2009 年福建省漳州市中考数学真题及答案
(满分:150 分;考试时间:120 分钟)
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姓名________________准考证号___________________
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.....重描
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一、选择题(共 10 小题,每小题 3 分,满分 30 分,每小题只有一个正确的选项,请在答题..
卡.的相应位置填涂)
1. 3 的倒数是(
A. 3
D.3
C.
B.
)
1
3
1
3
2.要调查某校九年级 550 名学生周日的睡眠时间,下列调查对象选取最合适的是(
A.选取该校一个班级的学生
B.选取该校 50 名男生
C.选取该校 50 名女生
D.随机选取该校 50 名九年级学生
3.一个几何体的三视图如图所示,这个几何体是(
A.圆柱
D.正方体
4.下列运算正确的是(
主(正)视图 左视图
C.圆锥
B.球
)
)
(第 3 题)
俯视图
)
A.
2
a a
2
a
2
B.
(
)a
2
a
2
C. 2 3
)a
(
5
a
D. 3
a
a
2
a
5.三角形在方格纸中的位置如图所示,则 tan 的值是(
A.
3
4
B.
C.
D.
3
5
4
5
4
3
)
α
(第 5 题)
6.据统计,2009 年漳州市报名参加中考总人数(含八年级)约为 102000 人,则 102000 用
科学记数法表示为(
)
0.102 10
6
A.
7.矩形面积为 4,它的长 y 与宽 x 之间的函数关系用图象大致可表示为(
B.
D.
C.
10.2 10
4
1.02 10
5
102 10
3
)
y
y
y
y
x
O
x
O
x
O
A.
B.
C.
8.如图,要使 ABCD
A. AB BC
C.
ABC
° D. 1
B. AC BD
2
成为矩形,需添加的条件是(
9.分式方程
的解是(
)
90
2
1x
1
x
O
D.
)
x
B
A
D
1
2
O
(第 8 题)
C
A.1
B. 1
C.
1
3
D.
1
3
110
10 . 如 图 , OAB△
A °,
O
(第 10 题)
A.30°
二、填空题(共 6 小题,每小题 4 分,满分 24 分,请将答案填入答题卡...的相应位置)
绕 点 O 逆 时 针 旋 转 80°得 到 OCD△
40
D °,则 的度数是(
B.40°
C.50°
D.60°
, 若
)
D
C
A
α
B
l1
l2
11.若分式
1
2x
l
12.如图,直线 1
无意义,则实数 x 的值是____________.
l∥ , 1 120
°,则 2 =_______________度.
2
13.若 2 2
m
m
,则 22
1
m
m
4
2007
的值是_______________.
14.已知一次函数
y
2
x
1
,则 y 随 x 的增大而_______________(填
1
2
(第 12 题)
“增大”或“减小”).
15.如图是第 29 届北京奥运会上获得金牌总数前六名国家的统计图,则这组金牌数的中位
数是____________枚.
60
16.如图,在菱形 ABCD 中,
则菱形 ABCD 的边长是_____________.
A °, E 、 F 分别是 AB 、 AD 的中点,若
EF ,
2
奥运金牌榜前六名国家
金牌数(枚)
60
50
40
30
20
10
0
51
36
中
国
美
国
(2008 年 8 月 24 日统计)
23
19
16
德
国
英
国
俄
罗
斯
(第 15 题)
14
澳
大
利
亚
国家
F
D
E
B
A
C
(第 16 题)
三、解答题(10 大题共 96 分,请将答案填入答题卡...的相应位置)
17.(满分 8 分)计算:
2
( 2)
0
1
1
3
.
18.(满分 8 分)给出三个多项式: 21
x
2
2
x
, 21
1
x
2
4
x
, 21
1
x
2
x .请选择你最
2
喜欢的两个多项式进行加法运算,并把结果因式分解.
19.(满分 8 分)如图,在等腰梯形 ABCD 中,E 为底 BC
的中点,连结 AE 、 DE .求证: ABE
DCE
≌△
△
.
A
D
B
E
(第 19 题)
C
20.(满分 8 分)漳浦县是“中国剪纸之乡”.漳浦剪纸以构图丰满匀称、细腻雅致著称.下
面两幅剪纸都是该县民间作品(注:中间网格部分未创作完成).
(1)请从“吉祥如意”中选一字填在图 1 网格中,使整幅..作品成为轴对称图形;
(2)请在图 2 网格中设计一个四边形图案,使整幅..作品既是轴对称图形,又是中心对称图
形.
图 1
图 2
(第 20 题)
30
21.(满分 8 分)如图,点 D 在 O⊙ 的直径 AB 的延长线上,点C 在 O⊙ 上, AC CD
D °,
(1)求证:CD 是 O⊙ 的切线;
(2)若 O⊙ 的半径为 3,求 BC 的长.(结果保留 π )
C
A
O
B
(第 21 题)
,
D
22.(满分 8 分)阅读材料,解答问题.
例 用图象法解一元二次不等式: 2 2
x
x
.
3 0
解:设
y
x
2 2
x
,则 y 是 x 的二次函数.
3
y
3
2
1
a
1 0
, 抛物线开口向上.
又当 0
y 时, 2 2
x
x
3 0
x
,解得 1
1
,
x
2
3
.
由此得抛物线
y
x
2 2
x
的大致图象如图所示.
3
观察函数图象可知:当
x 或 3
x 时, 0
y .
1
3
x
1 2
12
1
2
3
4
(第 22 题)
2 2
x
x
的解集是:
3 0
x 或 3
x .
1
(1)观察图象,直接写出一元二次不等式: 2 2
x
x
的解集是____________;
3 0
(2)仿照上例,用图象法解一元二次不等式: 2 1 0
x .(大致图象画在答题卡...上)
23.(满分 10 分)为了防控甲型 H1N1 流感,某校积极进行校园环境消毒,购买了甲、乙两
种消毒液共 100 瓶,其中甲种 6 元/瓶,乙种 9 元/瓶.
(1)如果购买这两种消毒液共用 780 元,求甲、乙两种消毒液各购买多少瓶?
(2)该校准备再次..购买这两种消毒液(不包括已购买的 100 瓶),使乙种瓶数是甲种瓶数的
2 倍,且所需费用不多于...1200 元(不包括 780 元),求甲种消毒液最多能再购买多少瓶?
24.(满分 11 分)小红与小刚姐弟俩做掷硬币游戏,他们两人同时各掷一枚壹元硬币.
(1)若游戏规则为:当两枚硬币落地后正面朝上时,小红赢,否则小刚赢.请用画树状图
或列表的方法,求小刚赢的概率;
(2)小红认为上面的游戏规则不公平,于是把规则改为:当两枚硬币正面都朝上时,小红
得 8 分,否则小刚得 4 分.那么,修改后的游戏规则公平吗?请说明理由;若不公平,请你
帮他们再修改游戏规则,使游戏规则公平(不必说明理由).
25.(满分 13 分)
几何模型:
条件:如下左图, A 、 B 是直线l 同旁的两个定点.
问题:在直线l 上确定一点 P ,使 PA PB 的值最小.
方法:作点 A 关于直线l 的对称点 A ,连结 A B 交 l 于点 P ,则 PA PB A B
(不必证明).
模型应用:
(1)如图 1,正方形 ABCD 的边长为 2,E 为 AB 的中点,P 是 AC 上一动点.连结 BD ,
由正方形对称性可知,B 与 D 关于直线 AC 对称.连结 ED 交 AC 于 P ,则 PB PE 的最
小值是___________;
(2)如图 2, O⊙ 的半径为 2,点 A B C、 、 在 O⊙ 上,OA OB ,
OB 上一动点,求 PA PC
45
PO ,Q R、 分别是OA OB、 上
°, P 是 AOB
(3)如图 3,
AOC
°,P 是
60
的最小值;
的值最小
AOB
内一点,
10
的动点,求 PQR△
周长的最小值.
A
A
′
P
B
l
A
B
E
P
D
图 1
C
(第 25 题)
A
O
P
图 2
C
B
O
B
P
R
Q
图 3
A
26.(满分 14 分)如图 1,已知:抛物线
交于点C ,经过 B C、 两点的直线是
y
x
与 x 轴交于 A B、 两点,与 y 轴
c
bx
21
x
2
,连结 AC .
2
y
1
2
(1) B C、 两点坐标分别为 B (_____,_____)、C (_____,_____),抛物线的函数关系
式为______________;
(2)判断 ABC△
(3)若 ABC△
上)?若能,求出在 AB 边上的矩形顶点的坐标;若不能,请说明理由.
内部能否截出面积最大的矩形 DEFC (顶点 D E F、 、 、G 在 ABC△
的形状,并说明理由;
各边
[抛物线
y
2
ax
bx
的顶点坐标是
c
4,
b
2
a
ac b
4
a
2
y
]
y
O
A
C
B
x
A
O
B
x
C
图 1
(第 26 题)
图 2(备用)
参考答案
一、选择题(共 10 题,每题 3 分,满分 30 分)
题号
答案
1
B
2
D
3
A
4
D
5
A
6
B
7
B
8
C
9
A
10
C
13.2009
12.120
16.4
二、填空题(共 6 小题,每题 4 分,满分 24 分)
11.2
15.21
三、解答题(10 大题,满分共 96 分)
17.解:原式= 2 1 3
·····················································································6 分
=0.·············································································································· 8 分
14.增大
18.解:情况一: 2
x
1
2
2
x
1
1
2
2
x
4
x
1
······················································2 分
= 2 6
x
x
······································································································· 5 分
= (
x x .··································································································· 8 分
6)
情况二: 2
x
1
2
2
x
1
1
2
2
x
2
x
······································································ 2 分
= 2 1
x ·········································································································· 5 分
= (
x
1)(
x
1)
.······························································································8 分
情况三: 2
x
2
x
4
1
2
1
1
2
1
··································································································· 5 分
······································································ 2 分
2
x
x
= 2
x
2
x
=
B
E
C
A
D
B
C
2
1)
,
≌△
DCE
BE EC
AB DC
.··································4 分
.·················································· 6 分
.······································· 8 分
x .·····································································································8 分
(
19.证明:四边形 ABCD 是等腰梯形,
E 为 BC 的中点,
ABE
△
20.(1)吉.(符合要求就给分)········································································3 分
(2)有多种画法,只要符合要求就给分.····························································8 分
21.(1)证明:连结 OC ,··································1 分
AC CD
,
D
°·············································2 分
OA OC
2
A
°,···········································3 分
1 60
°,
90
OCD
°.··························································································· 4 分
CD 是 O⊙ 的切线.······················································································5 分
D
30
(第 21 题)
(第 19 题)
°,
30
30
2
1
A
O
B
,
C
A
D
°,
(2) 1 60
π
n R
BC 的长=
180
60 π 3
180
.····································································7 分
π
答: BC 的长为 π .························································································· 8 分
22.(1) 1
.·········································2 分
y
1
y
x
2 1
3x
x
(2)解:设
y
1 0
a
y 时, 2 1 0
又当 0
2 1
,则 y 是 x 的二次函数.
, 抛物线开口向上.···························3 分
x ,解得 1
x
1
,
x
2
1
. 4 分
1
1
x
1
由此得抛物线
y
x
2 1
的大致图象如图所示.···· 6 分
观察函数图象可知:当
x 或 1x 时, 0
y .················································ 7 分
1
(第 22 题)
x 的解集是:
2 1 0
x 或 1x .····························································· 8 分
1
23.(1)解法一:设甲种消毒液购买 x 瓶,则乙种消毒液购买 (100
)x 瓶.···············1 分
依题意,得 6
x
9(100
x
) 780
.
解得: 40
x .······························································································ 3 分
100
x
(瓶).······································································4 分
答:甲种消毒液购买 40 瓶,乙种消毒液购买 60 瓶.·············································· 5 分
解法二:设甲种消毒液购买 x 瓶,乙种消毒液购买 y 瓶.········································ 1 分
100 40 60
依题意,得
x
6
100
y
9
y
x
,
780
·············································································· 3 分
.
解得:
x
y
40
60
,
······························································································ 4 分
.
答:甲种消毒液购买 40 瓶,乙种消毒液购买 60 瓶.·············································· 5 分
(2)设再次购买甲种消毒液 y 瓶,刚购买乙种消毒液 2y 瓶.································· 6 分
依题意,得 6
y
≤
9 2
y
1200
.······································································ 8 分
解得:
y ≤ .·····························································································9 分
50
答:甲种消毒液最多再购买 50 瓶.··································································· 10 分
26.(1) B (4,0), (0
C , .······································································ 2 分
2)
y
21
x
2
3
2
(2) ABC△
x
.························································································ 4 分
2
是直角三角形.············································································5 分