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2009年烟台市中考数学试卷



2009 年烟台市初中学生学业考试 年烟台市初中学生学业考试 数 学 试 题
一 选择题(共 12 小题,每小题 4 分,共 48 分) 1.|-3|的相反数是 A.3 B。-3 C。-1/3 D。1/3 2.视为表对我来说并不陌生。如图是视力表的一部分,其中开口向上的 两个“E”之间的变换是 A.平移 B。旋转 C。对称 D。位似 3.学完分式运算后,老师出了一道题“

化简: 小明的做法是:原式 =

x+3 2? x + ” x + 2 x2 ? 4

( x + 3)( x ? 2) x ? 2 x 2 + x ? 6 ? x ? 2 x 2 ? 8 ? 2 = = 2 ; x2 ? 4 x ?4 x2 ? 4 x ?4

2 2 小亮的做法是:原式 = ( x + 3)( x ? 2) + (2 ? x) = x + x ? 6 + 2 ? x = x ? 4 ;

小芳的做法是:原式 = 其中正确的是( ) A.小明 B.小亮
2

x+3 x?2 x+3 x + 3 ?1 1 ? = ? = = 1. x + 2 ( x + 2)( x ? 2) x + 2 x + 2 x+2
C.小芳 D.没有正确的
2

4.设 a,b 是方程 x + x ? 2009 = 0 的两个实数根,则 a + 2a + b 的值为( A.2006 B.2007 C.2008 D.2009 5.一个长方体的左视图、俯视图及相关数据如图所示, 则其主视图的面积为( ) A.6 B.8 C.12 D.24 6.如图,数轴上 A,B 两点表示的数分别为 ?1 和 3 , 点 B 关于点 A 的对称点为 C,则点 C 所表示的数为( A. ?2 ? 3 B. ?1 ? 3 C. ?2 + 3 ) D. 1 + 3 C 3 2 左视图

) 4

俯视图 (第 5 题图) A O B (第 6 题图)

7.某校初一年级有六个班,一次测试后,分别求得各个班级学生成绩的平均数,它们不完 全相同,下列说法正确的是( ) A.全年级学生的平均成绩一定在这六个平均成绩的最小值与最大值之间 B.将六个平均成绩之和除以 6,就得到全年级学生的平均成绩 C.这六个平均成绩的中位数就是全年级学生的平均成绩 y D.这六个平均成绩的众数不可能是全年级学生的平均成绩 8.如图,直线 y = kx + b 经过点 A( ?1, 2) 和点 B ( ?2, , ? 0) 直线 y = 2 x 过点 A,则不等式 2 x < kx + b < 0 的解集为( ) B A
第 1 页 共 11 页

O

x

(第 8 题图)

A. x < ?2 C. ?2 < x < 0

B. ?2 < x < ?1 D. ?1 < x < 0

9.现有四种地面砖,它们的形状分别是:正三角形、正方形、正六边形、正八边形,且它 们的边长都相等.同时选择其中两种地面砖密铺地面,选择的方式有( ) A A.2 种 B.3 种 C.4 种 D.5 种 10.如图,等边 △ ABC 的边长为 3, P 为 BC 上一点, 且 BP = 1 , D 为 AC 上一点,若 ∠APD = 60° ,则 CD 的长为( )

3 A. 2

2 B. 3

1 C. 2

3 D. 4

60° B

P (第 10 题图)

D C

11.二次函数 y = ax 2 + bx + c 的图象如图所示,则一次函数 y = bx + b 2 ? 4ac 与反比例函 数y= y y
?1

a+b+c 在同一坐标系内的图象大致为( x
y x



y x x

y x

O 1

x

O A.

O B.

O C.

O D.

(第 11 题图)

12. 利用两块长方体木块测量一张桌子的高度. 首先按图①方式放置, 再交换两木块的位置, 按图②方式放置.测量的数据如图,则桌子的高度是( ) A.73cm B.74cm C.75cm D.76cm 80cm 70cm



② (第 12 题图)

第Ⅱ卷
个小题, 二、填空题(本题共 6 个小题,每小题 4 分,满分 24 分) 填空题( 13.若 3 x m +5 y 2 与 x 3 y n 的和是单项式,则 n =
m

. .

14.设 a > b > 0 , a + b ? 6ab = 0 ,则
2 2

a+b 的值等于 b?a

15.如图,将两张长为 8,宽为 2 的矩形纸条交叉,使重叠部分是一个 (第 15 题图)
第 2 页 共 11 页

菱形,容易知道当两张纸条垂直时,菱形的周长有最小值 8,那么菱形周长的最大值 . 是

?x ? + a≥ 2 16.如果不等式组 ? 2 的解集是 0 ≤ x < 1 ,那么 a + b 的值为 ?2 x ? b < 3 ?
17.观察下表,回答问题: 序号 1 2 3 …



图形



第 个图形中“△”的个数是“○”的个数的 5 倍. 18.如图,△ ABC 与 △ AEF 中, AB = AE,BC = EF,∠B = ∠E,AB 交 EF 于 D .给 出下列结论: A ① ∠AFC = ∠C ; E ② DF = CF ; ③ △ ADE ∽△FDB ; D ④ ∠BFD = ∠CAF . C B F 其中正确的结论是 (填写所有正确结论的序号) . (第 18 题图) 解答题( 个小题, 三、解答题(本大题共 8 个小题,满分 78 分) 19. (本题满分 6 分) 化简: 18 ?

9 3+ 6 ? + ( 3 ? 2)0 + (1 ? 2)2 . 2 3

20. (本题满分 8 分) 将如图所示的牌面数字分别是 1,2,3,4 的四张扑克牌背面朝上,洗匀后放在桌面上. (1)从中随机抽出一张牌,牌面数字是偶数的概率是 ; (2)从中随机抽出二张牌,两张牌牌面数字的和是 5 的概率是 ; (3)先从中随机抽出一张牌,将牌面数字作为十位上的数字,然后将该牌放回并重新洗匀, 再随机抽取一张, 将牌面数字作为个位上的数字, 请用画树状图或列表的方法求组成的两位 数恰好是 4 的倍数的概率.

(第 20 题图)

第 3 页 共 11 页

21. (本题满分 8 分) 某市教育行政部门为了了解初一学生每学期参加综合实践活动的情况, 随机抽样调查了某校 初一学生一个学期参加综合实践活动的天数, 并用得到的数据绘制了下面两幅不完整的统计 图(如图) . 人数 4天 3天 15% 30% 2 天 10% 5% 7 天 15% a 5天 6天
60 50 40 30 20 10

2 天 3 天 4 天 5 天 6 天 7 天 时间

(第 21 题图) 请你根据图中提供的信息,回答下列问题: (1)求出扇形统计图中 a 的值,并求出该校初一学生总数; (2)分别求出活动时间为 5 天、7 天的学生人数,并补全频数分布直方图; (3)求出扇形统计图中“活动时间为 4 天”的扇形所对圆心角的度数; (4)在这次抽样调查中,众数和中位数分别是多少? (5)如果该市共有初一学生 6000 人,请你估计“活动时间不少于 4 天”的大约有多少人?

22.(本题满分 8 分) 腾飞中学在教学楼前新建了一座“腾飞”雕塑(如图①).为了测量雕塑的高度,小明在 ,底部 B 点的俯角为 45° ,小 二楼找到一点 C,利用三角板测得雕塑顶端 A 点的仰角为 30° 华在五楼找到一点 D,利用三角板测得 A 点的俯角为 60° (如图②).若已知 CD 为 10 米, 请求出雕塑 AB 的高度. (结果精确到 0.1 米,参考数据 3 = 1.73 ) . D

A C



B



(第 22 题图)

第 4 页 共 11 页

23.(本题满分 10 分) 某商场将进价为 2000 元的冰箱以 2400 元售出, 平均每天能售出 8 台, 为了配合国家 “家 电下乡”政策的实施,商场决定采取适当的降价措施.调查表明:这种冰箱的售价每降低 50 元,平均每天就能多售出 4 台. (1)假设每台冰箱降价 x 元,商场每天销售这种冰箱的利润是 y 元,请写出 y 与 x 之间 的函数表达式; (不要求写自变量的取值范围) (2)商场要想在这种冰箱销售中每天盈利 4800 元,同时又要使百姓得到实惠,每台冰 箱应降价多少元? (3)每台冰箱降价多少元时,商场每天销售这种冰箱的利润最高?最高利润是多少?

24.(本题满分 10 分) 如图,AB,BC 分别是 ⊙O 的直径和弦,点 D 为 BC 上一点,弦 DE 交 ⊙O 于点 E,交 AB 于点 F,交 BC 于点 G,过点 C 的切线交 ED 的延长线于 H,且 HC = HG ,连接 BH , 交 ⊙O 于点 M,连接 MD,ME . H 求证: (1) DE ⊥ AB ; (2) ∠HMD = ∠MHE + ∠MEH . D M C G B A O F E (第 24 题图) 25.(本题满分 14 分) 如图, 直角梯形 ABCD 中, AD ∥ BC ,∠BCD = 90° 且 CD = 2 AD,tan∠ABC = 2 , , 过点 D 作 DE ∥ AB ,交 ∠BCD 的平分线于点 E,连接 BE. (1)求证: BC = CD ; (2)将 △BCE 绕点 C,顺时针旋转 90° 得到 △DCG ,连接 EG.. 求证:CD 垂直平分 EG. A (3)延长 BE 交 CD 于点 P. 求证:P 是 CD 的中点. E B C (第 25 题图)

D

G

第 5 页 共 11 页

26.(本题满分 14 分) 如图,抛物线 y = ax 2 + bx ? 3 与 x 轴交于 A B 两点,与 y 轴交于 C 点,且经过点 ,

(2, 3a ) ,对称轴是直线 x = 1 ,顶点是 M . ?
(1) 求抛物线对应的函数表达式; (2) 经过 C,M 两点作直线与 x 轴交于点 N ,在抛物线上是否存在这样的点 P ,使 以点 P,A C,N 为顶点的四边形为平行四边形?若存在, , 请求出点 P 的坐标; 若不存在,请说明理由; (3) 设直线 y = ? x + 3 与 y 轴的交点是 D , 在线段 BD 上任取一点 E(不与 B,D 重 合) ,经过 A B,E 三点的圆交直线 BC 于点 F ,试判断 △ AEF 的形状,并说 , 明理由; (4) 当 E 是直线 y = ? x + 3 上任意一点时, (3)中的结论是否成立?(请直接写出 结论) . y

A O 1
?3 C

B

x

M (第 26 题图)

2009 年烟台市初中学生学业考试 数学试题参考答案及评分意见
本试题答案及评分意见,供阅卷评分使用.考生若写出其它正确答案, 本试题答案及评分意见,供阅卷评分使用.考生若写出其它正确答案,可参照评分意 见相应评分. 见相应评分. 选择题( 个小题, 一、选择题(本题共 12 个小题,每小题 4 分,满分 48 分) 题号 答案 1 B 2 D 14. ? 2 3 C 4 C 5 B 6 A 7 A 8 B 9 B 10 B 11 D 12 C

个小题, 二、填空题(本题共 6 个小题,每小题 4 分,满分 24 分) 填空题( 13.

1 4

15. 17

16.1

17.20

18.①,③,④

个小题, 三、解答题(本题共 8 个小题,满分 78 分) 解答题( 19. (本题满分 6 分) 解: 18 ?

9 3+ 6 ? + ( 3 ? 2)0 + (1 ? 2)2 2 3

第 6 页 共 11 页

3 2 ? (1 + 2) + 1+ |1 ? 2 | . ············································································· 2 分 2 3 =3 2? 2 ? 1 ? 2 + 1 + 2 ? 1 . ·················································································· 4 分 2 3 = 2 ? 1 ······························································································································ 6 分 2 =3 2?
20. (本题满分 8 分) 解: (1) (2)

1 ····························································································································· 1 分 2

1 ··································································································································· 3 分 3

(3)根据题意,画树状图: ································································································· 6 分 开始

第一次

1

2

3

4

第二次 1

2 3

4

1

2 3

4 1

2 3

4 1

2 3

4

(第 20 题图) 由树状图可知,共有 16 种等可能的结果:11,12,13,14,21,22,23,24,31,32,33, 34,41,42,43,44.其中恰好是 4 的倍数的共有 4 种:12,24,32,44. 所以, P (4 的倍数) =

4 1 = . ······················································································· 8 分 16 4

或根据题意,画表格: ··········································································································· 6 分 第一次 第二次 1 2 3 4 1 11 21 31 41 2 12 22 32 42 3 13 23 33 43 4 14 24 34 44

由表格可知,共有 16 种等可能的结果,其中是 4 的倍数的有 4 种,所以,

P (4 的倍数) =

4 1 = . ·································································································· 8 分 16 4

21. (本题满分 8 分) 解: (1) a = 1 ? (10% + 15% + 30% + 15% + 5%) = 25% . ·············································· 1 分 初一学生总数: 20 ÷ 10% = 200 (人) ············································································· 2 分 . (2)活动时间为 5 天的学生数: 200 × 25% = 50 (人) . 活动时间为 7 天的学生数: 200 × 5% = 10 (人) ····························································· 3 分 . 频数分布直方图(如图) 人数
60 50 40 30 20 10

第 7 页 共 11 页

························· 4 分 (3)活动时间为 4 天的扇形所对的圆心角是 360° 30% = 108° ·································· 5 分 × . (4)众数是 4 天,中位数是 4 天. ······················································································ 7 分 (5)该市活动时间不少于 4 天的人数约是

6000 × (30% + 25% + 15% + 5%) = 4500 (人) ······························································ 8 分 .
22. (本题满分 8 分) 解:过点 C 作 CE ⊥ AB 于 E . D

∵ ∠D = 90° 60° = 30° ∠ACD = 90° 30° 60° ? , ? = , ∴∠CAD = 90° . 1 ∵ CD = 10, AC = CD = 5 . ································ 3 分 ∴ 2 A 在 Rt△ ACE 中, B C 5 AE = AC isin ∠ACE = 5isin 30° , ···················· 4 分 = 2 5 B CE = AC icos ∠ACE = 5icos 30° = 3 , ················ 5 分 2 (第 22 题图) 在 Rt△BCE 中, 5 ∵ ∠BCE = 45° BE = CE i tan 45° , ∴ = 3 , ·································································· 6 分 2 5 5 5 ∴ AB = AE + BE = + 3 = ( 3 + 1) ≈ 6.8 (米) . 2 2 2 所以,雕塑 AB 的高度约为 6.8 米. ····················································································· 8 分
23. (本题满分 10 分) 解: (1)根据题意,得 y = (2400 ? 2000 ? x ) ? 8 + 4 × 即y=?

? ?

x ? ?, 50 ?

2 2 x + 24 x + 3200 .······························································································ 2 分 25 2 2 (2)由题意,得 ? x + 24 x + 3200 = 4800 . 25
2

整理,得 x ? 300 x + 20000 = 0 . ······················································································· 4 分 解这个方程,得 x1 = 100,x2 = 200 . ················································································· 5 分 要使百姓得到实惠,取 x = 200 .所以,每台冰箱应降价 200 元. ··································· 6 分 (3)对于 y = ?

2 2 x + 24 x + 3200 , 25

第 8 页 共 11 页

当x=?

24 = 150 时,······························································································ 8 分 ? 2 ? 2×? ? ? ? 25 ?

150 ? ? y最大值 = (2400 ? 2000 ? 150) ? 8 + 4 × ? = 250 × 20 = 5000 . 50 ? ?
所以,每台冰箱的售价降价 150 元时,商场的利润最大,最大利润是 5000 元. ············ 10 分 24. (本题满分 10 分) H (1)证明:连接 OC , ∵ HC = HG, HCG = ∠HGC . ································ 1 分 ∴∠ ∵ HC 切 ⊙O 于 C 点,∴∠1 + ∠HCG = 90° ·············· 2 分 , D M ∵ OB = OC, 1 = ∠2 , ················································ 3 分 ∴∠ C G ∵ ∠HGC = ∠3 ,∴∠2 + ∠3 = 90° ···························· 4 分 . B A O F ∴∠BFG = 90° ,即 DE ⊥ AB . ···································· 5 分 (2)连接 BE .由(1)知 DE ⊥ AB . E ∵ AB 是 ⊙O 的直径, (第 24 题图)

∴ BD = BE . ························································································································ 6 分

∴∠BED = ∠BME . ············································································································ 7 分 ∵ 四边形 BMDE 内接于 ⊙O ,∴∠HMD = ∠BED . ······················································· 8 分 ∴∠HMD = ∠BME . ∵ ∠BME 是 △HEM 的外角,∴∠BME = ∠MHE + ∠MEH . ······································ 9 分 ∴∠HMD = ∠MHE + ∠MEH .······················································································· 10 分
25. (本题满分 14 分) 证明: (1)延长 DE 交 BC 于 F . ∵ AD ∥ BC , AB ∥ DF , ∴ AD = BF,∠ABC = ∠DFC .···································· 1 分 在 Rt△DCF 中,∵ tan ∠DFC = tan ∠ABC = 2 ,

A

D

P CD E G = 2 ,即 CD = 2CF . CF ∵ CD = 2 AD = 2 BF ,∴ BF = CF . ···························· 3 分 B C F 1 1 ∴ BC = BF + CF = CD + CD = CD , (第 25 题图) 2 2 即 BC = CD .························································································································ 4 分 (2)∵ CE 平分 ∠BCD ,∴ ∠BCE = ∠DCE . 由(1)知 BC = CD, CE = CE ,∴△BCE ≌△DCE ,∴ BE = DE . ······················ 6 分 ∵ 由图形旋转的性质知 CE = CG,BE = DG, DE = DG . ··············································· 8 分 ∴ ∴ C,D 都在 EG 的垂直平分线上,∴ CD 垂直平分 EG . ··············································· 9 分 (3)连接 BD .由(2)知 BE = DE ,∴∠1 = ∠2 . ∵ AB ∥ DE .∴∠3 = ∠2 .∴∠1 = ∠3 . ······································································· 11 分 ∵ AD ∥ BC ,∴∠4 = ∠DBC . 由(1)知 BC = CD .∴∠DBC = ∠BDC ,∴∠4 = ∠BDP .····································· 12 分 又∵ BD = BD ,∴△BAD ≌△BPD ,∴ DP = AD .··················································· 13 分



第 9 页 共 11 页

1 1 ∵ AD = CD ,∴ DP = CD .∴ P 是 CD 的中点. ···················································· 14 分 2 2
28. (本题满分 14 分)

??3a = 4a + 2b ? 3, ? 解: (1)根据题意,得 ? b ················· 2 分 ?? 2a = 1. ?
解得 ?

y D E N

?a = 1, ?b = ?2.
2

∴ 抛物线对应的函数表达式为 y = x ? 2 x ? 3 . ·········· 3 分
(2)存在. 在 y = x 2 ? 2 x ? 3 中,令 x = 0 ,得 y = ?3 . 令 y = 0 ,得 x ? 2 x ? 3 = 0 ,∴ x1 = ?1,x2 = 3 .
2

A O

1

N

x

F C

P

M (第 26 题图)

∴ A(?1, , B (3, , C (0, 3) . 0) 0) ? ? 又 y = ( x ? 1) 2 ? 4 ,∴ 顶点 M (1, 4) . ·············································································· 5 分
容易求得直线 CM 的表达式是 y = ? x ? 3 . 在 y = ? x ? 3 中,令 y = 0 ,得 x = ?3 .

∴ N (?3, ,∴ AN = 2 . ···································································································· 6 分 0)
在 y = x 2 ? 2 x ? 3 中,令 y = ?3 ,得 x1 = 0,x2 = 2 .

∴ CP = 2, AN = CP . ∴
∵ AN ∥ CP ,∴ 四边形 ANCP 为平行四边形,此时 P (2, 3) .····································· 8 分 ?
(3) △ AEF 是等腰直角三角形. 理由:在 y = ? x + 3 中,令 x = 0 ,得 y = 3 ,令 y = 0 ,得 x = 3 .

∴ 直线 y = ? x + 3 与坐标轴的交点是 D (0, , B (3, . 3) 0) ∴ OD = OB ,∴∠OBD = 45° ························································································· 9 分 .
又∵ 点 C (0, 3) ,∴ OB = OC .∴∠OBC = 45° ······················································· 10 分 ? . 由图知 ∠AEF = ∠ABF = 45° ∠AFE = ∠ABE = 45° ············································· 11 分 , . ∴∠EAF = 90° ,且 AE = AF .∴△ AEF 是等腰直角三角形. ······