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2014南京市建邺区中考一模数学 2 - 副本

发布时间:2014-05-06 13:16:00  

2014年南京市建邺区中考一模

一、选择题(本大题共6小题,每小题2分,共计12分.)

1.在1,-1,-2这三个数中,任意两数之和的最大值是(▲).

A.1 B. 0 C.-1 D.-3

2.16的值等于(▲).

A.4 B.-4 C.±4 D3.计算(ab2)3的结果是(▲).

A.ab5 B.ab6 C.a3b5 D.a3b6

24.若反比例函数y=A(1,m),则m的值是(▲). x

A.?2 B.2 1 C.-2 1 D. 2

5.从正面观察下图所示的两个物体,看到的是(▲).

6.四个小朋友站成一排,老师按图中所示的规则数数,数到2014时对应的小朋友可得一朵红花.那么,得红花的小朋友是(▲).

A.小沈 B.小叶 C.小李 D.小王

小沈 小叶 小李 小王 (第6题)

二、填空题(本大题共10小题,每小题2分,共计20分.请把答案直接填写在答卷纸相应位置上) .......

7.计算: 3+1) (33)=

8.南京目前正全面推进9条轨道交通线的建设,在建线路超过150公里,总投资超800亿元.将800亿用科学记数法表示为 ▲ .

9.分解因式:a 2-9= ▲ .

110.在函数y=中,自变量x的取值范围是 ▲ . x-1

?x=2,11.已知?是方程2x+ay=5的解,则a= ▲ . ?y=1

12.一块长方形菜地的面积是150m2,如果它的长减少5m,那么菜地就变成正方形,若设原菜地的长为x m,则可列方程为: ▲ .

13.如图,在凸四边形ABCD中,AB=BC=BD,∠ABC =80°,则∠ADC等于°.

14.如图,大圆的半径等于小圆的直径,且大圆的半径为4,则图中阴影部分的面积是.

15.如图,在平面直角坐标系中,点A、B的坐标分别为(3,1)、(1,0),若将线段BA绕点B顺时针旋转90°

得到线段BA',则点A'的坐标为 ▲ .

16.如图,⊙C过原点并与坐标轴分别交于A、D两点.已知∠OBA =30°,点D的坐标为(0,23),则点C的

坐标为( ▲ , ▲ ).

三、解答题(本大题共有11小题,共计88分.请在答卷纸指定区域内作答,解答时应写出必要的文字说明、

证明过程或演算步骤)

a+ba2b2

17.(本题6分)计算: ( +) ÷. aba-bb-a

??2x+5≤3(x+2) ,

18.(本题6分)解不等式组?x-1x,并写出不等式组的整数解. <?23?

19.(本题7分)已知:如图,AD、BF相交于点O,点E、C在BF上,BE=FC,AC=DE,AB=DF.

求证:OA=OD,OB=OF.

B

20.(本题7分)某校为了组织一项球类对抗赛,在本校随机调查了若干名学生,对他们每人最喜欢的一项球类2 (第19题) D E O C F A

运动进行了统计,并绘制成如图①、②所示的条形和扇形统计图.

10学生人数2015

某校学生最喜欢的球类运动项目条形统计图某校学生最喜欢的球类运动项目扇形统计图篮球26%足球 20%羽毛球 16%其他乒乓球 32%5图①(第20题)

根据统计图中的信息,解答下列问题:

(1)求本次被调查的学生人数,并补全条形统计图;

(2)若全校有1 500名学生,请你估计该校最喜欢篮球运动的学生人数;

(3)根据调查结果,请你为学校即将组织的一项球类对抗赛提出一条合理化建议.

21.(本题8分)如图,为了测量停留在空中的气球的高度,小明先站在地面上某点处观测气球,测得仰角为27°,

然后他向气球方向前进了50 m,此时观测气球,测得仰角为45°.若小明的眼睛离地面1.6 m,求气球离地面的高度(精确到0.1 m).(下列数据供参考:sin27°≈0.45,cos27°≈0.89,tan27°≈0.51)

22. (本题8分)(1)甲、乙、丙三只不透明的口袋中都装有1个白球、1个红球,它们除颜色外都相同,搅匀后A (第21题) C

3

分别从三只口袋中任意摸出1个球,求从三只口袋摸出的都是红球的概率.

(2)甲、乙、丙、丁四位同学分别站在正方形场地的四个顶点A、B、C、D处,每个人都以相同的速度沿着正方形的边同时出发随机走向相邻的顶点处,那么甲、乙、丙、丁四位同学互不相遇的概率是 ▲ .① 1 1 1 1 ② ③ ④ 2 4 8 16

23.(本题8分)某物流公司有20条输入传送带,20条输出传送带.某日,控制室的电脑显示,每条输入传送

带每小时进库的货物流量如图a,每条输出传送带每小时出库的货物流量如图b,而该日仓库中原有货物8吨,在0时至4时,仓库中货物存量变化情况如图c.

(1)根据图像,在0时至2时工作的输入传送带和输出传送带的条数分别为(▲);

A.8条和8条 B.14条和12条 C.12条和14条 D.10条和8条

(2)如图c,求当2≤x≤4时,y与x 的函数关系式;

中把相应的图像补充完整.

(3)若4时后恰好只有4条输入传送带和4条输出传送带在工作(至货物全部输出完毕为止),请在图c时)23题)

(第图

24.(本题9分) 已知,在△ABC中,AD为∠BAC的平分线,点E在BC的延长线上,且∠EAC=∠B,以DE

4

为直径的半圆交AD于点F,交AE于点M.

(1)判断AF与DF的数量关系,并说明理由;

(2)只用无刻度的直尺画出△ADE的边DE上的高AH;(3)若EF=4,DF=3,求DH的长.

........

B D C

(第24题) E

25.(本题9分)已知二次函数y=

x2+bx+c的图像与x轴交于A、B两点,AB=4,其中点A的坐标为(1,0).

(1)求二次函数的关系式及顶点坐标;

(2)请设计一种平移方法,使(1)中的二次函数图像的顶点在一次函数y=x的图像上,并直接写出平移后

相应的二次函数的关系式.

26.(本题10分)如图,在△ABC中,AB=AC2,BC=8.⊙A的半径为2,动点P从点B出发沿BC方向以

每秒1个单位的速度向点C运动,以点P为圆心,以PB为半径作⊙P,设点P运动的时间为t秒.

(1)当⊙P与直线AC相切时,求t的值;(2)当⊙P与⊙A相切时,求t的值;

(3) 延长BA交⊙A于点D,连接AP交⊙A于点E,连接DE并延长交BC于点F.当△ABP与△FBD相

似时,求t的值.

(第26题) 27.(本题10分)已知△ABC中,∠C是其最小的内角,如果过

5

顶点B的一条直线把这个三角形分割成了两个三角形,其中一个为等腰三角形,另一个为直角三角形,则称这条直线为△ABC的关于点B的伴侣分割线.例如:如图1,在Rt△ABC中,∠C=20°,过顶点B的一条直线BD交AC于点D,且∠DBC=20°,显然直线BD是△ABC的关于点B的伴侣分割线.

(1)如图2,在△ABC中,∠C=20°,∠ABC=110°.请在图中画出△ABC的关于点B 的伴侣分割线, 并标注角度;

(2)在△ABC中,设∠B的度数为y,

最小内角∠C的度数为x.试探索y与xABC存在关于点B的伴侣分割线. A

B 图1 C B 图2 C

(第27题)

6

参考答案及评分标准

1.B 2.A 3.D 4.B .5.C 6.D 7.23 8.8?10 10 9.(a?3)(a?3)

10.x?1 11.1 12.x(x?5)?150 13.140 14.4? 15.(2,-2) 16.(?1,)

a2b2a?ba2?b2ab?)?17.(本题6分)解:原式=(= 3分 ?a?ba?bab(a?b)a?b

?(a?b)(a?b)ab?ab·?·············································································· 6分 (a?b)a?b

18.(本题6分)解:解不等式①,得x≥-1. ································································· 2分

解不等式②,得x<3. ·································································································· 4分

所以,不等式组的解集是-1≤x<3. ······································································· 5分

整数解为—1,0,1,2. ······························································································· 6分

19.(本题7分)证明:连接AF,BD,∵BE=CF,∴BC=FE.

又∵AC=DE,AB=DF,∴△ABC≌△DFE.……………………… 3分

∴∠ABF=∠DFB. ∴AB∥DF.又∵AB=DF,

∴四边形ABDF为平行四边形. ··················································································· 6分

∴ OA=OD, OB=OF. ································································································· 7分

20.(本题7分)(1)50,图略; 3分

(2)390; ····························································································································· 5分

(3)答案不唯一,例如:建议学校组织乒乓球和篮球比赛 ··············································· 7分

21.(本题8分)解:依题意得,BD=CD,设CD=x,则AD=x+50, ························ 1分

在Rt△ADC中,CDx?tan27?,∴?0.51. ················································ 4分 ADx?50

解得x?52.0. ············································································································· 6分

∴高度约为52.0?1.6?53.6(m). ············································································· 7分 答:气球离地面的高度约为53.6m.···················································································· 8分

22.(本题8分)(1)树状图或枚举法正确; ······································································· 3分

共有8种等可能结果 ·········································································································· 4分 ∴从三只口袋摸出的都是红球的概率是1. ······································································ 6分 8

(2)③ ··································································································································· 8分

7

23. (本题8分)(1)B.··························································································· 2分

(2)由图象可知:当2≤x≤4时,y是x的一次函数,设y?kx?b, 将(2,12)、(4,32)代入得:??2k?b?12?k?10,解得:? ?4k?b?32?b??8

∴当2≤x≤4时,y?10x?8 ··························································································· 6分

(3)画图正确 ····················································································································· 8分

B

24. (本题9分)解:(1)AF?DF.理由如下:

∵AD平分?BAC,∴∠BAD=∠CAD.又∵∠B=∠CAE,∴∠BAD+∠B=∠CAD+∠CAE.

即∠ADE=∠DAE ,∴AE?DE. ····················································································· 2分 ∵DE是直径,∴EF⊥AD,∴AF?DF.…………………………………3分

(2)画图正确…………………………………… 5分

(3)由勾股定理得AE?DE?5∵∠ADH=∠EDF,∠AHD=∠DFE=90°,

∴△ADH∽△EDF.∴DHAD?.∴DH?3.6.…………… 9分 DFDE

25.(本题9分)解:(1)∵A(1,0),AB=4,∴B(5,0)或(-3,0).

将A(1,0),B(5,0)或A(1,0),(-3,0)代入y?x?bx?c得?

222?b??6?b?2或?, ?c?5?c??3∴二次函数的关系式为y?x?6x?5或y?x?2x?3.………………………… 3分

顶点坐标分别为(3,-4)、(-1,-4) …………………………………………… 5分

(2)每一个结果正确各1分,平移方式正确各1分. ············································· 9分

26.(本题10分)解:(1)过点P作PK⊥AC,垂足为点K,

∵⊙P与直线AC相切,∴BP?PK?t.由AB=AC

=BC=8得△ABC是

等腰直角三角形,可得∠C=45°, ∴△PKC是等腰直角三角形.

∴PC=2PK=2t,∴t+2t=8.解得t=82?8 ······················································ 3分

(2)过点A作AM⊥BC,垂足为点M,则AP?AM?PM,AM=

222若⊙P与⊙A外切,则(2?t)?4?(4?t),解得t?222B1BC?4, PM= t-4或4-t, 27.………………………5分

3

8

B

若⊙P与⊙A内切,则(t?2)2?42?(t?4)2,解得t?7.

综上所述,当t?7或t?7时,⊙P与⊙A相切.……………………… ··················· 7分 3

(3)当△ABP∽△FBD时,∠D=∠BPA,又∠D=∠AED=∠FEP ,

∴∠D=∠AED=∠FEP =∠BPA.∴∠BFD=2∠D.∵?D??B??BFD?180?,

∴∠D=45°,∴∠BAP=90°.∴AP与AC重合,∴t?8.………………… ······· 10分

CB

27.(本题10分)解:(1)画图正确,角度标注正确 2分

(2)设BD为△ABC的伴侣分割线,分以下两种情况.

第一种情况:△BDC是等腰三角形,△ABD是直角三角形,易知∠C和∠DBC必为底角,∴ ∠DBC=∠C=x. 当∠A=90°时,△ABC存在伴侣分割线,此时y?90-x,当∠ABD=90°时,△ABC存在伴侣分割线,此时y?90?x,当∠ADB=90°时,△ABC存在伴侣分割线,此时x?45,且y?x;

第二种情况:△BDC是直角三角形,△ABD是等腰三角形,当∠DBC=90°时,若BD=AD,则△ABC存在伴侣分割线,此时180?x?y?y?90,∴y?135?

分割线,此时∠A=45°,∴y?135-x.

综上所述,当y?90-x或y?90?x或x?45,且y?x或y?135?

分割线. 1x,当∠BDC=90°时,若BD=AD,则△ABC存在伴侣21x或y?135-x时△ABC存在伴侣2

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