2007 年广东省韶关市中考数学真题及答案
说明:1.全卷共 4 页,分为选择题和非选择题两部分,考试时间为 100 分钟,满分为 120
分.
2.考生答题必须全部在答题卡上作答,写在试卷上的答案无效.
3.答卷前考生必须用黑色字迹的钢笔或签字笔将自己的姓名、准考证号按要求写、
涂在答题卡指定位置上,用 2B 铅笔将试室号、座位号填涂在答题卡指定位置上.
4.做选择题时,每小题选出答案后,用 2B 铅笔将答题卡上对应题目的答案标号涂
黑,如需改动,用橡皮擦干净后,再选涂其他答案,不能答在试卷上.做非选择题
时,用黑色字迹的钢笔或签字笔按各题要求将答案写在答题卡指定位置上,不能用
铅笔、圆珠笔或红笔作答.如需改动,先划掉原来的答案,再写上新的答案;不准
使用涂改液和涂改带.
5.考生必须保持答题卡的整洁.考试结束后,将本试卷和答题卡一并交回.
一、选择题(本题共 10 小题,每小题 3 分,共 30 分,每小题给出 4 个答案,其中只有一
个正确)
1.下列计算结果最小的是(
)
A.1 2
B.1 2
2.下列运算正确的是(
C.1 2
)
B. 2
a a
3
D.1 2
2
A.
(2 )
a
3.在 ABC
A.5
2
5
2
a
a
BC ,则 DE 等于(
中, D 、 E 分别是 AB 、 AC 边上的中点,若
D. 2 3
)a
6
C. 2
a
3
a
a
a
(
5
6
)
B. 4
C. 3
D. 2
4.2007 年 5 月份,某市市区一周空气质量报告中某项污染指数的数据是:31 35
31 34
30 32 31 ,这组数据的中位数、众数分别是(
)
A.32,31
B.31,32
C.31,31
D. 32,35
5.小明拿一个等边三角形木框在阳光下玩,等边三角形木框在地面上形成的投影不可能...是
(
)
A
B
A ,且 A 为锐角,则 A =(
C
)
D
6.已知
sin
A.30
1
2
B. 45
7.按下列程序计算,最后输出的答案是(
立方
1
a
a
a
C. 60
D. 75
)
答案
A. 3a
B. 2 1
a
C. 2a
D. a
8.一辆汽车由韶关匀速驶往广州,下列图象中大致能反映汽车距离广州的路程 s (千米)
和行驶时间t (小时)的关系的是(
)
t
s
o
s
t
o
t
s
s
o
A
t
o
9.如图 1,CD 是 Rt ABC
B
C
D
斜边上的高,则图中相似三角形的对数有 (
)
A. 0 对
B.1对
C. 2 对
D.3 对
10.有一个两位数,它的十位数字比个位数字大 2 ,并且这个两位数大于
40 且小于52 ,则这个两位数是(
A. 41
B. 42
C. 43
)
D. 44
D
B
图 1
二、填空题(本题共 5 小题,每小题 3 分,共 15 分)
11.据韶关市 2006 年国民经济和社会发展统计公报显示:我市 2006 年在校初中学生人数
A
C
约为15.9 万,用科学记数法表示为
12.因式分解: 3 4
a
.
13.如图 2, AD 是⊙O 的直径, AB ∥CD ,
a
=
.
AOC
,
60
则 BAD
度.
14.请你写一个图象在第二、四象限的反比例函数关系式:
15.按如下规律摆放三角形:
B
D
A
C
.
O
图 2
1
2
3
则第( 4 )堆三角形的个数为
;第( n )堆三角形的个数为
.
三、解答题(本大题共 5 小题,其中第 16、17 小题各 6 分,第 18、19、20 小题各 7 分,共
33 分)
16.计算:
2
17.解方程:
x
1
x
1
(3
)
0
9
1
2
2 1
x
.
18.如右图,方格纸中的每格都是边长为 1 的正方形,将 OAB
绕点O 按顺时针方向旋转90 得到 OA B
(1)在给定的方格纸中画出 OA B
(2)OA 的长为
,
.
;
A
B
O
AA 的长为
.
19.某中学准备搬迁新校舍,在迁入新校舍之前,同学们就该校学生如何到校问题进行了一
次调查,并将调查结果制成了表格、条形图和扇形统计图,请你根据图表信息完成下列各题:
(1)此次共调查了多少位学生?
(2)请将表格填充完整;
(3)请将条形统计图补充完整.
步行 骑自行车 坐公共汽车 其他
60
其他
3%
步行
20%
坐公共汽车
44%
骑自行车
33%
20.已知抛物线
y
x
2 2
x
与 x 轴的右交点为 A ,与 y 轴的交点为 B ,求经过 A 、B 两
3
点的直线的解析式.
四、解答题(本大题共 3 小题,每小题 8 分,共 24 分)
21.不透明的口袋里装有红、黄、蓝三种颜色的小球若干个(除颜色外其余都相同),其中
红球 2 个(分别标有1号、2 号),蓝球1个.若从中任意摸出一个球,它是蓝球的概率为
1
4
.
(1)求袋中黄球的个数;
(2)第一次任意摸一个球(不放回),第二次再摸一个球,请用画树状图或列表格的方
法,求两次摸到不同颜色球的概率.
22.如图 3, AB 是半⊙O 的直径,弦 AC 与 AB 成30 的角, AC CD
C
(1)求证:CD 是半⊙O 的切线;
(2)若
OA ,求 AC 的长.
2
.
A
O
图 3
B
D
23.为了改善小区环境,某小区决定要在一块一边靠墙(墙长 25m )的空地上修建一个矩
形绿化带 ABCD ,绿化带一边靠墙,另三边用总长为 40m 的栅栏围住(如图 4). 若设绿
化带的 BC 边长为 xm ,绿化带的面积为 2ym .
(1)求 y 与 x 之间的函数关系式,并写出自变量 x 的取值范围;
(2)当 x 为何值时,满足条件的绿化带的面积最大?
A
B
25m
图 4
五 、解答题(本大题共 2 小题,每小题 9 分,共 18 分)
24.如图 5, 四边形 ABCD 中,AD 不平行 BC ,现另给出三个条件:① CAB
② AC BD
后能够推出 ABCD 是等腰梯形,并加以证明(只需证明一种情况).
DBA
,
, ③ AD BC .请你从上述三个条件中选择两个条件,使得加上这两个条件
C
D
25. 如图 6, 在平面直角坐标系中, 四边形OABC 是矩形,
OA ,
x 与坐标轴交于 D 、E .设
AB , 直线
4
2
y
3
2
M 是 AB 的中点, P 是线段 DE 上的动点.
(1) 求 M 、 D 两点的坐标;
(2) 当 P 在什么位置时, PA PB ?求出此时 P 点的坐标;
(3) 过 P 作 PH BC
与 BC 相切于点 N 时, 求梯形 PMBH 的面积.
, 垂足为 H , 当以 PM 为直径的⊙ F
D
C
A
图 5
y
C
E
O
N
F
H
P
D
B
B
M
A
x
图 6
数学试题评分标准及参考答案
一、选择题:
B C C C B A C B D B
二、填空题:
11、
1.59 10
5
12、 (
a a
14、(答案不唯一) 15、14 ;3
三、解答题:16、原式= 2 3 2 1
a
2)
2)(
2n (第一空 1 分,第 2 空 2 分)
13、30
··································································· 4 分
=4 ··············································································· 6 分
x x ,·········································································1 分
1)
17、方程两边都乘以 (
得 2
x
2(
x
1)
(
x x
1)
·········································································· 3 分
解这个方程,得
x ············································································· 4 分
2
3
经检验,
x 是原方程的根
2
3
所以,原方程的根是
x . ····························· 6 分
2
3
18、(1) OA B
5
OA
AA
5 2
(2)
的位置如右图····························· 4 分
············································ 5 分
·········································· 7 分
19、(1)调查的学生数为:
60 20% 300
;······················ 3 分
(2)如下表······························· 6 分
(3)如右图······························· 7 分
140
120
100
80
60
40
20
0
O
步行 骑自行车 坐公共汽车 其他
99
132
9
行
步
车
行
自
骑
车
汽
共
公
坐
他
其
x
A
x
2
3 0
y ,得 2 2
x
………………1 分
1
20、令 0
3,
x
解得 1
则 (3,0)
······························································································· 2 分
又令 0
则 (0, 3)
设直线 AB 的解析式为 y
····························································································· 3 分
,·····························································4 分
x ,得
kx b
y
B
3
则
0
3
k b
3
b
························································································· 5 分
解得
k
1,
b
3
············································································· 6 分
所以直线 AB 的解析式为
y
x
3
··························································· 7 分
四、解答题:
21、(1)袋中黄球的个数为 1 个;······································································ 2 分
(2)法一、列表如下,··············································································· 6 分
所以两次摸到不同颜色球的概率为:
P
10
12
·······································8 分
5
6
红 1
红 2
黄
蓝
红 1
红 2
黄
蓝
(红 2,红 1)
(黄,红 1)
(蓝,红 1)
(红 1,红 2)
(红 1,黄)
(红 1,蓝)
(红 2,黄)
(红 2,蓝)
(黄,红 2)
(蓝,红 2)
(蓝,黄)
(黄,蓝)
法二、画数状图如下,·········································································· 6 分
所以两次摸到不同颜色球的概率为:
P
10
12
.······································8 分
5
6
开始
C
红1
红2
黄
蓝
红
2
黄 蓝
红
1
黄 蓝
红
1
红
2
蓝
红
1
红
2
黄
A
O
B
D
22、(1)连结OC ····························································································1 分
OA OC
,
A
ACO
30
COD
60
····································· 3 分
又 AC CD
CD 是半⊙O 的切线··············································································· 5 分
OCD
180
30
60
30
90
D
A
,
(2)连结 BC
在 Rt ABC
中, cos
A
AB 是直径,
AC
AB
3
2
4
A
ACB
90
················································· 6 分
·····································································7 分
2 3
······························································ 8 分
AC AB
cos
注意:其他解法参照本解法给分。
23、
y
(2)
1
2
x
40
2
2
x
x
20
x
自变量 x 的取值范围是 0
x
1
(
2
1
2
20
y
x
x
x
2
····································································· 3 分
25
20)
······························································4 分
2
200
··················································6 分
,所以当
x 时, y 有最大值 200 .
20
20 25
即当 20
x 时,满足条件的绿化带的面积最大················································ 8 分
DBA
CAB
AD BC
, AC BD
, ② AC BD
, AB BA
.············································1 分
五、解答题:
24、第一种选择:① CAB
DBA
证明:
△ ACB ≌△ BDA
ABC
作 DE ∥ BC 交 AB 于 E (如下图 1), 则 DEA
又 AD BC
AB ∥CD
···················································································· 8 分
又 AD 不平行 BC , ABCD 是等腰梯形·························································9 分
·········································································· 3 分
····································································· 4 分
DEA
, DE 平行且等于 BC , EBCD 是平行四边形
·································································· 6 分
, AD ED
BAD
CBA
DAE
,
D
C
E
D
C
A
E
图 1
B
A
图 2
B
第二种选择:② AC BD
···························································· 1 分
证明:延长 AD 、 BC 相交于 E (如上图 2)······················································ 2 分
, ③ AD BC
EA EB
DAB
DAB
, AD BC , AB BA
AC BD
≌ CBA
CBA
又 AD BC , DE CE
EBA
而 E
·················································································7 分
DC ∥ AB ······················································································· 8 分
又 AD 不平行 BC , ABCD 是等腰梯形···················································9 分
·············································································· 4 分
············································································· 5 分
, EDC
EAB
EDC
ECD
EAB
E
ECD
EDC
注意:其他解法参照本解法给分。
说明:由①、③不能推出 ABCD 是等腰梯形,反例见下图:
D
A
C
E
B
Y
25.
(1)
C
E
O
N
F
H
P
D
B
M
A
X
········································· 2 分
x 上, 所以 P 的坐标为
P x y , 因为 P 在
y
, 所以点 P 在线段 AB 的中垂线上 , 点 P 的纵坐标是1 ,又点 P 在
, ··················· 4 分
1( 1)
2
3
2
, 又 F 是 圆 心 , 所 以 N 是 线 段 HB 的 中 点 ,
, 连结 PN 、 MN 、NF , 依
x
( ,
P x
3
2
)
x 上, 所以
, FN BC
3(
2
,
D
M
(4,1)
, 0)
(2) 因为 PA PB
y
3
2
(3) 设 ( ,
)
题 意 , PN MN
x
HN NB
4
2
2 (
x
PH
,
x
BM ······························································· 6 分
1
)
3
2
HNP
,
1
2
HNP
HPN
BNM
90
HPN
BNM
, 又
PHN
B
90
△
Rt PNH
∽ Rt NMB△
HN
PH
BM BN
解得
x
6
22
(
x ,舍去)
3
2
x
4
2
1
1
2
x
x
4
2
x
2 12
x
14 0
x
6
22
················································· 8 分
S
PMBH
(
BM HP BH
)
2
(1 6
22
)(4 6
22)
1
2
2
37 19
2
4
22
········· 9 分