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2009年广东省中山市初中毕业生学业考试数学试题(word)

2009 年广东省中山市初中毕业生学业考试
数 学
说明: 说明 1.全卷共 4 页,考试用时 100 分钟,满分为 120 分. 2.答卷前,考生务必用黑色字迹的签字笔或钢笔在答题卡填写自己的准考证号、姓名、 试室号、座位号.用 2B 铅笔把对应该号码的标号涂黑. 3.选择题每小题选出答案后,用 2B 铅笔把答题卡上对应题目选项的答案信息点涂黑, 如需改动,用橡皮擦干净后,再选涂其他答案,答案不能答在试题上. 4.非选择题必须用黑色字迹钢笔或签字笔作答、答案必须写在答题卡各题目指定区域 内相应位置上;如需改动,先划掉原来的答案,然后再写上新的答案;不准使用铅笔和涂改 液.不按以上要求作答的答案无效. 5.考生务必保持答题卡的整洁.考试结束时,将试卷和答题卡一并交回. 一、选择题(本大题 5 小题,每小题 3 分,共 15 分)在每小题列出的四个选项中,只有一 个是正确的,请把答题卡上对应题目所选的选项涂黑. ) 1. 4 的算术平方根是( A. ±2 B. 2 C. ± 2 ) C. a
5

D. 2

2.计算 ( a 3 ) 2 结果是( A. a
6

B. a

9

D. a

8

3.如图所示几何体的主(正)视图是(



A. B. C. D. 4. 《广东省 2009 年重点建设项目计划 (草案) 显示, 》 港珠澳大桥工程估算总投资 726 亿元, 用科学记数法表示正确的是( ) A. 7.26 × 10 元
10

B. 72.6 × 10 元
9

C. 0.726 × 10 元
11

D. 7.26 × 10 元
11

5.方程组 ?

?3x + y = 0
2 2 ? x + y = 10

的解是(



A. ?

? x1 = 1 ? y1 = 3 ? x1 = 3 ? y1 = 1

? x2 = ?1 ? ? y2 = ?3 ? x2 = ?3 ? ? y 2 = ?1

B. ?

? x1 = 3 ? y1 = ?1 ? x1 = 1 ? y 1 = ?3

? x2 = ?3 ? ? y2 = 1 ? x2 = ?1 ? ? y2 = 3

C. ?

D. ?

二、填空题: (本大题 5 小题,每小题 4 分,共 20 分)请将下列各题的正确答案填写在答题 卡相应的位置上.
2 2 6.分解因式 x ? y ? 3 x ? 3 y



7.已知 ⊙O 的直径 AB = 8cm,C 为 ⊙O 上的一点,∠BAC = 30° , 则 BC = cm . 8.一种商品原价 120 元,按八折(即原价的 80%)出售,则现售价 A 应为 元. 9.在一个不透明的布袋中装有 2 个白球和 n 个黄球,它们除颜色不 同外,其余均相同.若从中随机摸出一个球,摸到黄球的概率是

C

B O

则 n = _____________. 10.用同样规格的黑白两种颜色的正方形瓷砖,按下图的方式铺地板,则第(3)个图形中 有黑色瓷砖 块,第 n 个图形中需要黑色瓷砖________块(用含 n 的代数式表示) .

4 , 5

第 7 题图

……

(1)

(2) 第 10 题图

(3)

三、解答题(一) (本大题 5 小题,每小题 6 分,共 30 分) 11. (本题满分 6 分)计算: ?

1 + 9 ? sin 30° π+3)0 . +( 2
A D

12. (本题满分 6 分)解方程

2 1 =? x ?1 x ?1 13. (本题满分 6 分) 如图所示,△ ABC 是等边三角形, D 点是 AC 的中点,延长 BC 到 E ,使 CE = CD , (1)用尺规作图的方法,过 D 点作 DM ⊥ BE ,垂足是 M (不写
2

作法,保留作图痕迹) ; (2)求证: BM = EM . 14. (本题满分 6 分)已知:关于 x 的方程 2 x + kx ? 1 = 0
2

B

C 第 13 题图

E

(1)求证:方程有两个不相等的实数根; (2)若方程的一个根是 ?1 ,求另一个根及 k 值.

15. (本题满分 6 分)如图所示, A 、 B 两城市相距 100km ,现 E 计划在这两座城市间修建一条高速公路(即线段 AB ) ,经测量, 森林保护中心 P 在 A 城市的北偏东 30° B 城市的北偏西 45° 30° 和 的方向上,已知森林保护区的范围在以 P 点为圆心, 50km 为半 径的圆形区域内,请问计划修建的这条高速公路会不会穿越保护 A 区,为什么?(参考数据: 3 ≈1.732,2 ≈1.414 )

P F 45°

B 第 15 题图

四、解答题(二) (本大题 4 小题,每小题 7 分,共 28 分) 16. (本题满分 7 分)某种电脑病毒传播非常快,如果一台电脑被感染,经过两轮感染后就 会有 81 台电脑被感染. 请你用学过的知识分析, 每轮感染中平均一台电脑会感染几台电脑? 若病毒得不到有效控制,3 轮感染后,被感染的电脑会不会超过 700 台? 17. (本题满分 7 分)某中学学生会为了解该校学生喜欢球类活动的情况,采取抽样调查的 方法,从足球、乒乓球、篮球、排球等四个方面调查了若干名学生的兴趣爱好,并将调查的 结果绘制成如下的两幅不完整的统计图(如图 1,图 2 要求每位同学只能选择一种自己喜欢 的球类; 图中用乒乓球、 足球、 排球、 篮球代表喜欢这四种球类中的某一种球类的学生人数) , 请你根据图中提供的信息解答下列问题: (1)在这次研究中,一共调查了多少名学生? (2)喜欢排球的人数在扇形统计图中所占的圆心角是多少度? (3)补全频数分布折线统计图. 人数 50 40 30 20 10 项目 O
足球 乒乓球 篮球 排球 篮球 40% 排球 乒乓球 20% 足球

图2

图1 第 17 题图

AD ∠ ,以 AB 为直径作 ⊙O , 18. (本题满分 7 分)在 ABCD 中, AB = 10, =m, D = 60° (1)求圆心 O 到 CD 的距离(用含 m 的代数式来表示) ; (2)当 m 取何值时, CD 与 ⊙O 相切. D A

?

O

C 第 18 题图

B

19. (本题满分 7 分)如图所示,在矩形 ABCD 中, AB = 12,AC =20 ,两条对角线相交 于点 O . OB 、OC 为邻边作第 1 个平行四边形 OBB1C , 以 对角线相交于点 A1 , 再以 A1 B1 、

A1C 为邻边作第 2 个平行四边形 A1 B1C1C ,对角线相交于点 O1 ;再以 O1 B1 、 O1C1 为邻边

作第 3 个平行四边形 O1 B1 B2C1 ……依次类推. (1)求矩形 ABCD 的面积; (2)求第 1 个平行四边形 OBB1C 、第 2 个平行四边形 A1 B1C1C 和第 6 个平行四边形的面 积. A D

O A1 O1 B1 A2 B2 第 19 题图 五、解答题(三) (本大题 3 小题,每小题 9 分,共 27 分) 20、 (本题满分 9 分) (1)如图 1,圆心接 △ ABC 中, AB = BC = CA , OD 、 OE 为 ⊙O 的半径, OD ⊥ BC 于点 F , OE ⊥ AC 于点 G, A A 求证: 阴影部分四边形 OFCG 的面积是 △ ABC 的面 E 1 G E 积的 . 3 O O (2)如图 2,若 ∠DOE 保持 120° 角度不变, B C B C F 求证:当 ∠DOE 绕着 O 点旋转时,由两条半径和 D D △ ABC 的两条边围成的图形(图中阴影部分)面积 图1 图2 1 始终是 △ ABC 的面积的 . 第 20 题图 C1 C2

B

C

3

21. (本题满分 9 分)小明用下面的方法求出方程 2 x ? 3 = 0 的解,请你仿照他的方法求 出下面另外两个方程的解,并把你的解答过程填写在下面的表格中. 方程 换元法得新方 程 令 x = t, 则 2t ? 3 = 0 解新方程 检验 求原方程的解

2 x ?3 = 0

t=

3 2

t=

3 >0 2

3 x= , 2 9 所以 x = 4

x+ 2 x ?3= 0 x+ x?2 ?4=0

22. (本题满分 9 分) 正方形 ABCD 边长为 4,M 、N 分别是 BC 、 CD 上的两个动点,当 M 点在 BC 上运动时,保持 AM 和 MN 垂 A 直, (1)证明: Rt△ ABM ∽ Rt△MCN ; (2)设 BM = x ,梯形 ABCN 的面积为 y ,求 y 与 x 之间的函数 关系式;当 M 点运动到什么位置时,四边形 ABCN 面积最大,并 求出最大面积; B (3) M 点运动到什么位置时 Rt△ ABM ∽ Rt△ AMN , x 的 当 求 值.

D

N C M 第 22 题图

广东省中山市 2009 年初中毕业生学业考试 数学试题参考答案及评分建议
小题, 一、选择题(本大题 5 小题,每小题 3 分,共 15 分) 选择题( 1.B 2.A 3.B 4.A 5.D 填空题( 小题, 二、填空题(本大题 5 小题,每小题 4 分,共 20 分) 6. ( x + y )( x ? y ? 3) 7.4 8.96 9.8 10.10, 3n + 1

(本大题 小题, 三、解答题(一) 本大题 5 小题,每题 6 分,共 30 分) 解答题( ( 11.解:原式=

1 1 + 3 ? + 1 ········································································································ 4 分 2 2

=4. ························································································································ 6 分 12.解:方程两边同时乘以 ( x + 1)( x ? 1) , ··············································································· 2 分

2 = ?( x + 1) , ······························································································································ 4 分
x = ?3 , ······································································································································· 5 分 经检验: x = ?3 是方程的解.····································································································· 6 分
13.解: (1)作图见答案 13 题图, A D M B C E

··········································································· 2 分 (2)∵ △ ABC 是等边三角形, D 是 AC 的中点, ∴ BD 平分 ∠ABC (三线合一) , ∴∠ABC = 2∠DBE . ················································································································ 4 分

答案 13 题图

∵ CE = CD , ∴∠CED = ∠CDE . 又∵ ∠ACB = ∠CED + ∠CDE , ∴∠ACB = 2∠E . ······················································································································ 5 分 又∵ ∠ABC = ∠ACB , ∴ 2∠DBC = 2∠E , ∴∠DBC = ∠E , ∴ BD = DE . 又∵ DM ⊥ BE , ∴ BM = EM .···························································································································· 6 分
14.解: (1) 2 x + kx ? 1 = 0 ,
2

? = k 2 ? 4 × 2 × (?1) = k 2 + 8 , ·································································································· 2 分
无论 k 取何值, k ≥ 0 ,所以 k + 8 > 0 ,即 ? > 0 ,
2 2

∴ 方程 2 x 2 + kx ? 1 = 0 有两个不相等的实数根. ······································································ 3 分
(2)设 2 x + kx ? 1 = 0 的另一个根为 x ,
2

k 1 , ( ?1)i x = ? , ·································································································· 4 分 2 2 1 解得: x = , k = 1 , 2 1 ∴ 2 x 2 + kx ? 1 = 0 的另一个根为 , k 的值为 1. ··································································· 6 分 2 15.解:过点 P 作 PC ⊥ AB , C 是垂足, P 则 ∠APC = 30° ∠BPC = 45° ··············································· 2 分 , , E F AC = PC i tan 30° BC = PC i tan 45° , , ∵ AC + BC = AB , ······································································· 4 分 ∴ PC i tan 30° PC i tan 45° 100 , + =
则 x ?1 = ?

? 3 ? A ∴? ? 3 + 1? PC = 100 , ································································· 5 分 ? ? ?
∴ PC = 50(3 ? 3) ≈ 50 × (3 ? 1.732) ≈ 63.4 > 50 ,

C 答案 15 题图

B

答:森林保护区的中心与直线 AB 的距离大于保护区的半径,所以计划修筑的这条高速公路 不会穿越保护区. ························································································································ 6 分 解答题( ( (本大题 小题, 四、解答题(二) 本大题 4 小题,每小题 7 分,共 28 分) 16.解:设每轮感染中平均每一台电脑会感染 x 台电脑, ························································ 1 分 依题意得: 1 + x + (1 + x ) x = 81 , ······························································································ 3 分

(1 + x) 2 = 81 ,

x + 1 = 9 或 x + 1 = ?9 ,

x1 = 8,x2 = ?10 (舍去) ········································································································ 5 分 , (1 + x)3 = (1 + 8)3 = 729 > 700 . ································································································ 6 分
答: 每轮感染中平均每一台电脑会感染 8 台电脑, 轮感染后, 3 被感染的电脑会超过 700 台. ······················································································································································· 7 分 17.解: (1) 20 ÷ 20% = 100 (人) ······················································································· 1 分 .

30 × 100% = 30% , ········································································································ 2 分 100 1 ? 20% ? 40% ? 30% = 10% , 360° 10% = 36° ···················································································································· 3 分 × . (3)喜欢篮球的人数: 40% × 100 = 40 (人) ······································································ 4 分 , 喜欢排球的人数: 10% × 100 = 10 (人) ················································································· 5 分 .
(2) 人数 50 40 30 20 10 项目 O
足球 乒乓球 篮球 排球

····························· 7 分

18.解: (1)分别过 A,O 两点作 AE ⊥ CD,OF ⊥ CD ,垂足分别为点 E ,点 F , ∴ AE ∥ OF,OF 就是圆心 O 到 CD 的距离. ∵ 四边形 ABCD 是平行四边形, ∴ AB ∥ CD, AE = OF .········································································································ 2 分 ∴ D A O F C B 答案 18 题图(1) 在 Rt△ ADE 中, ∠D = 60° ∠D = , sin D A O F C B 答案 18 题图(2)

答案 17 题图

E

E

AE AE , 60° sin = , AD AD

3 AE 3 3 = ,AE = m,OF = AE = m , ······································································· 4 分 2 m 2 2

圆心到 CD 的距离 OF 为

3 m . ······························································································· 5 分 2

(2)∵ OF =

3 m, 2

AB 为 ⊙O 的直径,且 AB = 10 , ∴ 当 OF = 5 时, CD 与 ⊙ O 相切于 F 点,


3 10 3 m = 5,m = , ········································································································· 6 分 2 3 10 3 时, CD 与 ⊙ O 相切. ····················································································· 7 分 3

∴当 m =

19.解: (1)在 Rt△ ABC 中,

BC = AC 2 ? AB 2 = 202 ? 122 = 16 ,
S矩形ABCD = ABi BC = 12 × 16 = 192 . ························································································ 2 分
(2)∵ 矩形 ABCD ,对角线相交于点 O ,

∴ S ABCD = 4 S△OBC . ···················································································································· 3 分
∵ 四边形 OBB1C 是平行四边形,

∴ OB ∥ CB1,OC ∥ BB1 , ∴∠OBC = ∠B1CB,∠OCB = ∠B1 BC .
又∵ BC = CB ,

∴△OBC ≌△B1CB ,
∴ SOBB1C = 2 S△OBC = 1 S ABCD = 96 , ························································································· 5 分 2 1 1 1 同理, S A1B1C1C = SOBB1C = × × S ABCD = 48 , ······································································ 6 分 2 2 2 1 第 6 个平行四边形的面积为 6 S ABCD = 3 . ··············································································· 7 分 2
(本大题 小题, 五、解答题(三) 本大题 3 小题,每小题 9 分,共 27 分) 解答题( ( 20.证明: (1)如图 1,连结 OA,OC , 因为点 O 是等边三角形 ABC 的外心, 所以 Rt△OFC ≌ Rt△OGC ≌ Rt△OGA .····································· 2 分 A G O B F C E

SOFCG = 2 S△OFC = S△OAC ,

D 答案 20 题图(1)

因为 S△OAC = 所以 SOFCG

1 S△ ABC , 3 1 = S△ ABC . ··············································································································· 4 分 3

(2)解法一: 连结 OA,OB 和 OC ,则 △ AOC ≌△COB ≌△BOA , ∠1 = ∠2 , ·································· 5 分 A 不妨设 OD 交 BC 于点 F , OE 交 AC 于点 G , E 2 ∠AOC = ∠3 + ∠4 = 120° ∠DOE = ∠5 + ∠4 = 120° , , 3 G ∴∠3 = ∠5 .·························································································· 7 分 O 4 5 在 △OAG 和 △OCF 中, 1 C B F ?∠1 = ∠2, D ?

?OA = OC, ? ?∠3 = ∠5,

答案 20 题图(2)

∴△OAG ≌△OCF , ················································································································ 8 分 1 ∴ SOFCG = S△ AOC = S△ ABC . ···································································································· 9 分 3
A 解法二: E 不妨设 OD 交 BC 于点 F , OE 交 AC 于点 G , G 3 K 作 OH ⊥ BC,OK ⊥ AC ,垂足分别为 H 、K , ······················· 5 分 O 2 在四边形 HOKC 中, ∠OHC = ∠OKC = 90° ∠C = 60° , , 1 C B HF ∴∠HOK = 360° 90 ? 90° ? 60° = 120° , ······························· 6 分 D 即 ∠1 + ∠2 = 120° . 答案第 20 题图(3) 又∵ ∠GOF = ∠2 + ∠3 = 120° , ∴∠1 = ∠3 . ································································································································ 7 分 ∵ AC = BC , ∴ OH = OK , ∴△OGK ≌△OFH , ··············································································································· 8 分

1 ∴ SOFCG = SOHCK = S△ ABC . ····································································································· 9 分 3
21.解: 方程 换元法得新方 程 令 解新方程 检验 求原方程的解

x = t ,则

x+ 2 x ?3= 0

t1 = 1,t2 = ?3
……2 分

t1 = 1 > 0, t2 = ?3 < 0 (舍去)
……3 分

x = 1 ,所以 x =1.
……4 分

t 2 + 2t ? 3 = 0
……1 分 令

x?2 =t,


x+ x?2 ?4=0

t1 = 1,t2 = ?2
……7 分

t1 = 1 > 0, t2 = ?2 < 0 (舍去)
……8 分

x ? 2 = 1 ,所以 x ? 2 = 1,x = 3 .
……9 分

t2 + t ? 2 = 0
……6 分

22.解: (1)在正方形 ABCD 中, AB = BC = CD = 4,∠B = ∠C = 90° , ∵ AM ⊥ MN , A ∴∠AMN = 90° , ∴∠CMN + ∠AMB = 90° . 在 Rt△ ABM 中, ∠MAB + ∠AMB = 90° , ∴∠CMN = ∠MAB , ∴ Rt△ ABM ∽ Rt△MCN . ······················································· 2 分 (2)∵ Rt△ ABM ∽ Rt△MCN ,

D

N C



AB BM 4 x = , ∴ = , MC CN 4 ? x CN

B M 答案 22 题图

∴ CN =

? x2 + 4 x , ···················································································································· 4 分 4

? 1 ? ? x2 + 4 x 1 1 ∴ y = S梯形ABCN = ? + 4 ?i4 = ? x 2 + 2 x + 8 = ? ( x ? 2) 2 + 10 , 2? 4 2 2 ?
当 x = 2 时, y 取最大值,最大值为 10.··················································································· 6 分 , (3)∵ ∠B = ∠AMN = 90°

∴ 要使 △ ABM ∽△ AMN ,必须有
由(1)知

AM AB = , ································································ 7 分 MN BM

AM AB = , MN MC ∴ BM = MC , ∴ 当点 M 运动到 BC 的中点时, △ ABM ∽△ AMN ,此时 x = 2 . ···································· 9 分
(其它正确的解法,参照评分建议按步给分)


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