2012 年福建省厦门市中考数学真题及答案
(试卷满分:150 分 考试时间:120 分钟)
准考证号
姓名
座位号
注意事项:
1.全卷三大题,26 小题,试卷共 4 页,另有答题卡.
2.答案一律写在答题卡上,否则不能得分.
3.可直接用 2B 铅笔画图.
一、选择题(本大题有 7 小题,每小题 3 分,共 21 分.每小题都有四个 选项,其中有且只
有一个选项正确)
1. -2 的相反数是
A.2
B.-2
C.±2
2.下列事件中,是必然事 件的是
A. 抛掷 1 枚硬币,掷得的结果是正面朝上
B. 抛掷 1 枚硬币,掷得的结果是反面朝上
C. 抛掷 1 枚硬币,掷得的结果不是正面朝上就是反面朝上
D.抛掷 2 枚硬币,掷得的结果是 1 个正面朝上与 1 个反面朝上
D.-
1
2
3.图 1 是一个立体图形的三视图,则这个立体图形是
A.圆锥
C.圆柱
B.球
D.三棱锥
4.某种彩票的中奖机会是 1%,下列说法正确的是
A.买 1 张这种彩票一定不会中奖
B.买 1 张这种彩票一定会中奖
C.买 100 张这种彩票一定会中奖
D.当购买彩票的数量很大时,中奖的频率稳定在 1%
5.若二次根式 x-1有意义,则 x的取值范围是
A.x>1
C.x<1
B.x≥1
D.x≤1
6.如图 2,在菱形 ABCD中,AC、BD是对角线,
若∠BAC=50°,则∠ABC等于
B.50°
A.40°
C.80°
D.100°
正
视
图
俯
视
图
B
7.已知两个变量 x和 y,它们之间的 3 组对应值如下表所示.
x
y
-1
-1
0
1
1
3
则 y 与 x之间的函数关系式可能是
左
视
图
D
图1
A
C
图2
A.y=x
B.y=2x+1
C.y=x2+x+1
3
D.y=
x
二、填空题(本大题有 10 小题,每小题 4 分,共 40 分)
8.计算: 3a-2a=
.
9.已知∠A=40°,则∠A的余角的度数是
.
10.计算: m3÷m2=
.
A
D
11.在分别写有整数 1 到 10 的 10 张卡片中,随机抽取 1 张
卡片,则该卡片上的数字恰好是奇数的概率是
.
12.如图 3,在等腰梯形 ABCD中,AD∥BC,对角线 AC
与 BD相交于点 O,若 OB=3,则 OC=
13.“x与 y的和大于 1”用不等式表示为
.
.
14.如图 4,点 D是等边△ABC内一点,如果△ABD绕点 A
逆时针旋转后能与△ACE重合,那么旋转了
度.
15.五边形的内角和的度数是
.
16.已知 a+b=2,ab=-1,则 3a+ab+3b=
;
a2+b2=
.
B
B
17.如图 5,已知∠ABC=90°,AB=πr,BC=
,半径为 r
πr
2
O
A
A
F
B
的⊙O从点 A出发,沿 A→B→C方向滚动到点 C时停止.
请你根据题意,在图 5 上画出圆心..O运动路径的示意图;
圆心 O运动的路程是
.
三、解答题(本大题有 9 小题,共 89 分)
18.(本题满分 18 分)
(1)计算:4÷(-2)+(-1)2×40;
(2)画出函数 y=-x+1 的图象;
(3)已知:如图 6,点 B、F、C、E在一条直线上,
∠A=∠D,AC=DF,且 AC∥DF.
求证:△ABC≌△DEF.
19.(本题满分 7 分)解方程组:
3x+y=4,
2x-y=1.
O
图3
A
D
图4
→
图5
图6
C
E
C
B
C
C
D
E
20.(本题满分 7 分)已知:如图 7,在△ABC中,∠C=90°,点 D、E分别在边 AB、AC
上,DE∥BC,DE=3, BC=9.
(1)求
AD
AB
的值;
(2)若 BD=10,求 sin∠A的值.
A
E
C
D
B
图7
21.(本题满分 7 分)已知 A 组数据如下:
0,1,-2,-1,0,-1,3.
(1)求 A 组数据的平均数;
(2)从 A 组数据中选取 5 个数据,记这 5 个数据为 B 组数据. 要求 B 组数据满足两个
条件:①它的平均数与 A 组数据 的平均数相等;②它的方差比 A 组数据的方差大.
你选取的 B 组数据是
,请说明理由.
【注:A 组数据的方差的计算式是
SA
2=
1
7
[(x1-
—
x)2+(x2-
—
x)2+(x3-
—
x)2+(x4-
—
x)2+(x5-
—
x)2+(x6-
—
x)2+(x7-
—
x)2]】
22.(本题满分 9 分)工厂加工某种零件,经测试,单独加工完成这种零件,甲车床需用
x小时,乙车床需用 (x2-1)小时,丙车床需用(2x-2)小时.
(1)单独加工完成这种零件,若甲车床所用的时间是丙车床的
2
3
,求乙车床单独加工
完成这种零件所需的时间;
(2)加工这种零件,乙车床的工作效率与丙车床的工作效率能否相同?请说明理由.
23.(本题满分 9 分)已 知:如图 8,⊙O是△ABC的外接圆,AB为⊙O的直径,弦 CD交 AB
于 E,∠BCD=∠BAC .
(1)求证:AC=AD;
C
(2)过点 C作直线 CF,交 AB的延长线于点 F,
若∠BCF=30°,则结论“CF一定是⊙O的切线”
是否正确?若正确,请证明;若不正确,请举反例.
F
B
E
O
A
D
图8
24.(本题满分 10 分)如图 9,在平面直角坐标系中,已知点 A(2,3)、B(6,3),连结 AB.
如果点 P在直线 y=x-1 上,且点 P到直线 AB的距离小于 1,那么称点 P是线段 AB
的“邻近点”.
(1)判断点C(
7
2
5
,
2
) 是否是线段 AB的“邻近点”,并说明理由;
(2)若点 Q (m,n)是线段 AB的“邻近点”,求 m的取值范围.
y
4
2
O
A
B
x
2
4
6
图9
25.(本题满分 10 分)已知□ABCD,对角线 AC与 BD相交于点 O,点 P在边 AD上,过点 P
分别作 PE⊥AC、PF⊥BD,垂足分别为 E、F,PE=PF.
(1)如图 10,若 PE= 3,EO=1,求∠EPF的度数;
(2)若点 P是 AD的中点,点 F是 DO的中点,
BF =BC+3 2-4,求 BC的长.
A
P
D
B
E
O
F
图10
C
26.(本题满分 12 分)已知点 A(1,c)和点 B (3,d )是直线 y=k1x+b与双曲线 y=
k2
x
(k2
>0)的交点.
(1)过点 A作 AM⊥x轴,垂足为 M,连结 BM.若 AM=BM,求点 B的坐标;
(2)设点 P在线段 AB上,过点 P作 PE⊥x轴,垂足为 E,并交双曲线 y=
k2
x
(k2>0)
于点 N.当
PN
NE
取最大值时,若 PN=
1
2
,求此时双曲线的解析式.
一、选择题(本大题共 7 小题,每小题 3 分,共 21 分)
参考答案
题号
选项
1
A
2
C
3
A
4
D
5
B
6
C
7
B
二、填空题(本大题共 10 小题,每题 4 分,共 40 分)
8. a.
9. 50°.
10. m.
11.
1
.
2
12. 3.
13. x+y>1.
14. 60.
15. 540°.
16. -1; 6.
17. 2πr.
三、解答题(本大题共 9 小题,共 89 分)
18.(本题满分 18 分)
(1)解:4÷(-2) +(-1)2×40
=-2+1×1···································································· 4 分
=-2+1·········································································5 分
=-1. ·········································································· 6 分
(2)解:正确画出坐标系································································8 分
正确写出两点坐标··························································· 10 分
画出直线······································································· 12 分
(3)证明:∵ AC∥DF,
∴ ∠ACB=∠DFE.
又∵ ∠A=∠D,
AC=DF,
∴ △ABC≌△EDF.
……13 分
……15 分
……16 分
……17 分
B
……18 分
A
F
E
C
D
19.(本题满分 7 分)
解 1:
3x+y=4,
2x-y=1.
①
②
①+②,得······································································1 分
5x=5,·········································································· 2 分
x=1. ············································································4 分
将 x=1 代入 ①,得
3+y=4,······································································· 5 分
y=1.············································································6 分
∴
x=1,
y=1.
······································································· 7 分
解 2:由①得 y=4-3x. ③········································ 1 分
将③代入②,得
2x-(4-3x) =1.··························································· 2 分
得 x=1. ········································································ 4 分
将 x=1 代入③ ,得
y=4-3×1······································································5 分
=1.············································································6 分
∴
x=1,
y=1.
······································································· 7 分
A
E
C
D
G
B
20.(本题满分 7 分)
(1)解:∵ DE∥BC ,∴ △ADE∽△ABC. ……1 分
∴
∴
AD
AB
AD
AB
=
=
DE
BC
1
.
3
AD
AB
AD
=
1
3
(2)解 1:∵
∴
.
……2 分
……3 分
,BD=10,
1
=
3
································································· 4 分
AD+10
∴ AD=5······································································· 5 分
经检验,符合题意. ∴ AB=15.
在 Rt△ABC中,································································6 分
sin∠A=
. ······························································ 7 分
解 2: ∵
3
=
5
BC
AB
1
,BD=10,
3
=
AD
AB
AD
∴
1
=
3
································································· 4 分
AD+10
∴ AD=5······································································· 5 分
经检验,符合题意.
∵ DE∥BC,∠C=90°
∴ ∠AED=90°
在 Rt△AED中,································································6 分
sin∠A=
ED
AD
3
=
5
. ······························································ 7 分
解 3:过点 D作 DG⊥BC,垂足为 G. ∴ DG∥AC.
∴∠A=∠BDG. ································································4 分
又∵ DE∥BC,∴四边形 ECGD是平行四边形.
∴ DE=CG.······································································5 分
∴ BG=6.
在 Rt△DGB中,································································6 分
∴ sin∠BDG=
BD
GB
3
=
5
. ···································7 分[来源:学科网]
∴ sin∠A=
3
.
5
21.(本题满分 7 分)
(1)解:A 组数据的平均数是
0+1-2-1+0-1+3
7
····························1 分
=0. ·················································· 3 分
(2)解 1:选取的 B 组 数据:0,-2,0,-1,3. ····························· 4 分
∵ B 组数据的平均数是 0. ················································ 5 分
∴ B 组数据的平均数与 A 组数据的平均数相同.
∴ SB
2=
∴
14
5
14
5
16
7
>
,SA
2=
16
7
. ····················································· 6 分
.···································································· 7 分
∴ B 组数据:0,-2,0,-1,3.
解 2:B 组数据:1,-2,-1,-1,3. ···································· 4 分
∵ B 组数据的平均数是 0. ················································ 5 分
∴ B 组数据的平均数与 A 组数据的平均数相同.
, SB
2=
. ····················································· 6 分
16
5
∵SA
2=
∴
16
5
>
16
7
16
7
······································································· 7 分
∴ B 组数据:1,-2,-1,-1,3.
22.(本题满分 9 分)
(1)解:由题意得,
x=
2
(2x-2) ···································································1 分
3
∴ x=4. ······································································· 2 分
∴ x2-1=16-1=15(小时). ·············································3 分
答:乙车床单独加工完成这种零件所需的时间是 15 小时. ······· 4 分
(2)解 1:不相同. ······································································ 5 分
若乙车床的工作效率与丙车床的工作效率相同,由题意得, ······6 分
. ·······························································7 分
1
x2-1
∴
1
=
1
2x-2
1
2
=
.
x+1
∴ x=1. ······································································8 分
经检验,x=1 不是原方程的解. ∴ 原方程无解. 9 分[来源:学科网 ZXXK]
答:乙车床的工作效率与丙 车床的工作效率不相同.
解 2:不相同. ······································································ 5 分
若乙车床的工作效率与丙车床的工作效率相同,由题意得, ······6 分
x2-1=2x-2. ································································ 7 分
解得,x=1. ···································································8 分