宁夏回族自治区2009年初中毕业暨高中阶段招生
数 学 试 题
注意事项:
1.考试时间120分钟,全卷总分120分. 2.答题前将密封线内的项目填写清楚.
3.答卷一律使用黑、蓝钢笔或圆珠笔.
4.凡使用答题卡的考生,答卷前务必将答题卡上的有关项目填写清楚.选择题的每小题选出答案后,用铅笔把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.不使用答题卡的考生,将选择题的答案答在试卷上.
一、选择题(下列每小题所给的四个答案中只有一个是正确的,每小题3分,共24分) 1.下列运算正确的是( )
A .3412
a a a =· B .623(6)(2)3a a a -÷-=
C .22(2)4a a -=-
D .23a a a -=-
2.某旅游景点三月份共接待游客25万人次,五月份共接待游客64万人次,设每月的平均增长率为x ,则可列方程为( )
A .225(1)64x +=
B .225(1)64x -=
C .264(1)25x +=
D .264(1)25x -= 3.把不等式组21123
x x +>-??+?≤的解集表示在数轴上,下列选项正确的是( )
A . B
.
C .
D .
4.某班抽取6名同学参加体能测试,成绩如下:85,95,85,80,
80,85.下列表述错误..
的是(
)
A .众数是85
B .平均数是85
C .中位数是80
D .极差是15 5.一次函数23y x =-的图象不经过( )
A .第一象限
B .第二象限
C .第三象限
D .第四象限
6.如图,是一个几何体的三视图,根据图中标注的数据可求得这个几何体的体积为( ) A .24π B .32π C .36π D .48π
1 0
1-
1 0 1- 1 0 1- 1 0 1- 主视图
左视图 俯视图
(6题图)
(7题图)
7.在44?的正方形网格中,已将图中的四个小正方形涂上阴影(如图),若再从其余小正方形中任选一个也涂上阴影,使得整个阴影部分组成的图形成轴对称图形.那么符合条件的小正方形共有( )
A .1个
B .2个
C .3个
D .4个
8.二次函数2
(0)y ax bx c a =++≠的图象如图所示,对称轴是直线1x =,则下列四个结论错误..的是( ) A .0c > B .20a b += C .240b ac -> D .0a b c -+> 二、填空题(每小题3分,共24分) 9.分解因式:32m mn -= .
10.在R t ABC △中,9032C A B B C ∠===°,,,则cos A 的值是 . 11.已知:32
a b +=
,1ab =,化简(2)(2)a b --的结果是 .
12.某商品的价格标签已丢失,售货员只知道“它的进价为80元,打七折售出后,仍可获利5%”.你认为售货员应标在标签上的价格为 元.
13.用一个半径为6,圆心角为120°的扇形围成一个圆锥的侧面,则圆锥的高为 . 14.如图,梯形A B C D 的两条对角线交于点E ,图中面积相等的三角形共有 对.
15.如图,A B C △的周长为32,且A
B A
C A
D B C =⊥,于D ,AC D △的周长为24,那么A D
的长为 .
16.如图,O ⊙是边长为2的等边三角形ABC 的内切圆,则图中阴影部分的面积为 .
三、解答题(共24分) 17.(6分)
1
01(2009)12-??
-++ ?
??
.
(8题图)
A D
B E (14题图) (15题图) A B C
D B (16题图)
18.(6分) 解分式方程:123
3x x x
+
=--.
19.(6分)
已知正比例函数1y k x =1(0)k ≠与反比例函数22(0)k y k x
=≠的图象交于A B 、两点,点A 的
坐标为(21),.
(1)求正比例函数、反比例函数的表达式; (2)求点B 的坐标.
20.(6分)
桌子上放有质地均匀,反面相同的4张卡片.正面分别标有数字1、2、3、4,将这些卡片反面朝上洗匀后放在桌面上,先从中任意抽出1张卡片,用卡片上所标的数字作为十位上的数字,将取出的卡片反面朝上放回洗匀;再从中任意抽取1张卡片,用卡片上所标的数字作为个位数字.试用列表或画树状图的方法分析,组成的两位数恰好能被3整除的概率是多少?
四、解答题(48分) 21.(6分)
在“首届中国西部(银川)房·车生活文化节”期间,某汽车经销商推出A B C D 、、、四种型号的小轿车共1000辆进行展销.C 型号轿车销售的成交率为50%,其它型号轿车的销售情况绘制在图1和图2两幅尚不完整的统计图中. (1)参加展销的D 型号轿车有多少辆?
(2)请你将图2的统计图补充完整;
(3)通过计算说明,哪一种型号的轿车销售情况最好?
(4)若对已售出轿车进行抽奖,现将已售出A B C D 、、、四种型号轿车的发票(一车一票)放到一起,从中随机抽取一张,求抽到A 型号轿车发票的概率.
22.(6分)
如图:在R t ABC △中,90A C B ∠=°,C D 是A B 边上的中线,将A D C △沿A C 边所在的直线折叠,使点D 落在点E 处,得四边形A B C E . 求证:EC AB ∥.
型号
D
C
20% B 20% A 35% 各型号参展轿车数的百分比
A B C D (图2)
(图1)
E
C
B
A
D
23.(8分)
已知:如图,A B 为O ⊙的直径,A B A C
B C =,交O ⊙于点D ,A C 交O ⊙于点
45E BAC ∠=,°. (1)求E B C ∠的度数; (2)求证:B D C D =.
24.(8分)
如图,抛物线2
1222
y x x =-
+
+与x 轴交于A B 、两点,与y 轴交于C 点.
(1)求A B C 、、三点的坐标;
(2)证明A B C △为直角三角形;
(3)在抛物线上除C 点外,是否还存在另外一个点P ,使A B P △是直角三角形,若存在,请求出点P 的坐标,若不存在,请说明理由.
C
25.(10分)
如图1、图2,是一款家用的垃圾桶,踏板A B (与地面平行)或绕定点P (固定在垃圾桶底部的某一位置)上下转动(转动过程中始终保持AP A P BP B P ''==,).通过向下踩踏点A 到A '(与地面接触点)使点B 上升到点B ',与此同时传动杆B H 运动到B H ''的位置,点H 绕固定点D 旋转(D H 为旋转半径)至点H ',从而使桶盖打开一个张角H D H '∠. 如图3,桶盖打开后,传动杆H B ''所在的直线分别与水平直线A B D H 、垂直,垂足为点M C 、,设H C '=B M '.测得6cm 12cm 8cm A P P B D H '===,,.要使桶盖张开的角度H D H '∠不小于60°,那么踏板A B 离地面的高度至少等于多少cm ?(结果保留两位有效数字)
(参考数据: 1.41
(图1)
D
(图2)
D
A ′
(图3)
26.(10分)
已知:等边三角形ABC的边长为4厘米,长为1厘米的线段M N在A B C
△的边A B上沿A B 方向以1厘米/秒的速度向B点运动(运动开始时,点M与点A重合,点N到达点B时运动终止),过点M N
△的其它边交于P Q
、分别作A B边的垂线,与A B C
、两点,线段M N运动的时间为t秒.
(1)线段M N在运动的过程中,t为何值时,四边形M NQP恰为矩形?并求出该矩形的面积;(2)线段M N在运动的过程中,四边形M NQP的面积为S,运动的时间为t.求四边形M NQP 的面积S随运动时间t变化的函数关系式,并写出自变量t的取值范围.
C
Q
P
B
A M N
宁夏回族自治区2009年初中毕业暨高中阶段招生
数学试卷参考答案
三、解答题(共24分) 17.(6分)计算: 解:原式=121++
····················································································· 4分
=··························································································································· 6分 18.(6分)解分式方程:
解:去分母得:12(3)x x -=- ···················································································· 3分 整理方程得:37x -=-
73
x =
··························································································································· 5分
经检验73
x =是原方程的解.
∴原方程的解为73
x =. ······························································································ 6分
19.(6分)
解:(1)把点(21)A ,分别代入1y k x =与2k y x
=
得
112
k =
,22k =. ········································································································ 2分
∴正比例函数、反比例函数的表达式为:122y x y x
==
,. ········································· 3分
(2)由方程组12
2
y x y x ?
=????=??
得1121x y =-??=-?,2221x y =??=?.
B ∴点坐标是(2,1)--. ······
························································································· 6分 20.(6分)
解:列表:
树状图:
···················································· 3分
∴能被3整除的两位数的概率是
516
.··········································································· 6分
四、解答题(共48分)
21(6分)
解:(1)100025%250?=(辆) ············································································· 1分 (2)如图,(100020%50%100??=) ············ 2分 (3)四种型号轿车的成交率:
16898A 100%48%B 100%49%350
200
?=?=:: C 50%: 130
D 100%52%250
?=:
∴D 种型号的轿车销售情况最好. ·
··························· 4分 (4)16816821
1689810013049662
==+++.
∴抽到A 型号轿车发票的概率为21
62. ········································································· 6分 22.(6分)
证明:C D 是A B 边上的中线,且90A C B ∠=°, C D A D ∴=.
C A
D A C D ∴∠=∠. ·································································································· 2分
又A C E △是由A D C △沿A C 边所在的直线折叠而成的,
E C A A C D ∴∠=∠. ··································································································· 4分 E C A C A D ∴∠=∠. ··································································································· 5分 E C A B ∴∥. ············································································································· 6分
1
1 2 3 4 14 13 12 11 1 2 3 4 24 23 22 21 2
1 2 3 4 34 33 32 31 3
1 2 3 4
44 43 42 41 4
开始
型号
A B C D
(1)解:A B 是O ⊙的直径, 90A E B ∴∠=°. 又45B A C ∠= °, 45ABE ∴∠=°. 又A B A C = ,
67.5ABC C ∴∠=∠=°.
22.5EBC ∴∠=°. ····································································································· 4分
(2)证明:连结A D . A B 是O ⊙的直径,
90AD B ∴∠=°. A D B C ∴⊥.
又A B A C = ,
B D
C
D ∴=. ············································································································· 8分
24.(8分)
解:(1) 抛物线2
1222
y x x =-
+
+与x 轴交于A B 、两点,
2
1202
2
x x ∴-
+
+=.
即2
40x --=.
解之得:12x x ==
∴点A B 、的坐标为(A B 、().······························································· 2分
将0x =代入2
122
2
y x x =-+
+,得C 点的坐标为(0,2)·
··································· 3分
(2)AC BC AB =
==
2
2
2
AB AC BC ∴=+,
则90A C B ∠=°,
A B C ∴△是直角三角形. ····························································································· 6分
(3)将2y =代入2
122
2
y x x =-
+
+
得2
1222
2
x x -
+
+=,
120x x ∴==
,
P
∴点坐标为. ································································································ 8分25.(10分)
过点A'作A N A B
'⊥垂足为N点,
在R t H C D
'
△中,
若H D H'
∠不小于60°,
则sin60
2
H C
H D
'
?=
'
≥
即
2
H C H D
''=
≥·······································5分
B M H C
''
=
≥··············································6分
R t R t
A N P
B M P
''
△∽△
A N A P
B M B P
''
∴=
''
6
3.5cm
12
A P
B M
A N
B P
''?
'
∴==
'
·
≥··························································· 9分
∴踏板A B离地面的高度至少等于3.5cm. ···········································10分26.(10分)(1)过点C作C D AB
⊥,垂足为D.
则2
AD=,
当M N运动到被C D垂直平分时,四边形M NQP是矩形,
即
3
2
A M=时,四边形M NQP是矩形,
3
2
t
∴=秒时,四边形M NQP是矩形.
tan60
PM AM
=
°=
M N Q P
S
∴=
四边形
··································································································· 4分(2)1°当01
t
<<时,
1
()
2
M N Q P
S PM Q N M N
=+
四边形
·
1
1)
2
t?
=++
?
2
=+·····························································6分
D
C
P Q
B
A M D N
C
P
Q
B
A M N
1()2
M N Q P S PM Q N M N =+四边形·
1
)12t ?=+-?·
=··································································· 8分 3°当23t <<时, 1()2
M N Q P S PM Q N M N =+四边形·
1
))2t t ?=-+-?
=+······················································ 10分
C
P
Q
B
A
M
N