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2011年全国各地100份中考数学试卷分类汇编(多边形与平行四边形)

2011年全国各地100份中考数学试卷分类汇编

第25章多边形与平行四边形

一、选择题

1.(2011安徽,6,4分)如图,D是△ABC内一点,BD⊥CD,AD=6,BD=4,CD=3,E、F、

G、H分别是AB、AC、CD、BD的中点,则四边形EFGH的周长是()

A.7 B.9 C.10 D.11

【答案】D

2. (2011广东广州市,2,3分)已知□ABCD的周长为32,AB=4,则BC=().

A.4

B.12

C.24

D.28

【答案】B

3. (2011山东威海,3,3分)在□ABCD中,点E为AD的中点,连接BE,交AC于点F,

则AF:CF=()

A.1:2 B.1:3 C.2:3 D.2:5

【答案】A

4. (2011四川重庆,9,4分)下面图形都是由同样大小的平行四边形按一定的规律组成,

其中,第①个图形一共有1个平行四边形,第②个图形一共有5个平行四边形,第③个图形一共有11个平行四边形,……,则第⑥个图形中平行四边形的个数为(

)

……

图①图②图③图④

A.55 B.42 C.41 D.29

【答案】C

5. (2011江苏泰州,7,3分)四边形ABCD中,对角线AC、BD相交于点O,给出下列四组条件:①AB∥CD,AD∥BC;②AB=CD,AD=BC;③AO=CO,BO=DO;④AB∥CD,AD=BC.其中一定能判定这个四边形是平行四边形的条件有

A.1组B.2组C.3组D.4组

【答案】C

6. (2011湖南邵阳,7,3分)如图(二)所示,A B C D

中,对角线AC,BD相交于点O,且AB≠AD,则下列式子不正确的是()

A.AC⊥BD

B.AB=CD

C. BO=OD

D.∠BAD=∠

BCD

【答案】A. 7. (2011重庆市潼南,9,4分)如图,在平行四边形 ABCD 中(AB≠BC ),直线EF

经过其对角线的交点O ,且分别交AD 、BC 于点M 、

N ,交BA 、DC 的延长线于点E 、F ,下列结论:

①AO=BO ;②OE=OF ; ③△EAM ∽△EBN ;

④△EAO ≌△CNO ,其中正确的是

A. ①②

B. ②③

C. ②④

D.③④

【答案】B

8. (2011广东东莞,5,3分)正八边形的每个内角为( )

A .120°

B .135°

C .140°

D .144°

【答案】B

9. (2011浙江省,8,3分)如图,在五边形ABCDE 中,∠BAE=120°, ∠B=∠E=90°,AB=BC ,AE=DE ,在BC ,DE 上分别找一点M,N ,使得△AMN 的周长最小时,则∠AMN+∠ANM 的度数为( )

A . 100°

B .110°

C . 120°

D . 130°

【答案】C 10. (2011台湾台北,33)图(十五)为一个四边形ABCD ,其中AC 与BD 交于E 点,且两灰色区域的面积相等。若AD =11,BC =10,则下列关系何者正确?

9题图A B C

D

E

F M N O

A .BCE DAE ∠<∠

B .BCE DAE ∠>∠

C .BE >DE

D .B

E <DE

【答案】A

11. (2011宁波市,7,3分)一个多边形的内角和是720°,这个多边形的边数是

A . 4

B . 5

C . 6

D . 7

【答案】C

12. (2011广东汕头,5,3分)正八边形的每个内角为( )

A .120°

B .135°

C .140°

D .144°

【答案】B

13. (2011内蒙古乌兰察布,10,3分)如图,已知矩形ABCD ,一条直线将该矩形 ABCD 分割成两个多边形,若这两个多边形的内角和分别为 M 和 N ,则 M + N 不可能是( )

A . 3600

B . 5400 C. 7200 D . 6300

【答案】D

14. (2011广东湛江2,3分)四边形的内角和为

A 180?

B 360?

C 540?

D 720?

【答案】B

15. (2011广东省,5,3分)正八边形的每个内角为( )

A .120°

B .135°

C .140°

D .144°

【答案】B

二、填空题 1. (

2011浙江金华,15,4分)如图,在□ABCD 中,AB =3,AD =4,∠ABC =60°,过BC 的中点E 作EF ⊥AB ,垂足为点F ,与DC 的延长线相交于点H ,则△DEF 的面积是 .

【答案】2 3

A C

B D

第10题H F

E

D C B A

2. (2011山东德州10,4分)如图,D ,E ,F 分别为△ABC 三边的中点,

则图中平行四边形的个数为___________.

【答案】3

3. (2011浙江丽水,15,4分)如图,在□ABCD 中,AB =3,AD =4,∠ABC =60°,过BC 的中点E 作EF ⊥AB ,垂足为点F ,与DC 的延长线相交于点H ,则△DEF 的面积是 .

【答案】23

4. (2011江苏苏州,12,3分)如图,在四边形ABCD 中,AB ∥CD ,AD ∥BC ,AC 、BD 相交于点O.若AC=6,则线段AO 的长度等于

___________.

【答案】3

5. (2011山东聊城,14,3分)如图,在□ABCD 中,AC 、BD 相交于点O ,点E 是AB 的中点,OE =3cm ,则AD 的长是__________cm .

【答案】6

6. (2011山东临沂,18,3分)如图,□ ABCD 中,E 是BA 延长线上一点,AB =AE ,连结

CE 交AD 于点F ,若CF 平分∠BCD ,AB =3,则BC 的长为 .

【答案】6 H F

E

D

C B A A

B C

D

E F 第10题图

7. (2011湖南常德,4,3分)四边形的外角和为__________.

【答案】360°

8. (2011四川广安,16,3分)若凸n 边形的内角和为1260°,则从一个顶点出发引的对角线条数是____

【答案】6

三、解答题

1. (2011浙江义乌,18,6分)如图,已知E 、F 是□ABCD 对角线AC 上的两点,

且BE ⊥AC ,DF ⊥AC .

(1)求证:△ABE ≌△CDF ;

(2)请写出图中除△ABE ≌△CDF 外其余两对全等

三角形(不再添加辅助线).

【答案】(1)∵四边形ABCD 是平行四边形

∴AB =CD AB ∥CD

∴∠BAE =∠FCD

又∵BE ⊥AC DF ⊥AC

∴∠AEB =∠CFD =90°

∴△ABE ≌△CDF (AAS )

(2)①△ABC ≌△CDA ②△BCE ≌△DAF

2. (2011湖南常德,21,7分)如图5,已知四边形ABCD 是平行四边形.

(1)求证:△MEF ∽△MBA ;

(2)若AF ,BE 分别,∠CBA 的平分线,求证DF =EC

【答案】

(1) 证明:在□ABCD 中,CD ∥AB

∴∠MEF =∠MBA ,∠MFE =∠MAB

∴△MEF ∽△MBA

(2) 证明:∵在□ABCD 中,CD ∥AB

∠DFA =∠FAB

又∵AF 是∠DAB 的平分线

∴∠DAF =∠FAB

∴∠DAF =∠DFA F E A C D A 图5 B

C D E F M

∴AD=DF

同理可得EC=BC

∵在□ABCD中,AD=BC

∴DF=EC

3. (2011四川成都,20,10分)如图,已知线段AB∥CD,AD与BC相交于点K,E是线段AD上一动点.

(1)若BK=5

2

KC,求

AB

CD

的值;

(2)连接BE,若BE平分∠ABC,则当AE=1

2

AD时,猜想线段AB、BC、CD三者之间有

怎样的等量关系?请写出你的结论并予以证明.再探究:当AE=1

n

AD(2

>

n),而其余条件

不变时,线段AB、BC、CD三者之间又有怎样的等量关系?请直接写出你的结论,不必证明.

K

E

C D

A B

【答案】解:(1)∵AB∥CD,BK=5

2

KC,∴

AB

CD

=

BK

CK

=

5

2

.

(2)如图所示,分别过C、D作BE∥CF∥DG分别交于AB的延长线于F、G三点,

∵BE∥DG,点E是AD的点,∴AB=BG;∵CD∥FG,CD∥AG,∴四边形CDGF是平行四边形,∴CD=FG;

∵∠ABE=∠EBC,BE∥CF,∴∠EBC=∠BCF,∠ABE=∠BFC,∴BC=BF,

∴AB-CD=BG-FG=BF=BC,∴AB=BC+CD.

K

E

C D

A B G

F

当AE=1

n

AD (2

>

n)时,(1

-

n)AB=BC+CD.

4. (2011四川宜宾,17⑶,5分)如图,平行四边形ABCD的对角线AC、BD交于点O,E、F 在AC上,G、H在BD上,AF=CE,BH=DG.

求证:GF∥HE.

【答案】证明:∵平行四边形ABCD 中,OA=OC ,

由已知:AF=CE

AF -OA=CE -OC ∴OF=OE

同理得:OG=OH

∴四边形EGFH 是平行四边形

∴GF ∥HE

5. (2011江苏淮安,20,8分)如图,四边形ABCD 是平行四边形,EF 分别是BC 、AD 上的点,∠1=∠2.

求证:△ABE ≌△CDF

.

【答案】∵四边形ABCD 是平行四边形,

∴∠B=∠D ,AB=DC ,

又∵∠1=∠2,

∴△ABE ≌△CDF (ASA ).

6. (2011四川凉山州,20,7分)如图,E F 、是平行四边形A B C D 的对角线A C 上的点,C E A F =,请你猜想:线段B E 与线段D F 有怎样的关系?并对你的猜想加以证明。

【答案】猜想:B E

D F 。 证明: ∵四边形ABCD 是平行四边形

∴C B A D =,C B ∥A D

∴B C E D A F ∠=

在B C E △和D AF △

B

C D

E

F

A 20题图 H A

(17(3)题图) C

B D O E

G

F

C B A D

B C E D A F

C E A F =??∠=∠??=?

∴B C E △≌D AF △

∴BE D F =,B E C D F A ∠=∠

∴B E ∥D F

即 B E D F 。

7. (2011江苏无锡,21,8分)(本题满分8分)如图,在□ABCD 中,E 、F 为对角线BD 上的两点,且∠BAE =∠DCF .

求证:BE = DF .

【答案】证明:∵□ABCD 中,AB = CD ,AB // CD ,…………………………………………(2

分)

∴∠ABE = ∠CDF ,……………………………………………………………(4分) 又∵∠BAE = ∠DCF ,∴△ABE ≌△CDF ,………………………………(6分) ∴BE = DF .…………………………………………………………………(8分)

8. (2011湖南永州,21,8分)如图,BD 是□ABCD 的对角线,∠ABD 的平分线BE 交AD 于点E ,∠CDB 的平分线DF 交BC 于点F .

求证:△ABE ≌△CDF .

【答案】证明:□ABCD 中,AB=CD ,∠A=∠C , AB ∥CD ∴∠ABD=∠CDB

∵∠ABE=21

∠ABD,∠CDF=21∠CDB ∴∠ABE=∠CDF

在△ABE 与△CDF 中 ??

???∠=∠=∠=∠C D F ABE CD AB C

A

∴△ABE ≌△CDF .

F E

C D

B A (第21题)

B C

D

A

E F

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全国中考数学试题分类汇编

A B C D P E 2015年全国中考数学试题分类汇编————压轴题 1. 在平面直角坐标系xOy 中,抛物线的解析式是y = 2 4 1x +1,点C 的坐标为(–4,0),平行四边形OABC 的顶点A ,B 在抛物线上,AB 与y 轴交于点M ,已知点Q (x ,y )在抛物线上,点P (t ,0)在x 轴上. (1) 写出点M 的坐标; (2) 当四边形CMQP 是以MQ ,PC 为腰的梯形时. ① 求t 关于x 的函数解析式和自变量x 的取值范围; ② 当梯形CMQP 的两底的长度之比为1:2时,求t 的值. (1)M(0,2)(2)1AC:y= 21x+1.PQ // MC.t x x --+0 14 12 =21 2. 如图,已知在矩形ABCD 中,AB =2,BC =3,P 是线段AD 边上的任意一点(不含端点 A 、D ),连结PC , 过点P 作PE ⊥PC 交A B 于E (1)在线段AD 上是否存在不同于P 的点Q ,使得QC ⊥QE ?若存在,求线段AP 与AQ 之间的数量关系;若不存在,请说明理由; (2)当点P 在AD 上运动时,对应的点E 也随之在AB 上运动,求BE 的取值范围. (3)存在,理由如下: 如图2,假设存在这样的点Q ,使得QC ⊥QE. 由(1)得:△PAE ∽△CDP , ∴ , ∴ ,

∵QC ⊥QE ,∠D =90 ° , ∴∠AQE +∠DQC =90 ° ,∠DQC +∠DCQ =90°, ∴∠AQE=∠DCQ. 又∵∠A=∠D=90°, ∴△QAE ∽△CDQ , ∴ , ∴ ∴ , 即 , ∴ , ∴ , ∴ . ∵AP≠AQ ,∴AP +AQ =3.又∵AP≠AQ ,∴AP≠ ,即P 不能是AD 的中点, ∴当P 是AD 的中点时,满足条件的Q 点不存在, 综上所述, 的取值范围8 7 ≤ <2; 3.如图,已知抛物线y =-1 2 x 2+x +4交x 轴的正半轴于点A ,交y 轴于点B . (1)求A 、B 两点的坐标,并求直线AB 的解析式; (2)设P (x ,y )(x >0)是直线y =x 上的一点,Q 是OP 的中点(O 是原点),以PQ 为对角线作正方形PEQF ,若正方形PEQF 与直线AB 有公共点,求x 的取值范围; (3)在(2)的条件下,记正方形PEQF 与△OAB 公共部分的面积为S ,求S 关于x 的函数解析式,并探究S 的最大值. (1)令x=0,得y=4 即点B 的坐标为(0,4) 令y=0,得(-1/2)x2+x+4=0 则x2-2x-8=0 ∴x=-2或x=4 ∴点A 的坐标为(4,0) 直线AB 的解析式为 (y-0)/(x-4)=(4-0)/(0-4) ∴y=-x+4 (2)由(1),知直线AB 的解析式为y=-x+4

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