2007 年内蒙古赤峰市中考数学真题及答案
注意事项:本试卷共 150 分,考试时间为 120 分钟.
一、选择题(每小题给出的四个选项中,只有一个正确选项,请将正确选项的标号填入题
后括号内.每小题 4 分,共 40 分)
)
1. 25 的相反数是(
A.5
2.改革开放二十多年来,赤峰市的经济得到了高效和谐的发展,2006 年我市地区生产总值
已达到 428 亿元,428 亿元用科学记数法表示为(
B. 5
C. 5
D. 25
)
A.
42.8 10 元
9
B.
4.28 10 元
9
C.
42.8 10 元
10
D.
4.28 10 元
10
3.下面右边的图形是由 8 个棱长为 1 个单位的小立方体组成的立体图形,这个立体图形的
主视图是(
)
A.
B.
C.
D.
4.如图, AB CD∥ ,点 E 在CB 的延长线上,若
则 ECD
的度数为(
)
ABE
60
,
C. 60
D. 20
B.100
A.120
5.如图,在菱形 ABCD 中,对角线 AC BD, 分别等于 8 和 6,
将 BD 沿CB 的方向平移,使 D 与 A 重合, B 与CB 延长线上的点
E 重合,则四边形 AECD 的面积等于(
A.36
6.一组数据 8,0,2, 4 ,4 的方差等于(
A.15
7.下列四副图案中,不是轴对称图形的是(
)
D.96
D.18
B.48
C.72
)
)
B.16
C.17
E
60
A
A
B
C
D
D
E
O
B
C
A.
B.
C.
D.
8.在一副扑克牌(54 张,其中王牌两张)中,任意抽取一张牌是“王牌”的概率是(
)
A.
1
54
B.
1
29
C.
1
27
D.
1
13
9.如图,在三角形纸片 ABC 中,
ACB
90
,
BC ,
3
B
C
D
A
E
6
AB ,在 AC 上取一点 E ,以 BE 为折痕,使 AB 的一部
分与 BC 重合, A 与 BC 延长线上的点 D 重合,则CE 的长度
为(
)
A.3
B. 6
C. 3
D. 2 3
10.如下图所示,半径为 1 的圆和边长为 3 的正方形在同一水平线上,圆沿该水平线从左向
右匀速穿过正方形,设穿过时间为t ,正方形除去圆部分的面积为 S (阴影部分),则 S 与t
的大致图象为(
)
s
O
A.
s
s
s
t
O
t
O
t
O
t
B.
C.
D.
二、填空题(本大题共 8 个小题,每小题 4 分,共 32 分,请把答案填在题中横线上)
11.分解因式: 23
x
12
.
12.如图,正方形 ABCD 的边长为 3cm,
ABE
15
,且 AB AE
,则 DE =
cm.
B
C
A
15
D
E
13.某同学的身高为 1.4 米,某一时刻他在阳光下的影长为 1.2 米,此时,与他相邻的一棵
小树的影长为 3.6 米,则这棵树的高度为
米.
.
BOC
14.如图,点 A B C, , 是 O 上的三点,若
则 A 的度数为
15.用正三角形作平面镶嵌,同一顶点周围,正三角形
的个数为
16.如图,半径为 2 的两圆 1O 和 2O 均与 y 轴相切于点O ,反
个.
,
50
比例函数
y
( 0
k )的图像与两圆分别交于点 A B C D
, , , ,
k
x
则图中阴影部分的面积是
.
17.已知
1
a
,则
4
1
b
a
a
3
ab b
7
2
b
ab
2
.
18.观察下列各式:
2
15
1 (1 1) 100 5
2
225
B
O
C
A
y
C
2
O O2
D
x
-2
O1
B
A
2
25
2 (2 1) 100 5
2
625
2
35
3 (3 1) 100 5
2
1225
……
依此规律,第 n 个等式( n 为正整数)为
三、解答题(本大题共 7 个题,满分 78 分,解答时应写出文字说明、证明过程或演算步骤)
19.(本题满分 6 分)
.
计算:
sin 30
1
2
2007
0
| 2 |
.
20.(本题满分 10 分)
“方程”是现实生活中十分重要的数学模型.请结合你的生活实际编写一道二元一次方程组
的应用题,并使所列出的二元一次方程组为
2
x
y
,
60
x
y
,并写出求解过程.
21.(本题满分 10 分)
OB ,如果将 Rt ABO△
2
在坐标平面内,绕原点O 按顺
30
,
中,
A
的位置.
如图 Rt ABO△
时针方向旋转到OA B
(1)求点 B 的坐标.
(2)求顶点 A 从开始到 A 点结束经过的路径长.
A
y
B
B
O
x
A
22.(本题满分 12 分)
有两个可以自由转动的均匀转盘 A B, 都被分成了 3 等份,并在每一份内均标有数字,如图
所示,规则如下:
①分别转动转盘 A B, ;②两个转盘停止后观察两个指针所指份内的数字(若指针停在等份
线上,那么重转一次,直到指针指向某一份内为止).
(1)用列表法(或树状图)分别求出“两个指针所指的数字都是..方程 2 5
x
的解”
6 0
x
的概率和“两个指针所指的数字都不是...方程 2 5
x
x
的解”的概率;
6 0
(2)王磊和张浩想用这两个转盘作游戏,他们规定:若“两个指针所指的数字都是..
x
2 5
x
6 0
x
的解”时,王磊得 1 分;若“两个指针所指的数字都不是...
2 5
x
的
6 0
解”时,张浩得 3 分,这个游戏公平吗?若认为不公平,请修改得分规定,使游戏对双方公
平.
1
2
2
3
3
A
4
B
23.(本题满分 13 分)
三角形中位线定理,是我们非常熟悉的定理.
①请你在下面的横线上,完整地叙述出这个定理:
②根据这个定理画出图形,写出已知和求证,并对该定理给出证明.
24.(本题满分 13 分)
某私立中学准备招聘教职员工 60 名,所有员工的月工资情况如下:
员工
管理人员
教学人员
.
人员结构
校长 副校长 部处主任 教研组长 高级教师 中级教师 初级教师
员工人数/人
1
2
4
10
3
2300
2200
900
2000
2500
17000
每人月工资/元 20000
请根据上表提供的信息,回答下列问题:
(1)如果学校准备招聘“高级教师”和“中级教师”共 40 名(其他员工人数不变),其中
高级教师至少要招聘 13 人,而且学校对高级、中级教师的月支付工资不超过 83000 元,按
学校要求,对高级、中级教师有几种招聘方案?
(2) (1)中的哪种方案对学校所支付的月工资最少?并说明理由.
(3)在学校所支付的月工资最少时,将上表补充完整,并求所有员工月工资的中位数和众
数.
25.(本题满分 14 分)
如图,一元二次方程 2
x
2
x
3 0
x
的二根 1
x, ( 1
x
2
x )是抛物线
2
y
2
ax
bx
与 x
c
轴的两个交点 B C, 的横坐标,且此抛物线过点 (3 6)
A , .
(1)求此二次函数的解析式.
(2)设此抛物线的顶点为 P ,对称轴与线段 AC 相交于点 Q ,求点 P 和点Q 的坐标.
(3)在 x 轴上有一动点 M ,当 MQ MA
取得最小值时,求 M 点的坐标.
y
A(3,6)
Q
C
O B
P
x
2007 年赤峰市初中毕业、升学统一考试试卷
数 学
参考答案及评分标准
一、选择题:(每小题 3 分,共 30 分)
1.B
二、填空:(每空 2 分,共 28 分)
2.D
3.D
4.A
5.A
6.B
7.A
8.C
9.C
10.A
11.3(
x
2)(
x
2)
12.3
13.4.2
14. 25° 15.6
16. 2π
(10
n
三、作图与实验探究题:(共 30 分)
17.1
18.
2
5)
(
n n
1) 100 5
2
1
2007
0
2
19.
sin 30
2
°
1 1 2
2
1
2
································································································4 分
2 ·············································································································· 6 分
20.应用题:我家里有 60 棵树,其中杨树是柳树的 2 倍,求杨树和柳树各有多少棵?
··················································································································· 5 分
解答过程:设杨树 x 棵,柳树 y 棵······································································ 6 分
依题意:
x
x
y
y
2
60
①
②
··················································································7 分
解得
x
y
40
20
·································································································· 9 分
答:我家有杨树 40 棵,柳树 20 棵.··································································10 分
21.解:(1)过点 B 作 B D x
轴于 D ······························································ 3 分
A OB
由题意知,
OB
A °,
OA
OD OB
30
4
………………………………2 分
cos60
° · …………………4 分
2
,
°,
60
∴
2
1
A
y
B
B O D
A
x
1
2
3
2
DB OB
sin 60
2
° ·
3
………………5 分
B∴ 的坐标为: (1 3)
B , ………………………6 分
60
AOA
180
°,
∴
(2)
∵
AOB
A∴ 由开始到结束所经过的路径长为:
22.解:(1)解方程 2 5
x
x
列表:
x
得 1
6 0
60
° ° °··········································7 分
120 π 4
180
2
,
·············································1 分
·········································· 10 分
120
8π
3
3
x
2
2
3
4
1
2
3
1,2
2,2
3,2
1,3
2,3
3,3
1,4
2,4
3,4
(或用树状图)····························································································· 4 分
由表知:指针所指两数都是该方程解的概率是:
4
9
·················································6 分
指针所指两数都不是该方程解的概率是:
·························································· 8 分
1
9
(2)不公平!
1
∵
3
4
9
1
9
············································································ 9 分
修改得分规则为:
指针所指两个数字都是该方程解时,王磊得 1 分.··············································· 10 分
指针所指两个数字都不是该方程解时,张浩得 4 分.············································ 11 分
此时
1
··························································································· 12 分
4
9
4
1
9
23.(1)三角形的中位线平行于第三边且等于第三边的一半.··································2 分
(2)
A
D
E
F
B
C
1
2
···················································································································· 3 分
已知: DE 是 ABC△
的中位线·········································································· 4 分
求证: DE
BC∥ ,
········································································ 5 分
DE
BC
CEF
ADE
, AED
CEF
≌△
ADE
,
····································································6 分
证明:延长 DE 到 F ,使 EF DE
连接CF
AE CE∵
········································································ 7 分
······················································································· 8 分
∴△
AD CF
········································································· 9 分
∴
AD CF
∴ ∥ ································································································ 10 分
AD BD∵
BD CF∴
································································································· 11 分
∴四边形 BCFD 是平行四边形········································································· 12 分
············································································ 13 分
∴ ∥ ,
CFE
DE
BC
DE
BC
1
2
24.解:(1)设高级教师招聘 x 人,则中级教师招聘 (40
)x 人······························· 1 分
依题意得: 2200
x
2000(40
≤
x
)
83000
·························································2 分
解此不等式得:
x ≤ ···················································································· 3 分
15
又
x∵ ≥
13
x
13
15
∴ ≤ ≤ ······························································································ 4 分
x∵ 是正整数,
∴学校对高级教师,中级教师有三种招聘方案
, , ·········································································· 5 分
1314 15
x ∴
方案一 高级教师 人 中级教师 人
方案二 高级教师 人 中级教师 人
方案三 高级教师 人 中级教师 人
13 ,
14 ,
15 ,
27
26
25
:
:
:
∵
2000
13 14 15
,即高级教师的月薪大于中级教师的月薪.······························· 7 分
(2) 2200
∴高级教师的招聘人数越小,学校所支付的月工资越少.········································8 分
·································································································· 9 分
∵
∴当高级教师招聘 13 人,中级教师招聘 27 人时,学校所支付的月工资最少.···········10 分
(3)补表:13、27·························································································11 分
在学校所支付的月工资最少时,中位数是 2100 元,众数是 2000 元·························· 13 分
25.解:(1)解方程 2
x
x
得 1
∴抛物线与 x 轴的两个交点坐标为: ( 3 0)
设抛物线的解析式为
y
A∵ , 在抛物线上
······························································································ 1 分
,, , ··············································· 2 分
····························································································· 3 分
(
a x
(3 6)
3
,
3 0
2
x
(1 0)
3)(
x
2
1)
C
B
1
x
∴
6
a
(3 3) (3 1)
·
a ∴
1
2
··········································································4 分
∴抛物线解析式为:
y
(2)由
y
1
2
2
x
x
3
2
21
x
2
1 (
2
x
···································································· 5 分
x
3
2
2
1)
·································································· 6 分
2
∴抛物线顶点 P 的坐标为: ( 1
, ,对称轴方程为:
设直线 AC 的方程为: y
kx b
, 在该直线上
∵ ,,
2)
C
( 3 0)
(3 6)
A
x ···································7 分
1
∴
3
6
k
b
3
k
b
0
解得
3
1
∴直线 AC 的方程为:
y
x ········································ 9 分
3
b
k
3
1
y
x 代入
x 得 2
y
, ························································································10 分
将
Q∴ 点坐标为 ( 1 2)
(3)作 A 关于 x 轴的对称点 (3
·················································································································· 11 分
设直线 A Q 方程为 y
, ,连接 A Q ; A Q 与 x 轴交于点 M 即为所求的点
kx b
A
6)
∴
3
6
k
b
2
k
b
解得
0
b
k
2
2
····················································································· 12 分
∴直线 A C :
y
y ···························································································13 分
令 0
M∴ 点坐标为 (0 0), ························································································ 14 分
x ,则 0
x
y
A(3,6)
Q
C
O B
P
x
(3
A ,
6)
说明:考生在解答第三题(19 -25 题)时,如有其它解法,只要正确,均可参照本标准合
理赋分.