泰安市2007年中等学校招生考试
数学试题
注意事项:
1.本试题分第I 卷和第II 卷两部分.第I 卷3页为选择题,36分;第II 卷8页为非选择题,84分;共120分.考试时间120分钟.
2.答第I 卷前务必将自己的姓名、考号、考试科目涂写在答题卡上.考试结束后,试题和答题卡一并收回.
3.第I 卷每题选出答案后,都必须用2B铅笔把答题卡上对应题目的答案标号(ABCD)涂黑,如需改动,必须先用橡皮擦干净,再改涂其他答案,不能答在试卷上. 第I 卷(选择题 共36分)
一、选择题(本大题共12小题,在每小题给出的四个选项中,只有一个是正确的,请把正确的选项选出来,每小题选对得3分,选错、不选或选出的答案超过一个,均记零分) 1.下列运算正确的是( ) A.42=±
B.2
142-??
=- ???
C.3
82-=-
D.|2|2--=
2.下列运算正确的是( ) A.3
3
6
2a a a +=
B.358()()a a a --=-
C.2
3
63
(2)424a b a a b -=-
D.2
21114416339
a b a b b a ????---=-
???????
3.若一个圆锥的母线长是它底面半径的3倍,则它的侧面展开图的圆心角等于( ) A.120
B.135
C.150
D.180
4.将(21)(2)1y x x =-++化成()y a x m n 2
=++的形式为( )
A.2
3252416y x ?
?=+- ???
B.2
317248y x ?
?=-- ???
C.2
317248y x ?
?=+- ??
?
D.2
317248y x ?
?=++ ??
?
5.计算211111a a ?
???
-- ???-????
的结果为( ) A.1a a +-
B.
1
a a
-
C.
1a a
- D.
1
1a a
+- 6.如图,在ABC △中,90ACB ∠=
,CD AB ⊥于D ,若23AC =,
32AB =,则tan BCD ∠的值为( )
A
C
B D
(第6题)
A.2
B.
22
C.
63
D.
33
7.如图,在正方形ABCD 中,E 是BC 的中点,F 是CD 上一点,
且14
CF CD =,下列结论:①30BAE ∠=
,②A B E A E F △∽△,
③AE EF ⊥,④ADF ECF △∽△.其中正确的个数为( )
A.1
B.2
C.3
D.4
8.如图,ABC △是等腰直角三角形,且90ACB ∠=
,曲线
CDEF 叫做“等腰直角三角形的渐开线”,其中 CD , DE , EF , 的圆心依次按A B C ,,循环.如果1AC =,那么由曲线
CDEF 和线段CF 围成图形的面积为( )
A.
(1272)π
4
+
B.
(952)π+2
4
+
C.
(1272)π2
4
++
D.
(952)π
4
+
9.已知三点11
1()P x y ,,222()P x y ,,3(12)P -,都在反比例函数k
y x
=的图象上,若10x <,20x >,则下列式子正确的是( )
A.120y y <<
B.120y y <<
C.120y y >> D.120y y >>
10.半径分别为13和15的两圆相交,且公共弦长为24,则两圆的圆心距为( ) A.
5
4
6或14 B.
65
4
或4
C.14
D.4或14
11.若1x ,2x 是方程2
240x x --=的两个不相等的实数根,则代数
式22
112223x x x -++的值是( )
A.19 B.15 C.11 D.3 12.如图,四边形ABCD 是边长为2cm 的正方形,动点P 在ABCD 的边上沿A B C D →→→的路径以1cm/s 的速度运动(点P 不与
A D ,重合).在这个运动过程中,APD △的面积2
(cm )S 随时间()
t s 的变化关系用图象表示,正确的为( )
A D C
B
E F
(第8题)
A B
C
F
D
E
(第7题)
A D
C
B
P
(第12题)
A .
O 1 2 3 4 5 6
t
s 1 2
B
O 1 2 3 4 5 6
t
s 1 2 C
O 1 2 3 4 5 6
t
s 1
2 D
O 1 2 3 4 5 6
t
s 1
2
第II 卷(非选择题 共84分)
注意事项:
1.答卷前将密封线内的项目填写清楚.
2.第II 卷共8页,用蓝黑钢笔或圆珠笔直接答在试卷上.
二、填空题(本大题共7小题,满分21分.只要求填写最后结果,每小题填对得3分)
13.方程(2)(3)20x x ++=的解是 .
14.如图,ABE △和ACD △是ABC △分别沿着AB AC ,边翻折180
形成的,若150BAC ∠=
,则θ∠的度数是 .
15.若关于x 的不等式组3(2)224
x x a x x --?
?+>??,有解,则实数a 的取
值范围是 .
16.如图,M 与x 轴相交于点(20)A ,
,(80)B ,,与y 轴相切于点C ,则圆心M 的坐标是 .
17.如图,图①,图②,图③,……是用围棋棋子摆成的一列具有一定规律的“山”字.则第n 个“山”字中的棋子个数是 .
18.如图,一游人由山脚A 沿坡角为30
的山坡AB 行走600m ,到达一个景点B ,再由B 沿山坡BC 行走200m 到达山顶C ,若在山顶C 处观测到景点B 的俯角为45
,则山高CD 等于 (结果用根号表
示)
19.为确保信息安全,信息需加密传输,发送方由明文→密文(加密),接收方由密文→明
文(解密).已知加密规则为:明文x y z ,,对应密文23343x y x y z ++,
,.例如:明文1,2,3对应密文8,11,9.当接收方收到密文12,17,27时,则解密得到的明文为 .
……
图①
图②
图③
图④
(第17题)
C
D
A E
B
θ
(第14题)
y
x
M B
A
O C
(第16题)
A
B
C D
(第18题)
三、解答题(本大题共7小题,满分63分.解答应写出必要的文字说明、证明过程或推演步骤) 20.(本小题满分6分)
某中学为了解毕业年级800名学生每学期参加社会实践活动的时间,随机对该年级60名学生每学期参加社会实践活动的时间进行了统计,结果如下表: 时间/天 4 5 6 7 8 9 10 11 12 13 人数
3
3
5
7
8
13
8
7
4
2
(1)补全右面的频率分布表;
(2)请你估算这所学校该年级的学生中,每学期
参加社会实践活动的时间大于7天的约有多少
人?
21.(本小题满分8分)
如图,在梯形ABCD 中,AD BC ∥,对角线BD 平分ABC ∠,BAD ∠的平分线AE 交BC 于E F G ,,分别
是AB AD ,的中点. (1)求证:EF EG =;
(2)当AB 与EC 满足怎样的数量关系时,EG CD ∥?
并说明理由.
22.(本小题满分9分)
某书店老板去图书批发市场购买某种图书.第一次用1200元购书若干本,并按该书定价7元出售,很快售完.由于该书畅销,第二次购书时,每本书的批发价比第一次提高了1元,他用1500元所购该书数量比第一次多10本.当按定价售出200本时,出现滞销,便以定价的4折售完剩余的书.试问该老板这两次售书总体上是赔钱了,还是赚钱了(不考虑其它因素)?若赔钱,赔多少?若赚钱,赚多少?
23.(本小题满分9分)
如图,在ABC △中,AB AC =,以AB 为直径的圆O 交
BC 于点D ,交AC 于点E ,过点D 作DF AC ⊥,垂
足为F .
(1)求证:DF 为O 的切线;
(2)若过A 点且与BC 平行的直线交BE 的延长线于G 点,连结CG .当ABC △是等边三角形时,求AGC ∠的
度数.
分组 频数 频率
3.5 5.5 6
0.1 5.57.5 12
0.2 7.59.5
9.511.5
11.513.5 6
0.1 合计 60 1 B E C D G A
F
(第21题) A G
F
E C B O
(第23题)
24.(本小题满分9分)
市园林处为了对一段公路进行绿化,计划购买A B ,两种风景树共900棵.A B ,两种树的相关信息如下表:
项目
品种
单价(元/棵)
成活率 A 80 92% B
100
98%
若购买A 种树x 棵,购树所需的总费用为y 元. (1)求y 与x 之间的函数关系式;
(2)若购树的总费用82000元,则购A 种树不少于多少棵?
(3)若希望这批树的成活率不低于94%,且使购树的总费用最低,应选购A B ,两种树各多少棵?此时最低费用为多少?
25.(本小题满分10分)
如图,在OAB △中,90B ∠= ,30BOA ∠=
,4OA =,将OAB △绕点O 按逆时针方向旋转至OA B ''△,C 点的坐标为
(0,4).
(1)求A '点的坐标;
(2)求过C ,A ',A 三点的抛物线2y ax bx c =++的解析式; (3)在(2)中的抛物线上是否存在点P ,使以O A P ,,为顶点的三角形是等腰直角三角形?若存在,求出所有点P 的坐标;若不存在,请说明理由. 26.(本小题满分12分)
如图,在ABC △中,90BAC ∠=
,AD 是BC 边上的高,E 是BC 边上的一个动点(不与B C ,重合),EF AB ⊥,EG AC ⊥,垂足分别为F G ,.
(1)求证:EG CG
AD CD
=; (2)FD 与DG 是否垂直?若垂直,请给出证明;若不垂直,请说明理由; (3)当AB AC =时,FDG △为等腰直角三角形吗?并说明理由.
A '
B ' O B
A
C
x
y (第25题)
F
A
G
C
E
D B
(第26题)
泰安市二〇〇七年中等学校招生考试
数学试题(A)(非课改区用)参考答案及评分标准
一、选择题(每小题3分,共36分) 题号 1 2 3 4 5 6 7 8 9 10 11 12 答案
C
D
A
C
A
B
B
C
D
D
A
B
二、填空题(本大题共7小题,每小题3分,共21分) 13.12x =,27x =- 14.60
15.4a >
16.(54),
17.52n +
18.(3001002)m +
19.329,
, 三、解答题(本大题共7小题,满分63分) 20.(本小题满分6分) 解:(1)21,0,35;15,0,25 ·········································································· 4分 (2)42
80056060
?
= ················································ 6分 21.(本小题满分8分) (1)证明:AD BC ∥ DBC ADB ∴∠=∠ 又ABD DBC ∠=∠
ABD ADB ∴∠=∠
AB AD ∴= ······························································· 2分
又12AF AB = ,1
2
AG AD =
AF AG ∴= ··················································································································· 3分 又BAE DAE ∠=∠ AE AE =
AFE AGE ∴△≌△ EF EG ∴= ···················································································································· 5分 (2)当2AB EC =时,EG CD ∥ ·············································································· 6分 2AB EC = 2AD EC ∴=
1
2
GD AD EC ∴== ····································································································· 7分
又GD EC ∥
∴四边形GECD 是平行四边形 EG CD ∴∥ ··················································································································· 8分
22.(本小题满分9分)
解:设第一次购书的进价为x 元,则第二次购书的进价为(1)x +元.根据题意得:
12001500
101
x x +=+ ·········································································································· 4分 B E C D
G
A F
(第21题)
去分母,整理得2
291200x x -+= 解之得:15x =,224x =
经检验15x =,224x =都是原方程的解
每本书的定价为7元
∴只取5x =
···················································································································· 6分 所以第一次购书为1200
2405
=(本). 第二次购书为24010250+=(本) 第一次赚钱为240(75)480?-=(元)
第二次赚钱为200(75 1.2)50(70.45 1.2)40?-?+??-?=(元) ························ 8分 所以两次共赚钱48040520+=(元)
答:该老板两次售书总体上是赚钱了,共赚了520元. ·············································· 9分 23.(本小题满分9分) (1)证明:连结AD OD ,
AB 是O 的直径
AD BC ∴⊥ ······························································ 2分
ABC △是等腰三角形 BD DC ∴=
又AO BO = OD AC ∴∥
DF AC ⊥ ······························································ 4分
OF OD ∴⊥
DF OD ∴⊥
DF ∴是O 的切线 ······································································································· 5分 (2)AB 是O 的直径 BG AC ∴⊥
ABC △是等边三角形 BG ∴是AC 的垂直平分线 GA GC ∴= ···················································································································· 7分
又AG BC ∥,60ACB ∠=
60CAG ACB ∴∠=∠= ACG ∴△是等边三角形
60AGC ∴∠= ·············································································································· 9分
24.(本小题满分9分)
解:(1)80100(900)y x x =+-
A G
F
E
C B
O
(第23题) D
2090000x =-+ ·························································································· 3分 (2)由题意得:
209000082000x -+≤ 45004100x -+≤ 400x ≥
即购A 种树不少于400棵 ······························································································· 5分 (3)92%98%(900)94%900x x +-?≥
92989009894900x x +?-?≥ 64900x --?≥ 600x ≤ ························································································································· 7分
2090000y x =-+ 随x 的增大而减小
∴当600x =时,购树费用最低为206009000078000y =-?+=(元)
当600x =时,900300x -=
∴此时应购A 种树600棵,B 种树300棵 ·
·································································· 9分 25.(本小题满分10分) 解:(1)过点A '作A D '垂直于x 轴,垂足为D 则四边形OB A D ''为矩形 在A DO '△中,
A D OA ''=sin 4sin 6023A OD '∠=?=
2OD A B AB ''===
∴点A '的坐标为(223), ·
··············································· 3分 (2)(04)C ,
在抛物线上,4c ∴= 24y ax bx ∴=++
(40)A ,,(223)A ',,
在抛物线2
4y ax bx =++上
1644042423
a b a b ++=??∴?++=??,
··································································································· 5分 解之得13
2
233a b ?-=???=-?
∴所求解析式为2
3(233)42
y x x 1-=
+-+. ······················································ 7分 (3)①若以点O 为直角顶点,由于4OC OA ==,点C 在抛物线上,则点(04)C ,
为满足A '
B ' O
B
A
C x
y
(第25题)
条件的点.
②若以点A 为直角顶点,则使PAO △为等腰直角三角形的点P 的坐标应为(44),或(44)-,,经计算知;此两点不在抛物线上.
③若以点P 为直角顶点,则使PAO △为等腰直角三角形的点P 的坐标应为(22),或(22)-,,经计算知;此两点也不在抛物线上.
综上述在抛物线上只有一点(04)P ,使OAP △为等腰直角三角形. ························· 10分
26.(本小题满分12分)
(1)证明:在ADC △和EGC △中Rt ADC EGC ∠=∠=∠ ,C C ∠=∠ ADC EGC ∴△∽△ EG CG
AD CD
∴
= ···································································· 3分 (2)FD 与DG 垂直 ······················································· 4分
证明如下:
在四边形AFEG 中,
90FAG AFE AGE ∠=∠=∠=
∴四边形AFEG 为矩形
AF EG ∴=
由(1)知EG CG
AD CD
= AF CG AD CD
∴= ·················································································································· 6分 ABC △为直角三角形,AD BC ⊥ FAD C ∴∠=∠ AFD CGD ∴△∽△ ADF CDG ∴∠=∠ ······································································································· 8分
又90CDG ADG ∠+∠=
90ADF ADG ∴∠+∠=
即90FDG ∠=
FD DG ∴⊥ ················································································································· 10分 (3)当AD AC =时,FDG △为等腰直角三角形,
理由如下:
AB AC = ,90BAC ∠= AD DC ∴=
由(2)知:AFD CGD △∽△
F
A
G
C
E
D B
(第26题)
1FD AD
GD DC ∴
== FD DG ∴=
又90FDG ∠=
FDG ∴△FDG ∴△为等腰直角三角形 ······································································ 12分