2020年上海市嘉定区初三数学一模试卷及详解答案(WORD版) 下载本文

2019 学年第一学期嘉定区九年级期终学业质量调研测试

数学试卷阅卷参考答案

(考试时间 100 分钟,总分 150 分)( 2020.1)

一、选择题 (本大题共 6 题,每题 4 分,满分 24 分) 1. D; 2. A; 3. D; 4. A; 5. B; 6. C.

二、填空题: (本大题共 12 题,每题 4 分,满分 48 分)

3

9. 54 ; 10. 7. ;8. 81;

2

2 ;11. ( 3

1,0) ;12. 0 ; 13. y x 2 1 ;14.

;15. 75; 16.

360 n

(不要求写出函数的定义域)

; 17. 32 ;18. 36 .

5

三、解答题( 本大题共 7 题,满分 58 分) 19. ( 本题满分 10 分)

解: 2cos30 tan 45 2sin 30 cot 30

31

= 2 × 1 2× - 3 ·······························8 分

2 2

= 3 1 1

3 0 . ································1+1 分

20. (本小题满分 10 分,第( 1)小题 6 分,第( 2)小题 4 分) 证明:在 Rt△ AOH 中,

∵ AHO 90 ,

AOH 30 , OH 0.6 ,

·······························2 分 ∴ AO 2OH 2 0.6 1.2 ( m). ·

∴ OB AB OA 3 1.2 1.8 ( m)

·····························2 分

在Rt△ BOH中,

∵ BHO 90 , OH

0.6 , OB 1.8 ,

·········································································· 0.6 1 ·1.8 3

2分

∴ sin ABH

OH OB

( 2)过点 A 向直线 BH 作垂线,垂足为 M ·1分 ·······························································

A

O

在Rt△ABM 中,

∵ AMB 90 , sin ABM ∴ AM

1

, AB 3 ,

M B

AB sin ABM

1 3

H 图 6-2 1

3× 1 ··································2分

3

3

答: ABH 的正弦值为

,点 A 到直线 BH 的距离是 1米. ············································· 1分

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21. ( 本题满分 10 分)

·································1 分 解:设 AOB m , COD n , ·

由题意,得 l1 ∵ l1

mr nr

, l 2

l 2 ,∴

180

nr mr

=

180

································2 分

.·······································1 分

180 180

···································2 分 COD . ·

∵ ∴ ············1 分 OA 、 OB 、 OC 、 OD 都是⊙ O 的半径 , OA OB OC OD .··

OA OC , AOB ∵

COD , OB OD ,

∴ m n ,即 AOB

∴△ AOB≌△ COD . ·············································2 分 ∴ AB CD . ·················································1 分

22. ( 本题满分 10 分)

解:过点 A 作直线 BC 的垂线,垂足为 D (如图 7 所示) ····················1 分

由题意,得 ∴ BAC 又∵ B 90

···························1 分 BAD 60 , CAD 30 . ·

BAD CAD 30 ········································································· 1 分

60

30 ,∴ B

BAC . ··················1 分

BAD 90

∴ AC BC .··············································1 分 ∵ BC 20,∴ AC 在 Rt△ ACD 中, AD

BC 20 (海里) ····························1 分

AC cos CAD 20

3 2

10 3 (海里) ·············2 分

由题意知:以海岛 A 为圆心,半径长为 10 海里范围内有暗礁 .这里, AD 10 3 10 ,

所以,如果货轮继续向东航行,没有触礁的危险

. ······················2 分

A

D

E

图 7

D

B

图 8

C

23. ( 本题满分 12 分,第( 1)小题 4 分,第( 2)小题 8 分)

证明:( 1)∵ DE ∥ BC , ∴ BED

CBE ··························································· 1 分

又∵ ABE ∴

···························1 分 C ,∴△ BDE ∽△ CBE. ·

.············································1 分

DE BE

BE BC

∴ BE2

DE BC . ·········································1 分

( 2)∵ DE ∥ BC ,∴ AED

又∵ EAD ∴

C .又 ABE

C ,∴ AED

ABE . ··········1 分

BAE ,∴△ ADE ∽△ ABE . ··························1 分

. ···········································1 分

AE AB

AD AE

第 6 页 共 4 页

AD,∵ DE ∥ BC ∴

BD

AE , 即 AD CE

AE

BD

.························1 分 CE

AE

BD

AB CE

∵ BE 平分 ABC ,∴ ABE

BD BE

. ···········································1 分

CBE ,又∵ ABE C ,∴ CBE C . ···1 分

∴ BE CE . ············································1 分 ∴

AE

.············································1 分 AB

24.( 本题满分 12 分,每小题 4 分)

解:( 1) 图像基本正确(开口方向、对称轴、顶点、大致光滑)

············2 分 y x2 2 ·······2 分

.

将图 9 中的抛物线 y x 2 向下平移 2 个单位长,可得抛物线

备注:如果使用“列表、描点、连线”的方式叙述,需要呈现列表使用的表格

2)由题意,得点 A( x, y) 的“关联点”为 A1( x, y x) ·( ···················1 分

由点 A( x, y) 在抛物线 y

x2 上,可得 A(x, x 2 ) , A1( x, x2 x 2 上,∴ x x

2

2

x) ······························ 1 分

又∵ A1 (x, y x) 在抛物线 y

x

2

2 ··················1 分

解得 x 2 .将 x 2 代入 A1( x, x 2 x) ,得 A1 (2,2) ·····················1 分

x 2 nx) , ·······················1 分 ( 3)点 A( x, y) 的“待定关联点”为 A ( x, 2

∵ A2 ( x, x 2 nx) 在抛物线 y

x2 n的图像上,∴ x2 nx x 2 n . ···········1 分

∴ n nx 0 , n(1 x) 0 .又∵ n 0 ,∴ x 1 . ·······················1 分

当 x 1 时, x

2

1 n) .··nx 1 n ,故可得 A2 (1,·····················1 分

25.( 本题满分 14 分,第( 1)小题 4 分,第( 2)、( 3)小题各 5 分)

证明: ( 1) ∵ ABP

∴ ABP 即

BAP APB APB

APB 180 , APB APB APB

BAP

·········1 分 BAC 180 , ·

BAP BAP

BAC . ·························1 分

CAP .

ABP

∴ ABP 又∵ APB

CAP. ··········································1 分

···························1 分 APC ,∴ △PAB∽ △ PCA.··

( 2) 如图 10-1,∵ APB ·······1 分 BAC 180 , APB 120 ,∴ BAC 60 . ·

1

············1 分 AC . ·在△ ABC 中,∵ ABC 90 , BAC 60 ,∴ AB

2

PB PAAB 1

又∵ △PAB∽ △PCA,∴ . ··························1 分

PA PC AC 2

PB PB PA 1 PC

,即 ∴ 4 . ································2 分

PB PC PA PC 4 A

A A A

P B

图 10-1

C

P

第 7 页 共 4 页

B

C

P

P

C

B

图 10-2

C B

( 3) ∵ BAC 45 , APB

BAC 180 , APB APC 360 135 135

APC ,∴ APB APC 135 .

∴ BPC 360

APB 90 .················1 分 PC PA PA PB

( AC) .

AB

2

PAPC AC PC

∵ △PCA∽ △PAB,∴ ,∴

PB PA AB PB

PC ( AC )2 1 .

①如图 10-2 ,当 △ABC 是等腰三角形,且 AB AC 时, tan PBC

PB AB

·····················································1 分

②如图 10-3,当 △ ABC 是等腰三角形,且 AB BC 时, ACB

BAC 45 , ABC 90 ,

易得

AC AB

2 ,∴ tan PBC

PC PB

(

AC2

) 2 ··························2 分 AB

BAC 45 , ACB 90 ,

③如图 10-4 ,当 △ABC 是等腰三角形,且 AC BC 时, ABC

易得

AC AB

AC 2

,∴ tan PBC ( )2 PB AB

2

PC

1 2

1 2

.··························1 分

2 分;两个

备注: 写出 tan PBC 2 , tan PBC

这两个答案之中的一个,即可得到

全部写出,得 3 分.

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