2009 年广西崇左市中考数学真题及答案
(全卷满分:120 分;考试时间:120 分钟)
一、填空题:本大题共 10 小题;每小题 2 分,共 20 分.请将答案填写在题中的横线上.
1. 5 的绝对值是
75
2.已知
A °,则 A 的余角的度数是
.
.
3.在函数
y
x
中,自变量 x 的取值范围是
3
.
4.分解因式: 22
x
4
x
2
.
5.写出一个图像位于第一、二、三象限内的一次函数表达式:
6.一元二次方程 2
x mx
的一个根为 1 ,则另一个根为
3 0
.
.
的母线长为
O
C
A
(第 8 题)
B
D
A
(第 10 题)
c
1
P
2
5
3
4
(第 11 题)
C
E
B
a
P
b
P
半径的半圆
为
的四个选项
小题选对得
(
)
7.已知圆锥的侧面积为
8πcm ,侧面展开图的圆心角为 45°,则该圆锥
2
cm.
8.如图,点O 是 O⊙ 的圆心,点 A B C、 、 在 O⊙ 上, AO BC∥ ,
AOB
的度数是
38
.
9.当 x ≤ 0 时,化简
10.如图,正方形 ABCD 中,E 是 BC 边上一点,以 E 为圆心、EC 为
与以 A 为圆心, AB 为半径的圆弧外切,则sin EAB
的结果是
的值
.
x
2
°,则 OAC
1 x
.
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
中,只有一项是正确的,请将正确答案前的字母填入题后的括号内,每
3 分,选错、不选或多选均得 0 分.
11.如图,直线 c 截二平行直线 a b、 ,则下列式子中一定成立的是
A. 1
C. 1
12.下列运算正确的是(
3
5
)
B. 1
D. 1
2
4
A. 2
x
2
3
6
x·
x
2
4
B. 2
x
2
2
3
x
1
C. 2
x
2
2
3
x
22
x
3
D. 2
x
2
2
3
x
4
5
x
13.一个等腰三角形的两边长分别为 2 和 5,则它的周长为(
A.7
D.9 或 12
B.9
)
14.不等式组
的整数解共有(
)
C.12
2
≤
2 1
x
x
B.4 个
A.3 个
\
15.如图,下列选项中不是..正六棱柱三视图的是(
C.5 个
D.6 个
)
A.
B.
C.
D.
16.某校九年级学生参加体育测试,一组 10 人的引体向上成绩如下表:
完成引体向上的个数
人 数
7
1
8
1
9
3
10
5
)
°,
E
A
B
1
F
(第 17 题)
D
C
则 AEF
=
绕点O 按逆
B.9.5 和 10
这组同学引体向上个数的众数与中位数依次是(
A.9 和 10
C.10 和 9
D.10 和 9.5
17.如图,把矩形 ABCD 沿 EF 对折后使两部分重合,若 1 50
(
A.110°
D.130°
B.115°
C.120°
)
18.已知点 A 的坐标为 (
a b, ,O 为坐标原点,连结OA ,将线段OA
)
时针方向旋转 90°得 1OA ,则点 1A 的坐标为(
)
A.(
a b
,
)
B. (
a
b,
)
C. (
b a
,
)
D. (
b
a,
)
三、解答题:本大题共 7 小题,共 76 分.
19.(本小题满分 6 分)
计算:
2sin 60
°
3tan 30
°
0
1
3
( 1)
2009
.
20.(本小题满分 8 分)
已知 2 2 0
x ,求代数式
(
x
x
2
1)
2
1
2
x
x
1
的值.
21.(本小题满分 10 分)
如图, ABC△
中, D E、 分别是边 BC AB、 的中点, AD CE、 相交于G .求证:
A
GE GD
AD
CE
.
1
3
E
G
B
D
(第 21 题)
C
22.(本小题满分 10 分)
一只口袋中放着若干只红球和白球,这两种球除了颜色以外没有任何其他区别,袋中的球已经搅匀,蒙上
眼睛从口袋中取出一只球,取出红球的概率是
1
4
.
(1)取出白球的概率是多少?
(2)如果袋中的白球有 18 只,那么袋中的红球有多少只?
23.(本小题满分 12 分)
五一期间某校组织七、八年级的同学到某景点郊游,该景点的门票全票票价为 15 元/人,若为 50~99 人可
以八折购票,100 人以上则可六折购票.已知参加郊游的七年级同学少于 50 人,八年级同学多于 50 人而少
于 100 人.若七、八年级分别购票,两个年级共计应付门票费 1575 元,若合在一起购买折扣票,总计应付
门票费 1080 元.问:
(1)参加郊游的七、八年级同学的总人数是否超过 100 人?
(2)参加郊游的七、八年级同学各为多少人?
24.(本小题满分 14 分)
如图,在等腰梯形 ABCD 中,已知 AD BC∥ ,
AB DC AD
,
2
,
BC
4
,延长 BC 到 E ,使CE AD
.
(1)证明: BAD
(2)如果 AC BD
△
DCE
≌△
,求等腰梯形 ABCD 的高 DF 的值.
;
A
D
B
CF
(第 24 题)
E
25.(本小题满分 16 分)
在平面直角坐标系中,现将一块等腰直角三角板 ABC 放在第二象限,斜靠在两坐标轴上,且点 (0 2)
A , ,
ax
经过点 B .
C , ,如图所示:抛物线
点 ( 1 0)
(1)求点 B 的坐标;
(2)求抛物线的解析式;
(3)在抛物线上是否还存在点 P (点 B 除外),使 ACP△
存在,求所有点 P 的坐标;若不存在,请说明理由.
2
y
2
ax
仍然是以 AC 为直角边的等腰直角三角形?若
y
A
(0,2)
B
(-1,0)
C
x
2009 年崇左市初中毕业升学考试
数 学 答 案
一、1.5
2.15°
3.
x
≥
3
4.
2(
x
1)
2
6. 3
7.8
8.19°
9.1
10.
3
5
(第 25 题)
y
x 等
1
5.
二、11.B
12.A
13.C
14.C
15A
16.D
17.B
18.C
三、19.原式=
2
3
2
3
3
3
1 1
··································································4 分
=0.····························································································6 分
20.原式=
1)
(
x
1)(
x
(
x
2
1)
2
x
x
1
·········································································· 2 分
=
x
x
1
1
2
x
x
1
····················································································4 分
2
x
=
1
x
1
x
·······················································································5 分
2 2 0
x
,
2
x ··························································································6 分
原式
1
··············································································· 7 分
2
2
1
x
x
原式=1··························································································· 8 分
A
G
E
B
D
C
AC
∽△
△
∥ ,
ACG
D E
DE
,··············································· 3 分
21.证明:连结 ED ,·················································· 1 分
、 分别是边 BC AB、 的中点,
1
DE
2
AC
DEG
,················································· 5 分
1
GE GD DE
2
GC AG AC
GE GD
CE
AD
P
22.(1) (
3
4
,············································ 7 分
1 P
(
1
4
1
3
取出白球
=
)
)
.························································································10 分
取出红球 ········································································ 3 分
x
18
3
4
(或
18
x
18
x
····································································································· 4 分
1
(2)设袋中的红球有 x 只,则有········································································ 5 分
)······································································ 8 分
1
4
x
解得 6
所以,袋中的红球有 6 只.·············································································· 10 分
23.(1)全票为 15 元,则八折票价为 12 分,六折票价为 9 元.·······························2 分
················································································· 4 分
参加郊游的七、八年级同学的总人数必定超过 100 人.········································5 分
(2)设七、八年级参加郊游的同学分别有 x 人、 y 人············································ 6 分
由(1)及已知, 50 50
,
.············································ 7 分
100 15 1500 1575
100
100
x
,
y
x
y
依题意可得:
15
x
9(
x
12
1575
y
) 1080
y
························································································ 10 分
解得
45
75
·······································································································11 分
x
y
答:参加郊游的七、八年级同学分别为 45 人和 75 人.········································· 12 分
24.(1)证明: AD BC
.··············································· 1 分
∥ ,
CDA
DCE
,
,
CDA
≌△
△
是等腰直角三角形,即
, DE BD
45
E °,
DCE
DF FE FC CE
.·············································································· 12 分
BAD
BAD
AB DC AD CE
,
≌△
AC DE
AC BD DE BD
,·········································· 2 分
DCE
.······················································································ 3 分
DCE
又四边形 ABCD 是等腰梯形, BAD
△
.·····················································································5 分
(2) AD CE AD BC
, ∥ , 四边形 ACED 是平行四边形,······························ 7 分
∥ .······························································································· 8 分
.··············································································9 分
由(1)可知, BAD
.··············································10 分
所以, BDE△
四边形 ABCD 是等腰梯形,而
FC .·································································································· 13 分
DF
25.(1)过点 B 作 BD x 轴,垂足为 D ,
CAO
BCD
BCD
又
BCD
△
,········································· 2 分
····························· 3 分
点 B 的坐标为 ( 31) , ;······································ 4 分
.································································································· 14 分
;···········································1 分
ACO
CAO
COA
CAO
≌△
1
,
N
O M
P1
CE AD
BD OC
CD OA
CB AC
2
,
BC
ACO
BDC
AD
90
90
90
°,
2
2
°;
4
,
1
3
B
D
y
A
°
,
C
P2
x
(2)抛物线
y
2
ax
ax
B , ,则得到1 9
a
3
a
,·····················5 分
2
解得
a ,所以抛物线的解析式为
x
;········································7 分
2
1
2
2
经过点 ( 31)
21
x
2
y
1
2
(3)假设存在点 P ,使得 ACP△
① 若以点C 为直角顶点;
仍然是以 AC 为直角边的等腰直角三角形:
则延长 BC 至点 1P ,使得 1PC BC
,得到等腰直角三角形
1ACP△
,······················· 8 分
过点 1P 作 1PM x 轴,
CP BC
1
,
MCP
1
BCD
,
PMC
1
BDC
90
°;
△
1MPC
≌△
DBC
······················································································10 分
CM CD
2
,
PM BD
1
1
,可求得点 1P(1,-1);···································· 11 分
② 若以点 A 为直角顶点;
则过点 A 作 2AP CA ,且使得 2AP
AC
,得到等腰直角三角形
2ACP△
,············ 12 分
过点 2P 作 2P N y 轴,同理可证
△
2AP N
≌△
CAO
;·········································13 分
NP OA
2
2
,
AN OC
1
P , ;·········································· 14 分
经检验,点 1(1 1)
P , 与点 2(2 1)
P , 都在抛物线
y
,可求得点 2(2 1)
21
x
2
1
2
x
上.························ 16 分
2