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九年级调研测试 数学答案

发布时间:2013-12-31 09:47:37  

九年级调研测试数学试题参考答案

一、选择题(本大题共有8小题,每小题3分,共24分) 1.B 5.D

二、填空题(本大题共有10小题,每小题3分,共30分)

9.x1=0,x2=1 14.2

三、解答题(本大题共有10小题,共96分) 19.(本题满分8分)

(1)x=-6 ······································································ (4分,未体现解法-2分); (2)CD=26 ·········································································································· (4分). 20.(本题满分8分)

证明:∵DE∥AC,CE∥BD,∴四边形OCED是平行四边形, ························· (3分) ∵四边形ABCD是矩形,∴OC=OD, ··········································· (3分,未证明-1分) ∴四边形OCED是菱形. ······················································································ (8分)

21.(本题满分8分)

31

解:(1-); ············ (3分,其中开口方向向上1分)

28(2)画出图象; ····································································································· (3分) (3)写出当y<0时,x的取值范围是x<1或x>3. ········································· (2分)

22.(本题满分8分)

解:(1)点D运动到点G所经过的路线长为:

10.50° 15.6

11.-4 16.22

12.略 17.>

13.24 187

2.A 6.B

3.A 7.C

4.D 8.C

23.(本题满分10分)

解:(1)36÷(1+80%)=20(元).

答:这种玩具的进价为每个20元; ······································································ (3分) (2)设平均每次降价的百分率为x.···································································· (4分) 由题意,得36(1-x)2=25 ······················································································ (7分)

111

解得,x1=,或x2=(不合题意,舍去) ··························································· (9分)

661

∴x

6答:平均每次降价的百分率16.7%. ··································································· (10分)

24.(本题满分10分)

解:(1)连接OE、OF.

∵⊙O是△ABC的内切圆,切点为D、E、F,∴∠EOF=2y°,∠OEA=∠OFA=90°,

1

∴∠A+∠EOF=360°-90°-90°=180°,∴y=90-x. ···································· (5分)

2(2)设⊙O的半径是r.由勾股定理得:ACBC-AB=6, ∵⊙O是△ABC的内切圆,切点为D、E、F,

∴AE=AF,CD=CF,BE=BD,∠OEA=∠OFA=∠A=90°,OE=OF, ∴四边形OEAF是正方形,∴OE=OF=AE=AF=r, ∴AC-r+AB-r=BC,∴6-r+8-r=10,∴r=2.

答:⊙O的半径是2. ·························································································· (10分) 25.(本题满分10分)

解:(1)根据题意得,(30+x-20)(230-10x)=2520, ········································· (2分) 即-10x2+130x+2300=2520,

解得x1=2,x2=11(不合题意,舍去) 当x=2时,30+x=32(元)

答:每件玩具的售价定为32元时,月销售利润恰为2520元. ·························· (4分) (2)y=(30+x-20)(230-10x)=-10x2+130x+2300=-10(x-6.5)2+2722.5, ∵a=-10<0,

∴当x=6.5时,y有最大值为2722.5.··································································· (8分) ∵0<x≤10且x为正整数,

∴当x=6时,30+x=36,y=2720(元);当x=7时,30+x=37,y=2720(元).

答:每件售价定为36元或37元时,每个月可获得最大利润2720元. ··········· (10分)

九年级调研测试 数学答案 第 1 页 共 2 页

2

2

90??190??290??3

++=3π. ··································································· (4分) 180180180

(2)直线GB⊥DF.

理由如下:∵CD=CB,∠DCF=∠BCG,CF=CG, ∴△FDC≌△GBC.∴∠F=∠G,

又∵∠F+∠FDC=90°,∴∠G+∠FDC=90°,

即∠GHD=90°,故GB⊥DF. ·············································································· (8分)

26.(本题满分10分)

解:(1)∵∠BCO=∠CBO=45°,∴OC=OB=3

又∵点C在y轴的正半轴上,∴点C的坐标为(0,3) ····································· (2分) (2)在Rt△OCP中,OPPC-OC=()?3=2 当点P在点B右侧时,如图,PQ=4-2=2 当点P在点B左侧时,如图,PQ=4+2=6

∴t的值为2 或6. ································································································· (6分)

(3)由题意知,若⊙P与四边形ABCD

①当⊙P与BC相切于点C时,有∠BCP=90° 从而∠OCP=45°,得到OP=3,此时t=1 ②当⊙P与CD相切于点C时,有PC⊥CD 即点P与点O重合,此时t=4

③当⊙P与AD相切时,由题意,∠DAO=90° ∴点A为切点,如图

PC =PA =( 9-t ),PO =( t-4 ) 于是( 9-t )=( t-4 )+3,解得:t=5.6

∴t的值为1或4或5.6 ························································································· (10分)

2

2

2

2

2

2

2

2

∠D1CM+∠ACH1+∠ACD1=180°,∠AH1C=∠ACD1 ∴∠H1AC=∠D1CM

∵AC=CD1,∠AGC=∠CMD1=90°,∴△ACG≌△CD1M ∴CG=D1M 同理可证CG=D2N

∴D1M=D2N ············································································································ (9分) ②如图(作图正确) ····························································································· (11分)

D1M=D2N仍成立 ································································································· (12分)

K K1 E1 1

K2 DE2 D2 FF2

A

H2 G H1 B 图2

H1 H2 B 图3

2

22

2

28.(本题满分12分)

解:(1)由题意,得A(1,4)

∵二次函数y=ax2+bx+3的图象经过点A和点C.

∴?

?a?b?3?4,?a??1,

解得?

9a?3b?3?0,b?2,??

2

∴抛物线的解析式为y=-x +2x+3 ···································································· (3分) (2)①∵A(1,4),C(3,0)

∴可求直线AC的解析式为y=-2x+6 ································································· (4分)

t

P(1,4-t),将y=4-t代入y=-2x+6中,解得点E的横坐标为x=1+ 2

2

tt∴点G的横坐标为1+ ,代入抛物线的解析式中,可求点G的纵坐标为4- 2 4

22tt∴GE=( 4- )-( 4-t )=t-

4 4

tt

又点A到GE的距离为 ,C到GE的距离为2- 2 2

1t1t

即S△ACG =S△AEG + S△CEG = EG + EG·( 2- )

2 2 2 2

2

1t12= 2( t- )=- t-2 )+1 2 4 4 当t=2时,S△ACG的最大值为1 ·············································································· (8分)

20②t= ······················································································ (12分) 或t=20-5

13

27.(本题满分12分)

解:(1)D1M=D2N ································································································ (1分) 证明:∵∠ACD1=90°,∴∠ACH+∠D1CK=90° ∵∠AHK+∠ACD1=90°,∴∠ACH+∠HAC=90° ∴∠D1CK=∠HAC

∵AC=CD1,∴△ACH≌△CD1M ∴D1M=CH 同理可证D2N=CH

∴D1M=D2N ············································································································ (4分) (2)①D1M=D2N仍成立 ······················································································ (5分) 证明:过点C作CG⊥AB,垂足为点G ∵∠H1AC+∠ACH1+∠AH1C=180°

九年级调研测试 数学答案 第 2 页 共 2 页

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