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图形的相似易错题汇编及答案

图形的相似易错题汇编及答案
图形的相似易错题汇编及答案

图形的相似易错题汇编及答案

一、选择题

1.如图,O 是AC 的中点,将面积为216cm 的菱形ABCD 沿AC 方向平移AO 长度得到

菱形OB C D ''',则图中阴影部分的面积是( )

A .28cm

B .26cm

C .24cm

D .22cm

【答案】C 【解析】 【分析】

根据题意得,?ABCD ∽?OECF ,且AO=OC=

1

2

AC ,故四边形OECF 的面积是?ABCD 面积的14

【详解】 解:如图,

由平移的性质得,?ABCD ∽?OECF ,且AO=OC=1

2

AC 故四边形OECF 的面积是?ABCD 面积14

即图中阴影部分的面积为4cm 2. 故选:C 【点睛】

此题主要考查了相似多边形的性质以及菱形的性质和平移性质的综合运用.关键是 应用相似多边形的性质解答问题.

2.如图,在平行四边形ABCD 中,点E 在边DC 上,DE :EC=3:1,连接AE 交BD 于点F ,则△DEF 的面积与△BAF 的面积之比为( )

A.3:4 B.9:16 C.9:1 D.3:1

【答案】B

【解析】

【分析】

可证明△DFE∽△BFA,根据相似三角形的面积之比等于相似比的平方即可得出答案.

【详解】

∵四边形ABCD为平行四边形,

∴DC∥AB,

∴△DFE∽△BFA,

∵DE:EC=3:1,

∴DE:DC=3:4,

∴DE:AB=3:4,

∴S△DFE:S△BFA=9:16.

故选B.

3.如图,在x轴的上方,直角∠BOA绕原点O按顺时针方向旋转.若∠BOA的两边分别与

函数

1

y

x

=-、

2

y

x

=的图象交于B、A两点,则∠OAB大小的变化趋势为()

A.逐渐变小B.逐渐变大C.时大时小D.保持不变【答案】D

【解析】

【分析】

如图,作辅助线;首先证明△BEO∽△OFA,,得到BE OE

OF AF

=;设B为(a,

1

a

-),A为

(b,2

b

),得到OE=-a,EB=

1

a

-,OF=b,AF=2

b

,进而得到222

a b=,此为解决问题的关

键性结论;运用三角函数的定义证明知tan ∠OAB=2

2

为定值,即可解决问题. 【详解】

解:分别过B 和A 作BE ⊥x 轴于点E

,AF ⊥x 轴于点F , 则△BEO ∽△OFA , ∴

BE OE

OF AF

=, 设点B 为(a ,1a

-),A 为(b ,2

b ),

则OE=-a ,EB=1a

-

,OF=b ,AF=2

b ,

可代入比例式求得222a b =,即2

2

2

a b =

, 根据勾股定理可得:OB=22221OE EB a a +=+

,OA=222

2

4OF AF b b

+=+, ∴tan ∠OAB=22

2

2

222

2

1

224

4b a OB a b OA b b b

b

++==++=222214()24b b b b ++=22

∴∠OAB 大小是一个定值,因此∠OAB 的大小保持不变. 故选D

【点睛】

该题主要考查了反比例函数图象上点的坐标特征、相似三角形的判定等知识点及其应用问题;解题的方法是作辅助线,将分散的条件集中;解题的关键是灵活运用相似三角形的判定等知识点来分析、判断、推理或解答.

4.如图所示,在△ABC 中,∠C =90°,AB =8,CD 是AB 边上的中线,作CD 的中垂线与CD 交于点E ,与BC 交于点F .若CF =x ,tanA =y ,则x 与y 之间满足( )

A .

2

2

44x y += B .

2

2

44x y -= C .

2

2

88x y -= D .

2

2

88x y += 【答案】A 【解析】 【分析】

由直角三角形斜边上的中线性质得出CD =1

2

AB =AD =4,由等腰三角形的性质得出∠A =∠ACD ,得出tan ∠ACD =

GE CE

=tan A =y ,证明△CEG ∽△FEC ,得出GE CE

CE FE =,得出y =2FE ,求出y 2=24FE ,得出24y

=FE 2,再由勾股定理得出FE 2=CF 2﹣CE 2=x 2

﹣4,即可得出答案. 【详解】 解:如图所示:

∵在△ABC 中,∠C =90°,AB =8,CD 是AB 边上的中线,

∴CD =1

2

AB =AD =4, ∴∠A =∠ACD , ∵EF 垂直平分CD ,

∴CE =

1

2

CD =2,∠CEF =∠CEG =90°, ∴tan ∠ACD =

GE

CE

=tanA =y , ∵∠ACD+∠FCE =∠CFE+∠FCE =90°, ∴∠ACD =∠FCE , ∴△CEG ∽△FEC , ∴

GE CE =CE

FE , ∴y =

2FE , ∴y 2=

2

4FE

24y

=FE 2, ∵FE 2=CF 2﹣CE 2=x 2﹣4,

∴24

y =x 2﹣4, ∴24

y

+4=x 2, 故选:A .

【点睛】

本题考查了解直角三角形、直角三角形斜边上的中线性质、等腰三角形的性质、相似三角形的判定与性质等知识;熟练掌握直角三角形的性质,证明三角形相似是解题的关键.

5.如图,在平面直角坐标中,正方形ABCD 与正方形BEFG 是以原点O 为位似中心的位似图形,且相似比为1

3

,点A ,B ,E 在x 轴上.若正方形ABCD 的边长为2,则点F 坐标为( )

A .(8,6)

B .(9,6)

C .19,62?? ???

D .(10,6)

【答案】B 【解析】 【分析】

直接利用位似图形的性质结合相似比得出EF 的长,进而得出△OBC ∽△OEF ,进而得出EO 的长,即可得出答案. 【详解】

解:∵正方形ABCD 与正方形BEFG 是以原点O 为位似中心的位似图形,且相似比为13

, ∴

1

3

BC OB EF EO ==,

∵BC =2, ∴EF =BE =6, ∵BC ∥EF , ∴△OBC ∽△OEF , ∴136

BO BO =+, 解得:OB =3, ∴EO =9,

∴F 点坐标为:(9,6), 故选:B . 【点睛】

此题主要考查了位似变换以及相似三角形的判定与性质,正确得出OB 的长是解题关键.

6.如图,正方形ABCD 中,点E 在边BC 上,BE EC =,将DCE ?沿DE 对折至

DFE ?,延长EF 交边AB 于点G ,连接DG ,BF .给出以下结论:

①DAG DFG ???;②2BG AG =;③EBF DEG ??:;④2

3

BFC BEF S S ??=.其中所有正确结论的个数是( )

A .1

B .2

C .3

D .4

【答案】B 【解析】 【分析】

根据正方形的性质和折叠的性质可得AD =DF ,∠A =∠GFD =90°,于是根据“HL”判定Rt △ADG ≌Rt △FDG ,可判断①的正误;设正方形ABCD 的边长为a ,AG =FG =x ,BG =a?x ,根据勾股定理得到x =

1

3

a ,得到BG =2AG ,故②正确;根据已知条件得到△BEF 是等腰三角形,易知△GED 不是等腰三角形,于是得到△EBF 与△DEG 不相似,故③错误;连接CF ,根据三角形的面积公式得到S △BFC =2S △BEF .故④错误. 【详解】

解:如图,由折叠和正方形性质可知,DF =DC =DA ,∠DFE =∠C =90°, ∴∠DFG =∠A =90°,

在Rt △ADG 和Rt △FDG 中,

AD DF

DG DG

??

?==, ∴Rt △ADG ≌Rt △FDG (HL ),故①正确;

设正方形ABCD 的边长为a ,AG =FG =x ,BG =a?x , ∵BE =EC , ∴EF =CE =BE

12

a

∴GE=

12

a+x 由勾股定理得:EG 2=BE 2+BG 2,

即:(12a+x)2=(1

2a)2+(a-x)2解得:x =13 ∴BG =2AG ,

故②正确; ∵BE =EF ,

∴△BEF 是等腰三角形,易知△GED 不是等腰三角形, ∴△EBF 与△DEG 不相似, 故③错误; 连接CF , ∵BE =CE , ∴BE =

1

2

BC , ∴S △BFC =2S △BEF . 故④错误,

综上可知正确的结论的是2个. 故选:B . 【点睛】

本题考查了相似三角形的判定和性质、图形的折叠变换的性质和正方形的性质,全等三角形的判定与性质,勾股定理,三角形的面积计算,有一定的难度.

7.如图,将ABC ?沿BC 边上的中线AD 平移到A B C '''?的位置.已知ABC ?的面积为16,阴影部分三角形的面积9.若1AA '=,则A D '等于( )

A .2

B .3

C .4

D .

32

【答案】B 【解析】 【分析】

由 S △ABC =16、S △A ′EF =9且 AD 为 BC 边的中线知 1922

A DE A EF S S '?'?=

=,182ABD

ABC S S ??== ,根据△DA ′E ∽△DAB 知2

A DE ABD S A D AD S ??'??

=' ???

,据此求解可得. 【详解】

16ABC S ?=Q 、9A EF S ?'=,且AD 为BC 边的中线,

1922A DE A EF S S ??''∴=

=,1

82

ABD ABC S S ??==, Q 将ABC ?沿BC 边上的中线AD 平移得到A B C '''?, //A E AB ∴', DA E DAB '∴?~?,

则2

A DE ABD S A D AD S ??'??

=' ???,即22991816A D A D ??== '?+??

', 解得3A D '=或3

7

A D '=-(舍), 故选:

B . 【点睛】

本题主要平移的性质,解题的关键是熟练掌握平移变换的性质与三角形中线的 性质、相似三角形的判定与性质等知识点.

8.如图,O 是平行四边形ABCD 的对角线交点,E 为AB 中点,DE 交AC 于点F ,若平行四边形ABCD 的面积为8,则DOE ?的面积是( )

A .2

B .

32

C .1

D .

94

【答案】C 【解析】 【分析】

由平行四边形的面积,找到三角形底边和高与平行四边形底边和高的关系,利用面积公式以及线段间的关系求解.分别作△OED 和△AOD 的高,利用平行线的性质,得出高的关系,进而求解. 【详解】

解:如图,过A 、E 两点分别作AN ⊥BD 、EM ⊥BD ,垂足分别为M 、N ,则EM ∥AN ,

∴EM :AN =BE :AB , ∵E 为AB 中点, ∴BE=

1

2AB , ∴EM =

1

2

AN , ∵平行四边形ABCD 的面积为8,

∴2×

1

2

×AN×BD =8, ∴AN×BD =8

∴S △OED =

12×OD×EM =12×12BD×1

2AN =18

AN×BD =1. 故选:C . 【点睛】

本题考查平行四边形的性质,综合了平行线分线段成比例以及面积公式.已知一个三角形的面积求另一个三角形的面积有以下几种做法:①面积比是边长比的平方比;②分别找到底和高的比.

9.如图,AB 为O e 的直径,C 为O e 上一点,弦AD 平分BAC ,交弦BC 于点E ,

4

CD=,2

DE=,则AE的长为()

A.2 B.4 C.6 D.8

【答案】C

【解析】

【分析】

根据角平分线的定义得到∠CAD=∠BAD,根据圆周角定理得到∠DCB=∠BAD,证明△DCE∽△DAC,根据相似三角形的性质求出AD,结合图形计算,得到答案.

【详解】

解:∵AD平分∠BAC,

∴∠CAD=∠BAD,

由圆周角定理得,∠DCB=∠BAD,

∴∠CAD=∠DCB,又∠D=∠D,

∴△DCE∽△DAC,

∴DE DC

DC DA

=,即

24

4AD

=,

解得,AD=8,

∴AE=AD-DE=8-2=6,

故选:C.

【点睛】

本题考查的是相似三角形的判定和性质、圆周角定理,掌握相似三角形的判定定理和性质定理是解题的关键.

10.在平面直角坐标系中,把△ABC的各顶点的横坐标都除以1

4

,纵坐标都乘

1

3

,得到

△DEF,把△DEF与△ABC相比,下列说法中正确的是()

A.横向扩大为原来的4倍,纵向缩小为原来的1 3

B.横向缩小为原来的1

4

,纵向扩大为原来的3倍

C.△DEF的面积为△ABC面积的12倍

D.△DEF的面积为△ABC面积的

1 12

【答案】A 【解析】【分析】

解:△DEF 与△ABC 相比,横向扩大为原来的4倍,纵向缩小为原来的1

3

;△DEF 的面积为△ABC 面积的169

, 故选A.

11.如图,在四边形ABCD 中,,90,5,10AD BC ABC AB BC ∠=?==P ,连接

,AC BD ,以BD 为直径的圆交AC 于点E .若3DE =,则AD 的长为( )

A .55

B .45

C .35

D .25

【答案】D 【解析】 【分析】

先判断出△ABC 与△DBE 相似,求出BD ,最后用勾股定理即可得出结论. 【详解】 如图1,

在Rt △ABC 中,AB=5,BC=10, ∴AC=55, 连接BE ,

∵BD 是圆的直径, ∴∠BED=90°=∠CBA , ∵∠BAC=∠EDB , ∴△ABC ∽△DEB , ∴

AB AC DE DB

= , ∴5355

DB

, ∴DB=35

在Rt △ABD 中,2225BD AB -, 故选:D .

此题考查勾股定理,相似三角形的判定和性质,正确作出辅助线是解题的关键.

12.如图,以正方形ABCD 的AB 边为直径作半圆O ,过点C 作直线切半圆于点E ,交AD 边于点F ,则

FE

EC

=( )

A .

12

B .

13

C .

14

D .38

【答案】C 【解析】 【分析】

连接OE 、OF 、OC ,利用切线长定理和切线的性质求出∠OCF =∠FOE ,证明△EOF ∽△ECO ,利用相似三角形的性质即可解答. 【详解】

解:连接OE 、OF 、OC . ∵AD 、CF 、CB 都与⊙O 相切,

∴CE =CB ;OE ⊥CF ; FO 平分∠AFC ,CO 平分∠BCF . ∵AF ∥BC ,

∴∠AFC+∠BCF =180°, ∴∠OFC+∠OCF =90°, ∵∠OFC+∠FOE =90°, ∴∠OCF =∠FOE , ∴△EOF ∽△ECO , ∴

=OE EF

EC OE

,即OE 2=EF?EC . 设正方形边长为a ,则OE =1

2

a ,CE =a . ∴EF =14

a . ∴

EF EC =14. 故选:C .

【点睛】

本题考查切线的性质、切线长定理、相似三角形的判定与性质,其中通过作辅助线构造相似三角形是解答本题的关键..

13.如图,在ABC V 中,//,,30DE BC AF BC ADE ⊥∠=?,2,33,DE BC BF ==则DF 的长为()

A .4

B .23

C .3

3D .3

【答案】D 【解析】 【分析】

先利用相似三角形的相似比证明点D 是AB 的中点,再解直角三角形求得AB ,最后利用直角三角形斜边中线性质求出DF . 【详解】 解:∵//DE BC , ∴ADE ~ABC V V , ∵2DE BC =, ∴点D 是AB 的中点,

∵,30AF BC ADE ⊥∠=?,33BF = ∴∠B =30°,

∴AB 6cos30BF

=

=?

∴DF=3, 故选:D . 【点睛】

此题主要考查相似三角形的判定与性质、解直角三角形和直角三角形斜边中线性质,熟练掌握性质的运用是解题关键.

14.如图,菱形ABCD 中,点P 是CD 的中点,∠BCD=60°,射线AP 交BC 的延长线于点E ,射线BP 交DE 于点K ,点O 是线段BK 的中点,作BM ⊥AE 于点M ,作KN ⊥AE 于点N ,连结MO 、NO ,以下四个结论:①△OMN 是等腰三角形;②tan ∠

OMN=33

;③BP=4PK ;④PM?PA=3PD 2,其中正确的是( )

A .①②③

B .①②④

C .①③④

D .②③④

【答案】B 【解析】 【分析】

根据菱形的性质得到AD ∥BC ,根据平行线的性质得到对应角相等,根据全等三角形的判定定理△ADP ≌△ECP ,由相似三角形的性质得到AD=CE ,作PI ∥CE 交DE 于I ,根据点P 是CD 的中点证明CE=2PI ,BE=4PI ,根据相似三角形的性质得到

1

=4

KP PI KB BE =,得到BP=3PK ,故③错误;作OG ⊥AE 于G ,根据平行线等分线段定理得到MG=NG ,又OG ⊥MN ,证明△MON 是等腰三角形,故①正确;根据直角三角形的性质和锐角三角函数求出∠3

②正确;然后根据射影定理和三角函数即可得到PM?PA=3PD 2,故④正确. 【详解】

解:作PI ∥CE 交DE 于I , ∵四边形ABCD 为菱形, ∴AD ∥BC ,

∴∠DAP=∠CEP ,∠ADP=∠ECP , 在△ADP 和△ECP 中,

DAP CEP ADP ECP DP CP ∠=∠??

∠=∠??=?

, ∴△ADP ≌△ECP , ∴AD=CE , 则PI PD CE DC =,又点P 是CD 的中点, ∴

1=2

PI CE ,

∵AD=CE ,

1

=4KP PI KB BE , ∴BP=3PK , 故③错误;

作OG ⊥AE 于G ,

∵BM 丄AE 于M ,KN 丄AE 于N , ∴BM ∥OG ∥KN , ∵点O 是线段BK 的中点, ∴MG=NG ,又OG ⊥MN , ∴OM=ON ,

即△MON 是等腰三角形,故①正确;

由题意得,△BPC ,△AMB ,△ABP 为直角三角形,

设BC=2,则CP=1,由勾股定理得,

根据三角形面积公式,BM=7

, ∵点O 是线段BK 的中点, ∴PB=3PO ,

∴OG=13BM=21

, MG=

23

MP=27,

tan ∠OMN=

OG MG ,故②正确; ∵∠ABP=90°,BM ⊥AP , ∴PB 2=PM?PA , ∵∠BCD=60°, ∴∠ABC=120°, ∴∠PBC=30°, ∴∠BPC=90°,

∴, ∵PD=PC , ∴PB 2=3PD ,

∴PM ?PA=3PD 2,故④正确. 故选B .

【点睛】

本题考查相似形综合题.

15.已知线段MN=4cm,P是线段MN的黄金分割点,MP>NP,那么线段MP的长度等于()

A.(25+2)cm B.(25﹣2)cm C.(5+1)cm D.(5﹣1)cm 【答案】B

【解析】

【分析】

根据黄金分割的定义进行作答.

【详解】

由黄金分割的定义知,

51

2

MP

MN

-

=,又MN=4,所以,MP=25- 2. 所以答案选B.

【点睛】

本题考查了黄金分割的定义,熟练掌握黄金分割的定义是本题解题关键.

16.如图,在直角坐标系中,有两点A(6,3)、B(6,0).以原点O为位似中心,相似比为1

3

,在第一象限内把线段AB缩小后得到线段CD,则点C的坐标为()

A.(2,1) B.(2,0) C.(3,3) D.(3,1)

【答案】A

【解析】

【分析】

根据位似变换的性质可知,△ODC∽△OBA,相似比是1

3

,根据已知数据可以求出点C的坐

标.【详解】

由题意得,△ODC∽△OBA,相似比是1

3

OD DC

OB AB

=, 又OB =6,AB =3, ∴OD =2,CD =1,

∴点C 的坐标为:(2,1), 故选A . 【点睛】

本题考查的是位似变换,掌握位似变换与相似的关系是解题的关键,注意位似比与相似比的关系的应用.

17.如图,已知AB ∥CD ∥EF ,它们依次交直线l 1、l 2于点A 、D 、F 和点B 、C 、E ,如果AD :DF =3:1,BE =10,那么CE 等于( )

A .

103

B .

203

C .

52

D .

152

【答案】C 【解析】 【分析】

根据平行线分线段成比例定理得到3AD BC

DF CE

==,得到BC=3CE ,然后利用BC+CE=BE=10可计算出CE 的长,即可. 【详解】

解:∵AB ∥CD ∥EF , ∴

3AD BC

DF CE

==, ∴BC=3CE , ∵BC+CE=BE , ∴3CE+CE=10,

∴CE=

52. 故选C . 【点睛】

本题考查了平行线分线段成比例:三条平行线截两条直线,所得的对应线段成比例.

18.如图,已知一组平行线////a b c ,被直线m 、n 所截,交点分别为A 、B 、C 和

D 、

E 、

F ,且 1.5AB =,2BC =, 1.8DE =,则EF =( )

A .4.4

B .4

C .3.4

D .2.4

【答案】D 【解析】 【分析】

根据平行线等分线段定理列出比例式,然后代入求解即可. 【详解】 解:∵////a b c

AB DE BC EF = 即1.5 1.8

2EF = 解得:EF=2.4 故答案为D . 【点睛】

本题主要考查的是平行线分线段成比例定理,利用定理正确列出比例式是解答本题的关键.

19.若△ABC 的每条边长增加各自的50%得△A 'B 'C ',若△ABC 的面积为4,则△A 'B 'C '的面积是( ) A .9 B .6

C .5

D .2

【答案】A 【解析】 【分析】

根据两个三角形三边对应成比例,这两个三角形相似判断出两个三角形相似,根据相似三角形的性质即可得到结论. 【详解】

解:∵△ABC 的每条边长增加各自的50%得△A ′B ′C ′, ∴△ABC 与△A ′B ′C ′的三边对应成比例, ∴△ABC ∽△A ′B ′C ′, ∴

214

()150%9

ABC A B C S S '''==+V V ,

∵△ABC 的面积为4,则△A'B'C'的面积是9. 故选:A . 【点睛】

本题考查了相似三角形的性质和判定,熟练掌握相似三角形的判定是解题的关键.

20.如图,在正方形ABCD 中,3AB =,点M 在CD 的边上,且1DM =,AEM ?与

ADM ?关于AM 所在直线对称,将ADM ?按顺时针方向绕点A 旋转90°得到ABF ?,连接EF ,则cos EFC ∠的值是 ( )

A 17

1365B 6

1365

C 7

1525

D .

617

【答案】A 【解析】 【分析】

过点E 作//HG AD ,交AB 于H ,交CD 于G ,作EN BC ⊥于N ,首先证明

AEH EMG V :V ,则有

1

3

EH AE MG EM == ,设MG x =,则3EH x =,1DG AH x ==+, 在Rt AEH V 中利用勾股定理求出x 的值,进而可求

,,,EH BN CG EN 的长度,进而可求FN ,再利用勾股定理求出EF 的长度,最后利用

cos FN

EFC EF

∠=

即可求解.

【详解】

过点E 作//HG AD ,交AB 于H ,交CD 于G ,作EN BC ⊥于N ,则

90AHG MGE ∠=∠=?,

∵四边形ABCD 是正方形,

∴3,90AD AB ABC C D ==∠=∠=∠=? , ∴四边形AHGD,BHEN,ENCG 都是矩形.

由折叠可得,90,3,1AEM D AE AD DM EM ∠=∠=?====,

90AEH MEG EMG MEG ∴∠+∠=∠+∠=? , AEH EMG ∴∠=∠, AEH EMG ∴V :V ,

1

3

EH AE MG EM ∴

== . 设MG x =,则3EH x =,1DG AH x ==+ 在Rt AEH V 中,

222AH EH AE +=Q ,

222(1)(3)3x x ∴++= ,

解得4

5

x =

或1x =-(舍去), 125EH BN ∴==

,65

CG CD DG EN =-== . 1BF DM ==Q

17

5

FN BF BN ∴=+=

. 在Rt EFN △ 中, 由勾股定理得,2213EF

EN FN =+=,

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A.6 B.8 C.10 D.12 【答案】D 【解析】 分析:根据正方形的性质可得出AB∥CD,进而可得出△ABF∽△GDF,根据相似三角形的性 质可得出AF AB GF GD ==2,结合FG=2可求出AF、AG的长度,由CG∥AB、AB=2CG可得出 CG为△EAB的中位线,再利用三角形中位线的性质可求出AE的长度,此题得解.详解:∵四边形ABCD为正方形, ∴AB=CD,AB∥CD, ∴∠ABF=∠GDF,∠BAF=∠DGF, ∴△ABF∽△GDF, ∴AF AB GF GD ==2, ∴AF=2GF=4, ∴AG=6. ∵CG∥AB,AB=2CG, ∴CG为△EAB的中位线, ∴AE=2AG=12. 故选D. 点睛:本题考查了相似三角形的判定与性质、正方形的性质以及三角形的中位线,利用相似三角形的性质求出AF的长度是解题的关键. 3.如图,在平面直角坐标中,正方形ABCD与正方形BEFG是以原点O为位似中心的位似 图形,且相似比为1 3 ,点A,B,E在x轴上.若正方形ABCD的边长为2,则点F坐标为 () A.(8,6)B.(9,6)C. 1 9,6 2 ?? ? ?? D.(10,6) 【答案】B

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