2008 年湖南省永州市中考数学真题及答案
考生注意:
1、本试卷共三道大题,25 个小题,满分 120 分,时量 120 分钟.
2、本试卷分Ⅰ卷和Ⅱ卷,Ⅰ卷为选择填空题 1-2 页;Ⅱ卷为解答题 3-8 页.
3、考生务必将Ⅰ卷的答案写在Ⅱ卷卷首的答案栏内,交卷时只交Ⅱ卷.
一、填空题(每小题 3 分,共 8 个小题,24 分.请将答案填在Ⅱ卷卷首的答案栏内.)
1. 若商品的价格上涨 5%,记为+5%,则价格下跌 3%,记作
2. 四川汶川地震发生以来,截至 6 月 4 日 12 时止,已接受国内外社会各界捐款 436.81 亿元,用科学记
.
数法(保留三个有效数字)记为
元.
3. 如图,直线 a、b被直线 c所截,若要 a∥b,需增加条件
(填一个即可).
4. 家家乐奥运福娃专卖店今年 3 月份售出福娃 3600 个,5 月份售
设 每 月 平 均 增 长 率 为 x, 根 据 题 意 , 列 出 关 于 x 的 方 程
.
5. 一棵树因雪灾于 A 处折断,如图所示,测得树梢触地点 B 到树
为 4 米,∠ABC 约 45°,树干 AC 垂直于地面,那么此树在未折
约为
米(答案可保留根号).
6. 一个角的补角是这个角的余角的 3 倍,则这个角为度
.
7. 右图是永州市几个主要景点示意图,根据图中信息可确定九疑
C 点的坐标为
.
8. 已知一组数据 1,2,0,-1,x,1 的平均数是 1,则这组数据
的极差为
.
二、选择题(每小题 3 分,共 8 个小题,24 分.每小题只有一个正确
选项的代号填入Ⅱ卷卷首的答案栏内.)
9. 如图,a、b、c分别表示苹果、梨、桃子的质量.同类水果质量
系正确的是(
)
出 4900 个 ,
为
根 C 处的距离
断之前的高度
山的中心位置
选项,请将正确
相等,则下列关
A.a>c>b
B.b>a>c
C.a>b>c
D.c>a>b
10.为悼念四川汶川地震中遇难同胞,在全国哀悼日第一天,某校升旗仪式中,先把国旗匀速升至旗杆顶
部,停顿 3 秒钟后再把国旗匀速下落至旗杆中部.能正确反映这一过程中,国旗高度 h(米)与升旗时
间 t(秒)的函数关系的大致图象是
11.下列判断正确的是(
)
A.
3 < 3 <2
2
C. 1< 5 - 3 <2
B. 2< 2 + 3 <3
D. 4< 3 · 5 <5
12.下图※是一种瑶族长鼓的轮廓图,其主视图正确的是(
)
13.6 张大小、厚度、颜色相同的卡片上分别画上线段、等边三角形、直角梯形、正方形、正五边形、圆. 在
看不见图形的条件下任意摸出 1 张,这张卡片上的图形是中心对称图形的概率是(
)
A.
1
6
B.
1
3
C.
1
2
D.
2
3
)
14.下列命题是假命题...的是(
A.两点之间,线段最短.
B.过不在同一直线上的三点有且只有一个圆.
C.一组对应边相等的两个等边三角形全等.
D.对角线相等的四边形是矩形.
15.一个圆锥的侧面展开图形是半径为 8cm,圆心角为 120°的扇形,则此圆锥的底面半径为 (
)
B.
16 cm
3
C.3cm
D.
4 cm
3
的式子叫做二阶行列式,它的运算法则用公式表示为
c
d
a
b
c
d
=ad-bc,依此法则计算
a
b
A.
8 cm
3
16.形如
2
3
1
4
的结果为(
)
A.11
B.-11
C.5
D.-2
永州市 2008 年初中毕业学业考试试卷
学
数
第 II 卷
题号 一 二
17
18
19
20
三
21
22
23
24
25
总分
合分
人
核分
人
得分
请将Ⅰ卷的答案填入下面答案栏内.
一、填空题
1.
5.
2.
6.
3.
7.
4.
8.
二、选择题
9
10
11
12
13
14
15
16
三、解答题:(本题 9 个小题,共 72 分,解答题要求写出证明步骤或解答过程)
17.(6 分)计算:cos45°·(-
1 )-2 -(2 2 - 3 )0+|- 32 |+
2
1
12
得 分 评卷人 复评人
18.(6 分)解方程:
x
2
x
x
+2=
2
x
x
1
19.(6 分)如图所示,左边方格纸中每个正方形的边长均为 a,右边方格纸中每个正方形的边长均为 b,将
左边方格纸中的图形顺时针旋转 90°,并按 b:a的比例画在右边方格纸中.
20.(8 分)如图,一次函数的图象经过 M 点,与 x
y轴交于 B 点,根据图中信息求:(1)这个函数
tan∠BAO.
轴交于 A 点,与
的解析式;(2)
21.(8 分)某物流公司,要将 300 吨物资运往某地,现有 A、B 两种型号的车可供调用,已知 A 型车每辆可
装 20 吨,B 型车每辆可装 15 吨,在每辆车不超载的条件下,把 300 吨物资装运完,问:在已确定调用
5 辆 A 型车的前提下至少还需调用 B 型车多少辆?
22.(8 分)如图△ABC 与△CDE 都是等边三角形,点 E、F
BC 上,且 EF∥AB
(1)求证:四边形 EFCD 是菱形;
(2)设 CD=4,求 D、F 两点间的距离.
分 别 在 AC 、
23.(10 分)为保护环境,节约资源,从今年 6 月 1 日起国家禁止超市、商场、药店为顾客提供免费塑料袋,
为解决顾客购物包装问题,心连心超市提供了 A.自带购物袋;B.租借购物篮;C.购买环保袋;D.徒
手携带,四种方式供顾客选择.该超市把 6 月 1 日、2 日两天的统计结果绘成如下的条形统计图和 6 月
1 日的扇形统计图,请你根据图形解答下列问题:
(1)请将 6 月 1 日的扇形统计图补充完整.
(2)根据统计图求 6 月 1 日在该超市购物总人次和 6 月 1 日自带购物袋的人次.
(3)比较两日的条形图,你有什么发现?请用一句话表述你的发现.
24.(10 分)如图,已知⊙O 的直径 AB=2,直线 m 与⊙O 相切
上一动点(与点 A、点 B 不重合),PO 的延长线与⊙O 相交
的切线与直线 m相交于点 D.
(1)求证:△APC∽△COD.
(2)设 AP=x,OD=y,试用含 x的代数式表示 y.
(3)试探索 x为何值时,△ACD 是一个等边三角形.
于点 A,P 为⊙O
于点 C,过点 C
25.(10 分)如图,二次函数 y=ax2+bx+c(a>0)与坐
标轴交于
点 A、B、C 且 OA=1,OB=OC=3 .
(1)求此二次函数的解析式.
(2)写出顶点坐标和对称轴方程.
(3)点 M、N 在 y=ax2+bx+c的图像上(点 N 在点 M 的
MN∥x轴,求以 MN 为直径且与 x轴相切的圆的半径.
右边),且
永州市 2008 年初中毕业学业考试试卷
数学参考答案及评分标准
一、填空题(每小题 3 分,共 24 分)
1. 3%
2.
10
4.37 10
3. 1
或 1
或 1
4
3
2 180
4.
3600(1
)
x
2
4900
5. 4 4 2
6.45°
7.(3,1)
8.4
二、选择题(每小题 3 分,共 24 分)
题号
答案
9
C
三、解答题
10
B
11
A
12
D
13
C
14
D
15
A
16
A
17.(6 分)解:原式
2
2
4 1
32
1
2 1
················································· 2 分
2 2 1 4 2
2 1
····················································· 4 分
18.(6 分)解:
7 2
1
1
x
2
···············································································6 分
2
x
1
x
········································································1 分
方程两边同乘以 (
x
1)(
x
1)
,得
x
1 2(
x
1)(
x
1)
2 (
x x
1)
········································································ 3 分
x ······························································································· 4 分
x 代入 (
x
1)(
x
1)
得
1
0
·························································································· 5 分
x 是原方程的根.··················································································· 6 分
1
3
1
3
解之,得
检验:把
1
3
1
1
3
1
3
19.(6 分)
20.(8 分)
(1)设一次函数的解析式为 y
kx b
( 0
k )
将点 (0 6)
,,
B
M
( 1 4)
, 代入,得
k
k
0
6
b
,
( 1)
4
b
解之,得 2
k
b
,
解析式为 2
y
x
6
6
······························································································ 2 分
····················································································· 4 分
(2)令 0
y ,代入 2
x
y
,得
6
x
3
可知点 A 的坐标 ( 3 0)
, ·····················································································6 分
···························································································· 8 分
2
tan
BAO
21.(8 分)
解:设还需要 B 型车 x 辆,根据题意,得:
20 5 15
300
x
113
3
≥ ··························································································3 分
解得:
x ≥ ······························································································ 5 分
由于 x 是车的数量,应为整数,所以 x 的最小值为 14.·········································· 7 分
答:至少需要 14 台 B 型车.·············································································· 8 分
22.(8 分)
(1)证明:
ED CD
△
ABC
与 CDE△
都是等边三角形
A
DCE
BCA
AB CD DE CF
DCE
60
·····························································1 分
AB∥
∥ , ∥ ··················································································2 分
又 EF
EF CD∥ ·································································································· 3 分
四边形 EFCD 是菱形···················································································· 4 分
(2)解:连结 DF ,与CE 相交于点G ·······························································5 分
CG ················································································· 6 分
由
CD ,可知
4
2
DG
2
4
2
2
2 3
················································································· 7 分
DF
4 3
··································································································8 分
22%”·········································3 分
23.(10 分)(1)在扇形统计图的空白处填上“D
(2)6 月 1 日在该超市购物的总人次为 1250(人次)············································ 6 分
6 月 1 日自带购物袋的有 225 人次······································································· 8 分
(3)答案不唯一,如“自带购物袋的人增多”
“租借购物篮的人减少”等·············································································· 10 分
24.(10 分)(1)∵ PC 是⊙O的直径,CD是⊙O的切线
∠PAC=∠OCD=90°,显然△DOA≌△DOC····························································· 1 分
∴∠DOA=∠DOC·······························································································2 分
∴∠APC=∠COD·······························································································3 分
△
························································································ 4 分
(2)由 APC
COD
··························································6 分
COD
APC
∽△
∽△
,得
△
AP OC
PC OD
,
························································································· 7 分
y
x
2
1
y
30
60
,
OD
ADC
ODC
······················· 8 分
(3)若 ACD△
2
OC
于是
故,当 1x 时, ACD△
25.(1)依题意 ( 1 0)
解方程组得所求解析式为
3 (
x
y ,
是一个等边三角形························································10 分
(3 0)
(0
C
B
····················· 1 分
2 2
3
y
x
x
····························································· 4 分
2
2
2
4
1)
····································································· 5 分
(2)
4), ,对称轴 1x ·······································································7 分
顶点坐标 (1
(3)设圆半径为 r ,当 MN 在 x 轴下方时, N 点坐标为 (1
, ·························· 8 分
,, ,, , 分别代入
ax
bx
3)
A
r
)
y
y
x
2
x
c
r
把 N 点代入
y
x
2 2
x
得
3
r
17
1
2
·························································9 分
同理可得另一种情形
圆的半径为
1
2
r
17
1
2
1
或
17
17
2
2
x
是一个等边三角形,则
1x
,可得 2