2009 年广西桂林市百色市中考数学真题及答案
(考试用时:120 分钟 满分: 120 分)
注意事项:
1.本试卷分选择题和非选择题两部分.在本试卷上作答无效.
..........
2.考试结束后,将本试卷和答题卷一并交回.
3.答题前,请认真阅读答题卷上的注意事项
...............
一、选择题(共 12 小题,每小题 3 分,共 36 分.在每小题给出的四个选项中只有一项是符合要求的,用
2B 铅笔把答题卷...上对应题目的答案标号涂黑)
1. 8 的相反数是(
).
1
8
A. 8
1
8
2.下面的几个有理数中,最大的数是(
B.8
C.
D.
A.2
B.
1
3
C.-3
D.
1
5
3.如图,在所标识的角中,同位角是(
A. 1 和 2
C. 1 和 4
4.右图是一正四棱锥,它的俯视图是(
B. 1 和 3
D. 2 和 3
).
).
).
1
2
3
4
(第 3 题图)
A.
B.
C.
D.
(第 4 题图)
5.下列运算正确的是(
).
A. 2
a b
2
ab
B.
(
)ab
2
2 2
a b
C. 2a · 2a =
22a
D. 4
a
2
a
2
6.二次函数
y
(
x
1)
2
的最小值是(
2
).
A.2
2
3
7.右图是一张卡通图,图中两圆的位置关系是(
C.-3
B.1
D.
).
A.相交
B.外离 C.内切
D.内含
(第 7 题图)
8.已知
x
y
2
1
是二元一次方程组
ax by
ax by
7
1
的解,则 a b 的值为(
).
A.1
9.有 20 张背面完全一样的卡片,其中 8 张正面印有桂林山水,7 张正面印有百色风光,5 张正面印有北
B.-1
C. 2
D.3
海海景;把这些卡片的背面朝上搅匀,从中随机抽出一张卡片,抽中正面是桂林山水卡片的概率是(
).
A
D
A.
1
4
B.
7
20
C.
2
5
D.
5
8
10.如图,□ABCD中,AC、BD为对角线,BC=6,
B
第 3 题图
C
BC边上的高为 4,则阴影部分的面积为(
).
A.3
B.6
C.12
D.24
11.如图所示,在方格纸上建立的平面直角坐标系中,
将△ABO绕点 O按顺时针方向旋转 90°,
△
得 A B O
,则点 A 的坐标为(
).
A
y
B
4
3
2
1
A.(3,1)
B.(3,2)
-3
-2
-1
0
1
2
3
C.(2,3)
D.(1,3)
第 11 题图
12.如图,正方形 ABCD的边长为 2,将长为 2 的线段 QR的两端放
在正方形的相邻的两边上同时滑动.如果 Q点从 A点出发,沿
图中所示方向按 A→B→C→D→A滑动到 A止,同时点 R从 B点
出发,沿图中所示方向按 B→C→D→A→B滑动到 B止,在这个
过程中,线段 QR的中点 M所经过的路线围成的图形的面积为
(
).
A.2
B. 4 π
C. π
D. π 1
A
Q
B
x
M
R
第 12 题图
D
C
二.填空题(共 6 道小题,每小题 3 分,共 18 分,请将答案填在答题卷上)
13.因式分解: 2 3
x
x
.
14.据统计,去年我国粮食产量达 10570 亿斤,用科学记数法表示为
亿斤.
15.如图,在一次数学课外活动中,测得电线杆底部 B与钢缆固定
A
点 C的距离为 4 米,钢缆与地面的夹角为 60º,则这条钢缆在电
线杆上的固定点 A到地面的距离 AB是 米.(结果保留根号).
16.在函数
y
2
x
1
中,自变量 x 的取值范围是
.
B
第 15 题图
C
y
2
O
-1
第 17 题图
x
A
B
C
第 18 题图
A2
A1
D
17.如图,是一个正比例函数的图像,把该图像
向左平移一个单位长度,得到的函数图像的
解析式为
.
18.如图,在△ABC中,∠A=.∠ABC与∠ACD的
平分线交于点 A1,得∠A1;∠A1BC与∠A1CD的平分线相
交于点 A2,得∠A2; ……;∠A2008BC与∠A2008CD的平
分线相交于点 A2009,得∠A2009 .则∠A2009=
.
三、解答题(本大题共 8 题,共 66 分,请将答案写在答题卷上.)
19.(本题满分 6 分)计算:
1(
2
)
1
(2009
0
3)
4sin 30
º- 2
20.(本题满分 6 分)先化简,再求值:
其中
x
2
,
y
3
.
1
2
x
1 (
y
x
2
x
2
y
y
)
x
2
x
,
21.(本题满分 8 分)如图:在等腰梯形 ABCD中,AD∥BC,对角线 AC、BD相交于 O.
(1)图中共有
(2)写出你认为全等的一对三角形,并证明.
对全等三角形;
A
D
O
B
第 21 题图
C
22. (本题满分 8 分)2008 年 11 月 28 日,为扩大内需,国务院决定在全国实施“家电下乡”政策.第一
批列入家电下乡的产品为彩电、冰箱、洗衣机和手机四种产品.某县一家家电商场,今年一季度对以上
四种产品的销售情况进行了统计,绘制了如下的统计图,请你根据图中信息解答下列问题:
数量(台)
200
150
100
50
手机
40% 洗衣机
冰箱
20%
彩电
彩电
冰箱
洗衣机
手机 品种
(1)该商场一季度彩电销售的数量是
(2) 请补全条形统计图和扇形统计图.
台.
23. (本题满分 8 分)在保护地球爱护家园活动中,校团委把一批树苗分给初三(1)班同学去栽种.如
果每人分 2 棵,还剩 42 棵;如果前面每人分 3 棵,那么最后一人得到的树苗少于 5 棵(但至少分得一
棵).
(1)设初三(1)班有 x 名同学,则这批树苗有多少棵?(用含 x 的代数式表示).
(2) 初三(1)班至少有多少名同学?最多有多少名
24. (本题满分 8 分)在我市某一城市美化工程招标时,有甲、乙两个工程队投标.经测算:甲队单
独完成这项工程需要 60 天;若由甲队先做 20 天,剩下的工程由甲、乙合做 24 天可完成.
(1)乙队单独完成这项工程需要多少天?
(2)甲队施工一天,需付工程款 3.5 万元,乙队施工一天需付工程款 2 万元.若该工程计划在 70 天内
完成,在不超过计划天数的前提下,是由甲队或乙队单独完成该工程省钱?还是由甲乙两队全程合
作完成该工程省钱?
25. (本题满分 10 分)如图,△ABC内接于半圆,AB是直径,过 A作直线 MN,
若∠MAC=∠ABC .
(1)求证:MN是半圆的切线;
(2)设 D是弧 AC的中点,连结 BD交 AC 于 G,
过 D作 DE⊥AB于 E,交 AC于 F.
求证:FD=FG.
(3)若△DFG的面积为 4.5,且 DG=3,GC=4,
试求△BCG的面积.
M
D
C
G
A
N
F
E
第 25 题图
B
26.(本题满分 12 分)如图,已知直线
分别为 A、B两点.
:
l y
3
x
4
,它与 x 轴、 y 轴的交点
3
(1)求点 A、点 B的坐标;
(2)设 F是 x 轴上一动点,用尺规作图作出⊙P,使⊙P经过点 B且与 x 轴相切于点 F(不写作法和证明,
保留作图痕迹);
(3)设(2)中所作的⊙P的圆心坐标为 P( x
(4)是否存在这样的⊙P,既与 x 轴相切又与直线l 相切于点 B,若存在,求出圆心 P的坐标;若不存在,
y, ),求 y 与 x 的函数关系式;
请说明理由.
y
B
V
A
O ·
F
x
2009 年桂林市、百色市初中毕业升学考试
第 26 题图
数学参考答案及评分标准
一、选择题:
题号 1
答案 B
2
A
3
C
4
C
5
B
6
A
7
D
8
B
9
C
10
C
11
D
12
B
二、填空题:
13. (
x x
3)
14.1.057×104
17.
y
2
x
或
2
y
2(
x
1)
16. x ≥
1
2
15. 4 3
20092
18.
三、解答题:
19.解:原式=2-1+4×
1
2
20.解:原式
1
2
x
y
( )
x
1
2
x
(
x
x
-2··············································································· 4 分
=1································································································ 6 分
(
x
)(
y x
y
)
y
1
x
y
y
x
2
x
·········································2 分
·················································································· 3 分
1
x
1
2
x
y
)
····························································································· 4 分
··································································································· 5 分
y
把
x
2
, 代入上式,得原式=3
3
y
2
·················································· 6 分
21.解:(1)3 …………………………………………………………………………………3 分
(写 1 对、2 对均不给分)
(2)△ABC≌△DCB··············································································· 4 分
证明:∵四边形 ABCD是等腰梯形
∴AB=DC,∠ABC=∠DCB····························································· 6 分
又 BC=CB
∴△ABC≌△DCB ······································································ 8 分
(注:选其它两对证明的,按以上相应步骤给分,全等三角形对应点不对应不扣分)
22.解(1)150 ······················································································· (2 分)
(2)10%·························································································· (2 分)
(3)每正确补全一个图形给 2 分,其中扇形统计图每补全一个扇形给 1 分.
数量(台)
200
150
100
50
手机
10%
40% 洗衣机
冰箱
20%
彩电
30%
彩电
23.解(1)这批树苗有( 2
x )棵································································1 分
2
x
2
x
··················································· 5 分
1) 5
1
1)
≥
42
冰箱
42 3(
42 3(
(2)根据题意,得
手机 品种
洗衣机
x
x
(每列对一个不等式给 2 分)
解这个不等式组,得 40< x ≤44································································· 7 分
答:初三(1)班至少有 41 名同学,最多有 44 名同学.········································· 8 分
24.解:(1)设乙队单独完成需 x 天··································································· 1 分
············································ 3 分
根据题意,得
) 24 1
20 (
1
60
1
x
1
60
解这个方程,得 x =90······································································ 4 分
经检验, x =90 是原方程的解
∴乙队单独完成需 90 天··································································· 5 分
(2)设甲、乙合作完成需 y 天,则有
(
1
60
1
90
)
y
1
解得 36
y (天)··················································································6 分
甲单独完成需付工程款为 60×3.5=210(万元)
乙单独完成超过计划天数不符题意(若不写此行不扣分).
甲、乙合作完成需付工程款为 36(3.5+2)=198(万元)······························· 7 分
答:在不超过计划天数的前提下,由甲、乙合作完成最省钱.······················· 8 分
25.证明(1):∵AB是直径
∴∠ACB=90º ,∴∠CAB+∠ABC=90º················································· 1 分
∵∠MAC=∠ABC
∴∠MAC+∠CAB=90º,即 MA⊥AB
∴MN 是半圆的切线.······························ 2 分
M
(2)证法 1:
∵D是弧 AC的中点, ∴∠DBC=∠2············ 3 分
∵AB是直径,∴∠CBG+∠CGB=90º
∵DE⊥AB,∴∠FDG+∠2=90º··················· 4 分
∵∠DBC=∠2,∴∠FDG=∠CGB=∠FGD
∴FD=FG················································ 5 分
证法 2:连结 AD,则∠1=∠2························· 3 分
D
H
3
1
F
E
A
N
∵AB是直径,∴∠ADB=90º
∴∠1+∠DGF=90º
C
G
2
B
又∵DE⊥AB ∴∠2+∠FDG=90º·························································· 4 分
∴∠FDG=∠FGD, ∴FD=FG····························································· 5 分
(3)解法 1:过点 F作 FH⊥DG于 H,························································ 6 分
1
2
1
2
9
4
又∵DF=FG ∴S△FGH=
S△DFG=
×4.5=
··········································· 7 分
∵AB是直径,FH⊥DG ∴∠C=∠FHG=90º·········································· 8 分
∵∠HGF=∠CGB,∴△FGH∽△BGC
∴
S
S
FGH
(
BGC
HG
CG
2
)
(
1.5
4
2
)
9
64
·····················································9 分
∴S△BCG=
9 64 16
4
9
······································································10 分
解法 2:∵∠ADB=90º,DE⊥AB,∴∠3=∠2················································ 6 分
∵∠1=∠2, ∴∠1=∠3
∴AF=DF=FG·············································································· 7 分
∴S△ADG=2S△DFG=9··········································································· 8 分
∵∠ADG=∠BCG,∠DGA=∠CGB
∴△ADG∽△BCG········································································ 9 分
)
2
(
24
)
3
16
9
∴
S
S
△
BCG
△
ADG
(
CG
DG
16 9 16
9
∴S△BCG=
····································································· 10 分
解法 3:连结 AD,过点 F作 FH⊥DG于 H,
∵S△FDG=
DG×FH=
×3FH=4.5
1
2
1
2
∴FH=3······················································································· 6 分
∵H是 DG的中点,FH∥AD
∴AD=2FH=6 ················································································· 7 分
∴S△ADG=
1
2
AD DG
1 6 3 9
···················································· 8 分
2
(以下与解法 2 同)
26.解(1)A( 4 ,0),B(0,3)········································2 分(每对一个给 1 分)
(2)满分 3 分.其中过 F作出垂线 1 分,作出 BF中垂线 1 分,找出圆心并画出⊙P给 1 分.
(注:画垂线 PF不用尺规作图的不扣分)
(3)过点 P作 PD⊥ y 轴于 D,则 PD= x ,BD= 3 y ,··············6 分
PB=PF= y ,∵△BDP为直角三形,
∴
2
PB
2
PD BD
2
∴ 2
BP
2
PD BD
2
························· 7 分
y
B
D
O
P
F
x
即
2
y
2
x
3
y
2
即 2
y
2
x
(3
2
y
)
∴ y 与 x 的函数关系为
y
21
x
6
A
································································ 8 分
3
2
(4)存在
解法 1:∵⊙P与 x 轴相切于点 F,且与直线l 相切于点 B
∴ AB AF
································································································· 9 分
∵ 2
AB OA OB
2
2
25
∴ 2
AF
25
∵AF=
4x
, ∴
(
x
4)
2
2
5
···································································10 分
∴ 1
x
或
x
9
·························································································· 11 分
把 1
x
或
x
9
代入
∴点 P的坐标为(1,
21
x
6
,得
3
2
y
5
3
或
y
15
)或( 9,15)·························································· 12 分
y
5
3