2007 年广西玉林市中考数学真题及答案
(本卷共 8 页,满分 120 分;考试时间 120 分钟)
友情提示:答题前,请你先通览全卷;答题时,请你认真审题,做到先易后难;答题
后,要注意检查.祝你成功!
一、填空题:本大题共 10 小题,每小题 2 分,共 20 分.请将答案直接写在题中的横线上.
1.若向南走 2m 记作 2m ,则向北走3m ,记作
2.某部门要了解一批药品的质量情况,常用的调查方式是
“全面”或“抽样”).
3.如图 1 的圆柱体,它的左视图是
4.某学校积极响应上级的号召,举行了“决不让一个学生因贫困而失学”的
捐资助学活动,其中 6 个班同学的捐款平均数如下表:
(填图形的名称即可).
调查(填
m .
图 1
班别
捐款平均数(元)
一班
6
二班
4.6
三班
4.1
四班
3.8
五班
4.8
六班
5.2
则这组数据的中位数是
元.
5.不等式组
1
a
,
4 0
a
的解集是
.
.
6.已知O 是 ABCD 的对称中心, E 是 AB 的中心,请写出一个
与OE 有关的结论:
7.如图 2,要制作底边 BC 的长为 44cm ,顶点 A 到 BC 的距离
与 BC 长的比为1: 4 的等腰三角形木衣架,则腰 AB 的长至少需
要
8.要组织一次篮球联赛,赛制为单循环形式(每两队之间都赛一
场),计划安排 21 场比赛,应邀请
个球队参加比
赛.
cm (结果保留根号的形式).
9.瑞士的一位中学教师巴尔末从光谱数据
9
5
,
16
12
,
25
21
,
36
32
,
中,成功地发现了其规律,从而得到了巴尔末公式,继而打开了
光谱奥妙的大门.请你根据这个规律写出第 9 个数
.
A
图 2
B
A
E
D
P
C
C
B
10.如图 3,在等腰梯形 ABCD 中, AD BC∥ ,
AD AB .点 E F, 分别在 AD ,AB 上,AE BF ,DF 与CE
相交于 P ,则 DPE
二、选择题:本大题共 8 小题,每小题 3 分,共 24 分.在每小题给出的四个选项中,只有
一项是符合题意的,请将你认为正确答案的序号填在题后的括号内.
11.下列文字图案中,是轴对称图形的是(
图 3
.
)
60
,
F
B
高 山 流 水
A.
B.
12.某个多面体的平面展开图如图 4 所示,那么这个多面体是(
A.三棱柱
B.四棱柱
C.
D.
)
图 4
C.三棱锥
D.四棱锥
13.因式分解
a ab ,正确的结果是(
2
)
A.
a
(1
b
2
)
B. (1
a
b
)(1
b
)
C.
(
a b
2
)
D.
a
(1
b
)
2
5
)
x
x 的解是(
4
B. 2
D. 0
x
x
OA AB BC CD
x
x 或 0
,那么周长是接近100 的圆是(
B.以OB 为半径的圆
D.以OD 为半径的圆
14.方程 2
x
A. 4
C. 4
15.如图 5 是一个由四个同心圆构成的靶子示意图,点 O 为圆心,
且
)
A.以OA 为半径的圆
C.以OC 为半径的圆
16.如图 6 是张亮、李娜两位同学零花钱全学期各项支出
的统计图.根据统计图,下列对两位同学购买书籍支出占
全学期总支出的百分比作出的判断中,正确的是(
)
A.张亮的百分比比李娜的百分比大
B.张娜的百分比比张亮的百分比大
C.张亮的百分比与李娜的百分比一样大
D.无法确定
17.甲、乙两个清洁队共同参与了城中垃圾场的清运工
作.甲队单独工作 2 天完成总量的三分之一,这时增加了
乙队,两队又共同工作了 1 天,总量全部完成.那么乙队
单独完成总量需要(
A.6 天
200
150
100
50
0
C.3 天
D.2 天
y
y
,它们的共同点是:①在每一个象
18.已知函数
限内,都是函数 y 随 x 的增大而增大;②都有部分图象在第一象限;
③都经过点 (1 4), ,其中错误..的有(
)
C.2 个
B.1 个
A.0 个
友情提示:三 八题为解答题,满分共 76 分.解答应写出文字说明,
证明过程或演算步骤.
三、本大题共 2 个小题,满分共 16 分.
19.(本小题满分 8 分)
D.3 个
计算:
4sin 45
( 2007)
0
8
.
20.(本小题满分 8 分)
)
B.4 天
x ,
5
4
x
AO
B C D
图 5
全学期支出/元
零
食
书
籍
日
用
品
张亮
其
它
项
目
零食
25%
其它
23%
日用品
20% 书籍
32%
李娜
图 6
先化简,后求值: 3
x
2
x
(
x
2)
,其中
x .
3
2
四、本大题共 2 小题,满分共 16 分.
21.(本小题满分 8 分)
如图 7,A 是直角边长等于 a 的等腰直角三角形,B 是直径为 a
的 圆 . 圆 8 是 选 择 基 本 图 形 A B, 用 尺 规 画 出 的 图 案 :
S
阴影
2
a
2
a
4
π
.
A
B
图 7
. 证 明 如 下 : 四 边 形 ABCD 是 平 行 四 边 形 ,
.”
∽△
(1)请你以图 7 的图形为基本图形,按给定图形的大小设计画
.....,还要选择恰当的图形部分涂上阴影,并直接写出其
一个新图案
面积(尺规作图,不写作法,保留痕迹,作直角三角形时可使用
三角板).
(2)请你写出一句在解答本题的过程中体会最深且与数学有关
的话.
22.(本小题满分 8 分)
在数学课堂上,老师讲解“相似三角形”之后,接着出了一
道题目让同学练习,题目是:“如图 9,四边形 ABCD 是平
行四边形,E 是 BA 延长线上一点,CE 与 AD 相交于 F .请
写出与 EBC△
聪聪看后,迅速写出了下面解答:
“ 与 EBC△
相 似 的 只 有 EAF△
EAF
∥ . EBC
AD BC
△
你对聪聪的解答有何意见?为什么?
五、本大题共 1 小题,满分 10 分.
23.(本小题满分 10 分)
在物理试验中,当电流在一定时间段内正常通过电子元
件
通电或断开,并且这两种状态的可能性相等.
时,每个电子元件的状态有两种可能:
相似的三角形,并加以证明.”
E
(1)如图10 1 ,当只有一个电子元件时, P Q, 之间电
流通过的概率是
(2)如图10 2 ,当有两个电子元件 a b, 并联时,请你
用树状图(或列表法)表示图中 P Q, 之间电流能否通过
.
的所有可能情况,求出 P Q, 之间电流通过的概率;
P
(3)如图10 3 ,当有三个电子元件并联时,请你猜想
P Q, 之间电流通过的概率是
.
图 8
D
C
F
A
图 9
B
P
P
Q
Q
Q
图 10-1
a
b
图 10-2
图 10-3
六、本大题共 1 小题,满分 10 分.
24.(本小题满分 10 分)
某化妆公司每月付给销售人员的工资有两种方案.
方案一:没有底薪,只拿销售提成;
方案二:底薪加销售提成.
设 x(件)是销售商品的数量,y(元)是销售人员的月工资.如
图 11 所示, 1y 为方案一的函数图象, 2y 为方案二的函数图
象.已知每件商品的销售提成方案二比方案一少 7 元.从图中
信息解答如下问题(注:销售提成是指从销售每件商品得到的
销售费中提取一定数量的费用):
y(元)
2y
1y
560
420
O
30
图 11
x(件)
(1)求 1y 的函数解析式;
(2)请问方案二中每月付给销售人员的底薪是多少元?
(3)如果该公司销售人员小丽的月工资要超过 1000 元,那么小丽选用哪种方案最好,至少
要销售商品多少件?
七、本大题共 1 小题,满分 12 分
25.(本小题满分 12 分)
如图 12,在锐角 ABC△
于
D ,以 AD 为直径的 O 分别交 AB , AC 于 E F, ,
连结 DE DF, .
中, AB AC
, AD BC
A
E
O
B
D
图 12
F
C
P
, APB
,那
EDF
EAF
(1)求证:
(2)已知 P 是射线 DC 上一个动点,当点 P 运动到
PD BD
时,连结 AP ,交 O 于G ,连结 DG .设 EDG
180
;
么 与 有何数量关系?
试证明你的结论[在探究 与 的数量关系时,必要时可直接运
用(1)的结论进行推理与解答].
八、本大题共 1 小题,满分 12 分.
26.(本小题满分 12 分)
如图 13,在直角坐标系中,O 为原点,抛物线
y
2
x
bx
与 x 轴
3
的负半轴交于点 A ,与 y 轴的正半轴交于点 B ,
tan
ABO
顶点为 P .
(1)求抛物线的解析式;
,
1
3
y
B
A
P
O
x
图 13
(2)若抛物线向上或向下平移 k 个单位长度后经过点 ( 5 6)
C , ,试求 k 的值及平移后抛物
线的最小值;
(3)设平移后的抛物线与 y 轴相交于 D ,顶点为Q ,点 M 是平移的抛物线上的一个动点.请
探究:当点 M 在何位置时, MBD△
的面积是 MPQ△
面积的 2 倍?求出此时点 M 的坐
标 . [ 友 情 提 示 : 抛 物 线
y
2
ax
bx
(
c a
的 对 称 轴 是
0)
x
, 顶 点 坐 标 是
b
2
a
b
2
a
4
,
2
ac b
4
a
]
数学参考答案及评分标准
,OE
4. 4.7
BC∥ ,OE
10.120
5. 1a
AB
BC
2.抽样
一、填空题:(每小题 2 分,共 20 分)
1.3
3.矩形
2
OE
6.答案不唯一,参考举例:
121
7.11 5
117
二、选择题:(每小题 3 分,共 24 分)
11.B
17.D
12.A
18.B
8. 7
9.
13.B
14.C
15.C
16.A
三、解:19.原式 2 2 1 2 2
································································· 6 分
1 .········································································································· 8 分
20.先化简原式,原式 3(
······························································· 3 分
2)
x
x
x
3
2
6x
································································································ 5 分
6x
.··································································································6 分
3
2
································································· 7 分
x 时,原式
3
2
6
2
当
3 .·········································································································8 分
四、21.解:(1)正确画出图形······································································ 3 分
涂上阴影并写出阴影面积··············································································· 6 分
答案不唯一,参考举例:
S
2
a
2
a
16
π
S
2
a
2
a
8
π
S
2
a
2
a
8
π
S
2
a
2
a
4
π
S
2
a
2
a
16
π
S
2
a
4
π
S
2
a
(2)写出与要求相符的话··············································································8 分
答案不唯一,参考举例:①这两个图形的关系很密切,能组合设计出许多美丽的图案来装
点我们的生活;②运用圆的半径可作出等腰直角三角形三边的中点;③作数学图形需要一
丝不苟,否则会产生误差影响图案的美观,····························································
22.解:聪聪的解答不全面,还有 CDF△
与 EBC△
相似.································ 2 分
应补上如下证明:四边形 ABCD 是平行四边形,
∥ , CDF
.································································· 4 分
E
.························································································6 分
EBC
△
.·················································································· 8 分
∽△
五、23.解:(1)0.5 ··················································································· 2 分
(2)用树状图表示是:
ECD
CDF
AB DC
ABC
a
b
通电
断开
通电 断开
通电 断开
或用列表法表示为:
b
a
通电
断开
通电
断开
(通电,通电)
(通电,断开)
(断开,通电)
(断开,断开)
·················································································································6 分
从上可以看到 P Q, 之间电流通过的概率是
.·················································8 分
3
4
···································································································· 10 分
(3)
7
8
六、24.解(1)设 1y 的函数解析式为
y
(
kx x
≥ .·······································1 分
0)
1y 经过点 (30 420), , 30
k
k .··································································································2 分
1y 的函数解析式为 14 (
x x
≥ .···························································· 3 分
420
14
0)
.
y
≥ ,它经过点 (30 560), ,
0)
.······················································································· 4 分
y
(
ax b x
560 30a b
(2)设 2y 的函数解析式为
每件商品的销售提成方案二比方案一少 7 元,
a
b
(3)由(2),得 2y 的函数解析式为 7
14 7
350
560 30 7 b
7
y
x
350(
x
≥ .
0)
.························································································5 分
.
,即方案二中每月付给销售人员的底薪为350 元.································6 分
联合 14
y
x
与 7
x
y
350
组成方程组,解得 50
x , 700
y
.······················ 7 分
1000 700
,小丽选择方案一最好.························································· 8 分
371
7
,得
x
x
1000
由14
.·········································································9 分
x 为正整数, x 取最小整数 72 .故小丽至少要销售商品 72 件.····················· 10 分
七、25.(1)证明: AD
.·················· 2 分
是 C 的直径,
AED
AFD
90
AED
AFD
EAF
EDF
360
.················································· 3 分
EAF
EDF
180
.···········································································5 分
(2)答:
.················································································6 分
2
证法一: DP BD
APB
.··········································· 7 分
.················································································ 8 分
, AB AP
, AD BC
B
由结论(1)可知,
BAP
EDG
180
.····················································9 分
BAP
B
APB
180
,
BAP
180
2
.············································································ 10 分
180
2
180
.······································································· 11 分
.·························································································12 分
2
AGD
90
.·································7 分
.··················································································· 8 分
.················································································· 9 分
.··················································································10 分
是 O 的直径,
AED
, AB AP
.
BAD
DP BD
证法二: AD
△
, AD BC
PAD
AD AD
,
≌△
ADG
ADG
ADE
ADE
由
90
AGD
ADG
2
EDG
2
,得 DG AP .
APB
APB
.·················································································· 11 分
,
即
.·························································································12 分
八、26.解:(1)令 0
x ,则 3
y . B 点坐标为 (0 3), ,
OB .··················1 分
3
tan
AOB
OA OA
3
AB
AO
. A 点坐标为 ( 1 0)
1
,····································································· 2 分
1
3
, .·································································· 3 分
0 ( 1)
2
.求得 4
( 1) 3
b
b .····························································· 4 分