算正确的是(
)
A. 3
a a
2
6
a
C. 8
a
2
a
4
a
B. 2 3
)a
(
5
a
D. 2 2
)ab
(
2 4
a b
12.在直角坐标系中,点M(sin50°,-cos70°)所在的象限是(
)
A. 第一象限
B. 第二象限
C. 第三象限
D. 第四象限
13.在下列命题中,是真命题的是(
)
A.两条对角线相等的四边形是矩形
B.两条对角线互相垂直的四边形是菱形
C.两条对角线互相平分的四边形是平行四边形
D.两条对角线互相垂直且相等的四边形是正方形
14. O⊙ 的半径为 10cm,弦 AB=12cm,则圆心到 AB的距离为(
)
A. 2cm
B. 6cm
C. 8cm
D. 10cm
15.一个不透明的袋中装有除颜色外均相同的 3 个红球和 2 个黄球,从中随机摸出一个,摸到黄球的概率
是(
A. 2
3
l
16.如图,1
//
A.20°
C.50°
)
B. 1
5
C. 2
5
l ,∠1=120°,∠2=100°,则∠3= (
2
)
D. 3
5
B.40°
D.60°
3
2
1
l1
l2
三、解答题(本大题 9 小题,共 72 分,每个题目都要求在答题卡的相应位置写出计算或证明的主要步骤)
17.(本题 5 分)先化简再计算:
2
2
x
x
y
y
2
x
y
,其中 x =3, y =2.
18.(本题 5 分)解方程:
2
2
x
x
y
y
7
5
①
②
19.(本题 6 分)如图,在△ABC中,DE∥BC,
EF∥AB,求证:△ADE∽△EFC.
A
D
B
E
F
C
20.(本题 6 分)吉首某中学九年级学生在社会实践中,向市区的中小学教师调查他们的学历情况,并将
调查结果分别用下图的扇形统计图和折线统计图(不完整)表示.
(1) 求这次调查的教师总数;
研究生 中师 专科 本科 其它 学历结构
(2) 补全折线统计图.
21.(本题 6 分)在反比例函数
y 的图像的每一条曲线上, y 都随 x 的增大而减小.
k
x
(1) 求 k 的取值范围;
(2) 在曲线上取一点 A,分别向 x 轴、 y 轴作垂线段,垂足分别为 B、C,坐标原
点为 O,若四边形 ABOC面积为 6,求 k 的值.
22.(本题 6 分)如图,在离水面高度为 5 米的岸上有人用绳
岸,开始时绳子与水面的夹角为 30°,此人以每秒 0.5
(1) 未开始收绳子的时候,图中绳子 BC的长度是
多少米?
(2) 收绳 8 秒后船向岸边移动了多少米?(结果保留根号)
子 拉 船 靠
米收绳.问:
23.(本题 8 分)2009 年 5 月 22 日,“中国移动杯”中美篮球对抗赛在吉首进行.为组织该活动,中国移
动吉首公司已经在此前花费了费用 120 万元.对抗赛的门票价格分别为 80 元、200 元和 400 元.已知
2000 张 80 元的门票和 1800 张 200 元的门票已经全部卖出.那么,如果要不亏本,400 元的门票最低
要卖出多少张?
24.(本题 10 分)如图,等腰直角△ABC腰长为 a,现分别按图 1、图 2 方式在△ABC内内接一个正方形 ADFE
和正方形 PMNQ.设△ABC的面积为 S,正方形 ADFE的面积为 S1,正方形 PMNQ的面积为 S2,
(1) 在图 1 中,求 AD∶AB的值;在图 2 中,求 AP∶AB的值;
(2) 比较 S1+S2 与 S的大小.
A
D
E
A
P
Q
B
F
图 1
C
B
M
N
图 2
C
25.(本题 20 分)在直角坐标系 xoy中,抛物线
y
2
x
bx
与 x轴交于两点 A、B,与 y轴交于点 C,
c
其中 A在 B的左侧,B的坐标是(3,0).将直线 y
kx 沿 y轴向上平移 3 个单位长度后恰好经过点 B、
C.
(1) 求 k的值;
(2) 求直线 BC和抛物线的解析式;
(3) 求△ABC的面积;
(4) 设抛物线顶点为 D,点 P在抛物线的对称轴上,且∠APD=∠ACB,求点 P的坐标.
2009 年湘西自治州初中毕业学业考试数学参考答案
一、(本题 8 小题,每题 3 分,共 24 分,填错记 0 分)
1.3; 2.a+b; 3.<; 4.16π; 5.0; 6.4; 7.26.5; 8.1/2
二、(本题 8 小题,每题 3 分,共 24 分,选错记 0 分)
9.A
10.C
11.D
12.D
13.C
14.C
15.C
16.B
三、(本题 9 个题,共 72 分)
17.(本题 5 分)
)(
xy
y
x
解:原式=
(
x
(
y
)
)
2
x
y
····································································· 2 分
=x+y-2x+y
=-x+2y ······················································································· 4 分
因为 x=3,y=2
所以原式=-3+4=1 ············································································ 5 分
18.(本题 5 分)
解:①+② 得 4x=12,即 x=3
代入① 有 6-y=7,即 y=-1
··································································2 分
···························································· 4 分
所以原方程的解是:
x
y
3
1
··········································································5 分
19.(本题 6 分)
证明:∵DE∥BC,∴DE∥FC,∴∠AED=∠C ······················································· 3 分
又∵EF∥AB,∴EF∥AD,∴∠A=∠FEC ······················································· 5 分
∴△ADE∽△EFC ················································································· 6 分
20.(本题 6 分)
解:(1)总人数= 275 55% 500
人
(2)教师中专科学历的人数=500 10% 50
················································· 3 分
人 ········································· 4 分
作图:
····························································································· 6 分
注:第(2)问虽然没明确指出专科人数
为 50,但只要作图准确就可得 6 分.
研究生 中师 专科 本科 其它 学历结构
21.(本题 6 分)
解(1)因为 y的值随 x的增大而减小,所以 k>0
············································· 2 分
(2)设 A(x0,y0)
································································ 4 分
则由已知,应有|x0y0|=6
即|k|=6
而 k>0
所以 k=6. ······················································································ 6 分
22.(本题 6 分)
解(1)如图,在 Rt△ABC中,
∴ BC=
5
sin30
=10 米
AC =sin30° ···················································· 2 分
BC
·······················································3 分
(2)收绳 8 秒后,绳子 BC缩短了 4 米,只有 6 米, ······································ 4 分
这时,船到河岸的距离为
2
6
2
5
36
25
11
米.·································· 6 分
23.(本题 8 分)
解:2000 张 80 元的门票收入为 2000×80=160000 元; ······································· 2 分
1800 张 200 元的门票收入为 1800×200=360000 元;······································· 4 分
1200000-160000-360000=680000 元,························································ 5 分
故 400 元的门票至少要卖出:680000÷400=1700 张.
答:400 元的门票最少要卖出 1700 张.························································ 8 分
24.(本题 10 分)
解(1)图 1 中,∵AD=DF,∠B=45°,从而 DF=DB,∴AD=DB,
∴AD∶AB=1∶2
··································································· 2 分
图 2 中,∵PM=MN,∠B=45°,从而 PM=MB,∴MN=MB,
∴MN=MB=NC,
∴AP∶AB=PQ∶BC=MN∶BC=1∶3 ····························································4 分
(2)图 1 中,S1=
1
2
a
2
1
4
2
a
···································································· 6 分
又 PQ∶BC=AP∶AB=1∶3,
∴PQ=
2
3
a
,∴S2=
2
3
a
2
2
9
2
a
································································· 8 分
从而 S1+S2=
1
4
2
9
2
a
17
36
2
a
又 S=
1
2
2
a
18
36
2
a
∴S1+S2<S·····································································································10 分
25.(本题 20 分)
解(1)直线 y
kx 沿 y轴向上平移 3 个单位后,过两点 B,C
从而可设直线 BC的方程为
y
kx
···································································2 分
3
x ,得 C(0,3)···················································································· 3 分
k ······································································································· 5 分
x ·························································7 分
(2)由(1),直线 BC 的方程为
3
y
令 0
又 B(3,0)在直线上,
∴ 0 3
∴
3k
1
又抛物线
y
2
x
bx
过点 B,C
c
∴
3
c
39
cb
0
b
c
4
3
∴抛物线方程为
y
x
2 4
x
········································································ 10 分
3
(3)由(2),令 2 4
x
x
3 0
x
得 1
1
,
x
2
3
···························································································· 12 分
即 A(1,0),B(3,0),而 C(0,3)
∴△ABC的面积 S△ABC=
(3-1)·3=3 平方单位····················································15 分
1
2
(4)由(2),D(2, 1 ),设对称轴与 x轴交于点 F,与 BC交于 E,可得 E(2,1),
连结 AE.
∵
AF FB FE
1
∴AE⊥CE,且 AE= 2 ,CE=
22
(或先作垂线 AE⊥BC,再计算也可)
在 Rt△AFP与 Rt△AEC中,
∵∠ACE=∠APE(已知)
1 =
2
PF
22
即
∴
PF
CE
2
AF
AE
PF ·····························································18 分
∴
∴点 P的坐标为(2,2)或(2, 2 )······················· 20 分
(x轴上、下方各一个)
(注:只有一个点扣 1 分)