四边形易错题汇编附答案解析

四边形易错题汇编附答案解析
四边形易错题汇编附答案解析

四边形易错题汇编附答案解析

一、选择题

1.如图,ABCD Y 的对角线AC 与BD 相交于点O ,AD BD ⊥,30ABD ∠=?,若23AD =.则OC 的长为( )

A .3

B .3

C 21

D .6

【答案】C

【解析】

【分析】 先根据勾股定理解Rt ABD △求得6BD =,再根据平行四边形的性质求得3OD =,然后根据勾股定理解Rt AOD △、平行四边形的性质即可求得21OC OA ==

【详解】

解:∵AD BD ⊥

∴90ADB ∠=?

∵在Rt ABD △中,30ABD ∠=?,23AD =∴243AB AD ==

∴226BD AB AD =-=

∵四边形ABCD 是平行四边形

∴132

OB OD BD ===,12OA OC AC == ∴在Rt AOD △中,23AD =3OD = ∴2221OA AD OD +=

∴21OC OA ==

故选:C

【点睛】

本题考查了含30°角的直角三角形的性质、勾股定理、平行四边形的性质等知识点,熟练掌握相关知识点是解决问题的关键.

2.如图1,点F 从菱形ABCD 的项点A 出发,沿A -D -B 以1cm/s 的速度匀速运动到点B .图2是点F 运动时,△FBC 的面积y (m 2)随时间x (s)变化的关系图象,则a 的值为( )

A .5

B .2

C .52

D .25

【答案】C

【解析】

【分析】 过点D 作DE BC ⊥于点E 由图象可知,点F 由点A 到点D 用时为as ,FBC ?的面积为2acm .求出DE=2,再由图像得5BD =,进而求出BE=1,再在DEC Rt △根据勾股定理构造方程,即可求解.

【详解】

解:过点D 作DE BC ⊥于点E

由图象可知,点F 由点A 到点D 用时为as ,FBC ?的面积为2acm .

AD BC a ∴==

∴1

2

DE AD a =g 2DE ∴=

由图像得,当点F 从D 到B 时,用5s

5BD ∴=

Rt DBE V 中,

2222(5)21BE BD DE =-=-=

∵四边形ABCD 是菱形,

1EC a ∴=-,DC a =

DEC Rt △中,

2222(1)a a =+-

解得52

a =

故选:C .

【点睛】

本题综合考查了菱形性质和一次函数图象性质,要注意函数图象变化与动点位置之间的关系,解答此题关键根据图像关键点确定菱形的相关数据.

3.如图,点M 是正方形ABCD 边CD 上一点,连接AM ,作DE ⊥AM 于点E ,BF ⊥AM 于点F ,连接BE ,若AF =1,四边形ABED 的面积为6,则∠EBF 的余弦值是( )

A .21313

B .31313

C .23

D .1313

【答案】B

【解析】

【分析】

首先证明△ABF ≌△DEA 得到BF=AE ;设AE=x ,则BF=x ,DE=AF=1,利用四边形ABED 的面

积等于△ABE 的面积与△ADE 的面积之和得到

12

?x?x+?x×1=6,解方程求出x 得到AE=BF=3,则EF=x-1=2,然后利用勾股定理计算出BE ,最后利用余弦的定义求解.

【详解】

∵四边形ABCD 为正方形,

∴BA =AD ,∠BAD =90°,

∵DE ⊥AM 于点E ,BF ⊥AM 于点F ,

∴∠AFB =90°,∠DEA =90°,

∵∠ABF+∠BAF =90°,∠EAD+∠BAF =90°,

∴∠ABF =∠EAD ,

在△ABF 和△DEA 中 BFA DEA ABF EAD AB DA ∠=∠??∠=??=?

∴△ABF ≌△DEA (AAS ),

∴BF =AE ;

设AE =x ,则BF =x ,DE =AF =1,

∵四边形ABED 的面积为6, ∴

111622

x x x ??+??=,解得x 1=3,x 2=﹣4(舍去), ∴EF =x ﹣1=2, 在Rt △BEF 中,222313BE +

∴313cos 13BF EBF BE ∠=

==. 故选B .

【点睛】 本题考查了正方形的性质:正方形的四条边都相等,四个角都是直角;正方形具有四边形、平行四边形、矩形、菱形的一切性质.会运用全等三角形的知识解决线段相等的问题.也考查了解直角三角形.

4.如图,在菱形ABCD 中,60ABC ∠=?,1AB =,点P 是这个菱形内部或边上的一点,若以点P ,B ,C 为顶点的三角形是等腰三角形,则P ,D (P ,D 两点不重合)两点间的最短距离为( )

A .12

B .1

C 3

D 31

【答案】D

【解析】

【分析】

分三种情形讨论①若以边BC 为底.②若以边PC 为底.③若以边PB 为底.分别求出PD 的最小值,即可判断.

【详解】

解:在菱形ABCD 中, ∵∠ABC=60°,AB=1,

∴△ABC ,△ACD 都是等边三角形,

①若以边BC 为底,则BC 垂直平分线上(在菱形的边及其内部)的点满足题意,此时就转化为了“直线外一点与直线上所有点连线的线段中垂线段最短“,即当点P 与点A 重合时,PD 值最小,最小值为1;

②若以边PC 为底,∠PBC 为顶角时,以点B 为圆心,BC 长为半径作圆,与BD 相交于一点,则弧AC (除点C 外)上的所有点都满足△PBC 是等腰三角形,当点P 在BD 上时,PD 31

③若以边PB 为底,∠PCB 为顶角,以点C 为圆心,BC 为半径作圆,则弧BD 上的点A 与点D 均满足△PBC 为等腰三角形,当点P 与点D 重合时,PD 最小,显然不满足题意,故此种情况不存在;

上所述,PD 的最小值为31

故选D .

【点睛】

本题考查菱形的性质、等边三角形的性质、等腰三角形的判定和性质等知识,解题的关键是学会用分类讨论的思想思考问题,属于中考常考题型.

5.正九边形的内角和比外角和多( )

A .720?

B .900?

C .1080?

D .1260?

【答案】B

【解析】

【分析】

根据多边形的内角和公式求出正九边形的内角和,减去外角和360°即可.

【详解】

∵正九边形的内角和是(92)1801260-?=o o ,

∴1260360-=o o 900?,

故选:B.

【点睛】

此题考查多边形的内角和公式、外角和,熟记公式是解题的关键.

6.如图,在矩形ABCD 中,AB =5,AD =3,动点P 满足S △PAB =13

S 矩形ABCD ,则点P 到A 、B 两点距离之和PA +PB 的最小值为( )

A 29

B 34

C .2

D 41【答案】D 【解析】 解:设△ABP 中AB 边上的高是h .∵S △PAB =

13S 矩形ABCD ,∴12 AB ?h =13AB ?AD ,∴h =23

AD =2,∴动点P 在与AB 平行且与AB 的距离是2的直线l 上,如图,作A 关于直线l 的对称点E ,连接AE ,连接BE ,则BE 就是所求的最短距离.

在Rt △ABE 中,∵AB =5,AE =2+2=4,∴BE 22AB AE +2254+41PA +PB 的

41.故选D .

7.一个多边形的每一个外角都是72°,那么这个多边形的内角和为( ) A.540°B.720°C.900°D.1080°【答案】A

【解析】

【详解】

解:∵多边形的每一个外角都是72°,∴多边形的边数为:360

5 72

=,

∴该多边形的内角和为:(5-2)×180°=540°.故选A.

【点睛】

外角和是360°,除以一个外角度数即为多边形的边数.根据多边形的内角和公式可求得该多边形的内角和.

8.如图,正方形ABDC中,AB=6,E在CD上,DE=2,将△ADE沿AE折叠至△AFE,延长EF交BC于G,连AG、CF,下列结论:①△ABG≌△AFG;②BG=CG;③AG∥CF;

④S?FCG=3,其中正确的有().

A.1个B.2个C.3个D.4个

【答案】C

【解析】

【分析】

利用折叠性质和HL定理证明Rt△ABG≌Rt△AFG,从而判断①;设BG=FG=x,则CG=6-x,GE=x+2,根据勾股定理列方程求解,从而判断②;由②求得△FGC为等腰三角形,由此推

180

2

FGC

FCG

-∠

∠=

o

,由①可得

180

2

FGC

AGB

-∠

∠=

o

,从而判断③;过点F作

FM⊥CE,用平行线分线段成比例定理求得FM的长,然后求得△ECF和△EGC的面积,从而求出△FCG的面积,判断④.

【详解】

解:在正方形ABCD中,由折叠性质可知DE=EF=2,AF=AD=AB=BC=CD=6,∠B=∠D=∠AFG=

∠BCD=90°

又∵AG=AG

∴Rt △ABG ≌Rt △AFG ,故①正确;

由Rt △ABG ≌Rt △AFG

∴设BG=FG=x ,则CG=6-x ,GE=GF+EF=x+2,CE=CD-DE=4

∴在Rt △EGC 中,222

(6)4(2)x x -+=+

解得:x=3

∴BG =3,CG=6-3=3

∴BG =CG ,故②正确;

又BG =CG , ∴1802FGC FCG -∠∠=o 又∵Rt △ABG ≌Rt △AFG

∴1802FGC AGB -∠∠=o ∴∠FCG=∠AGB

∴AG ∥CF ,故③正确; 过点F 作FM ⊥CE ,

∴FM ∥CG

∴△EFM ∽△EGC

∴FM EF GC EG =即235

FM = 解得65FM =

∴S ?FCG =116344 3.6225

ECG ECF S S -=

??-??=V V ,故④错误 正确的共3个

故选:C .

【点睛】 本题考查正方形的性质,全等三角形的判定和性质,相似三角形的判定和性质,等腰三角形的判定和性质,综合性较强,掌握相关性质定理正确推理论证是解题关键.

9.下列命题错误的是( )

A .平行四边形的对角线互相平分

B .两直线平行,内错角相等

C .等腰三角形的两个底角相等

D .若两实数的平方相等,则这两个实数相等

【答案】D

【解析】

【分析】

根据平行四边形的性质、平行线的性质、等腰三角形的性质、乘方的定义,分别进行判断,即可得到答案.

【详解】

解:A 、平行四边形的对角线互相平分,正确;

B 、两直线平行,内错角相等,正确;

C 、等腰三角形的两个底角相等,正确;

D 、若两实数的平方相等,则这两个实数相等或互为相反数,故D 错误;

故选:D.

【点睛】

本题考查了判断命题的真假,以及平行四边形的性质、平行线的性质、等腰三角形的性质、乘方的定义,解题的关键是熟练掌握所学的性质进行解题.

10.如图,在矩形ABCD 中, 3,4,AB BC ==将其折叠使AB 落在对角线AC 上,得到折痕,AE 那么BE 的长度为( )

A .1

B .2

C .32

D .85

【答案】C

【解析】

【分析】 由勾股定理求出AC 的长度,由折叠的性质,AF=AB=3,则CF=2,设BE=EF=x ,则CE=4x -,利用勾股定理,即可求出x 的值,得到BE 的长度.

【详解】

解:在矩形ABCD 中,3,4AB BC ==,

∴∠B=90°, ∴22345AC =+=,

由折叠的性质,得AF=AB=3,BE=EF ,

∴CF=5-3=2,

在Rt △CEF 中,设BE=EF=x ,则CE=4x -,

由勾股定理,得:2222(4)x x +=-, 解得:32x =

; ∴32

BE =. 故选:C .

【点睛】

本题考查了矩形的折叠问题,矩形的性质,折叠的性质,以及勾股定理的应用,解题的关键是熟练掌握所学的性质,利用勾股定理正确求出BE 的长度.

11.如图,ABC V 中,5AB AC ==,AE 平分BAC ∠交BC 于点E ,点D 为AB 的中点,连接DE ,则DE 的长为( )

A .2

B .2.5

C .3

D 5【答案】B

【解析】

【分析】 根据等腰三角形三线合一可得AE ⊥BC ,再根据直角三角形斜边上的中线是斜边的一半即可求得DE 的长度.

【详解】

解:∵5AB AC ==,AE 平分BAC ∠,

∴AE ⊥BC ,

又∵点D 为AB 的中点,

∴1 2.52

DE AB =

=, 故选:B .

【点睛】

本题考查等腰三角形三线合一和直角三角形斜边上的中线.熟练掌握相关定理,并能正确识图,得出线段之间的关系是解题关键.

12.如图,点P是矩形ABCD的对角线AC上一点,过点P作EF∥BC,分别交AB,CD于E、F,连接PB、PD.若AE=2,PF=8.则图中阴影部分的面积为()

A.10 B.12 C.16 D.18

【答案】C

【解析】

【分析】

首先根据矩形的特点,可以得到S△ADC=S△ABC,S△AMP=S△AEP,S△PFC=S△PCN,最终得到S矩形EBNP= S 矩形MPFD

,即可得S△PEB=S△PFD,从而得到阴影的面积.

【详解】

作PM⊥AD于M,交BC于N.

则有四边形AEPM,四边形DFPM,四边形CFPN,四边形BEPN都是矩形,

∴S△ADC=S△ABC,S△AMP=S△AEP,S△PFC=S△PCN

∴S矩形EBNP= S矩形MPFD ,

又∵S△PBE= 1

2

S矩形EBNP,S△PFD=

1

2

S矩形MPFD,

∴S△DFP=S△PBE=1

2

×2×8=8,

∴S阴=8+8=16,

故选C.

【点睛】

本题考查矩形的性质、三角形的面积等知识,解题的关键是证明S△PEB=S△PFD.

13.如图11-3-1,在四边形ABCD中,∠A=∠B=∠C,点E在边AB上,∠AED=60°,则一定有()

A.∠ADE=20°B.∠ADE=30°C.∠ADE=1

2

∠ADC D.∠ADE=

1

3

∠ADC

【答案】D

【解析】

【分析】

【详解】

设∠ADE=x,∠ADC=y,由题意可得,

∠ADE+∠AED+∠A=180°,∠A+∠B+∠C+∠ADC=360°,即x+60+∠A=180①,3∠A+y=360②,

由①×3-②可得3x-y=0,

所以

1

3

x y

,即∠ADE=

1

3

∠ADC.

故答案选D.

考点:三角形的内角和定理;四边形内角和定理.

14.如图,四边形ABCD的对角线为AC、BD,且AC=BD,则下列条件能判定四边形ABCD 为矩形的是()

A.BA=BC

B.AC、BD互相平分

C.AC⊥BD

D.AB∥CD

【答案】B

【解析】

试题分析:根据矩形的判定方法解答.

解:能判定四边形ABCD是矩形的条件为AC、BD互相平分.

理由如下:∵AC、BD互相平分,

∴四边形ABCD是平行四边形,

∵AC=BD,

∴?ABCD是矩形.

其它三个条件再加上AC=BD均不能判定四边形ABCD是矩形.

故选B.

考点:矩形的判定.

15.如图,□ABCD的对角线AC、BD交于点O,AE平分BAD交BC于点E,且∠ADC=

60°,AB=1

2

BC,连接OE.下列结论:①AE=CE;②S△ABC=AB?AC;③S△ABE=2S△AOE;

④OE=1

4

BC,成立的个数有()

A.1个B.2个C.3个D.4

【答案】C

【解析】

【分析】

利用平行四边形的性质可得∠ABC=∠ADC=60°,∠BAD=120°,利用角平分线的性质证明

△ABE是等边三角形,然后推出AE=BE=1

2

BC,再结合等腰三角形的性质:等边对等角、三

线合一进行推理即可.

【详解】

∵四边形ABCD是平行四边形,

∴∠ABC=∠ADC=60°,∠BAD=120°,∵AE平分∠BAD,

∴∠BAE=∠EAD=60°

∴△ABE是等边三角形,

∴AE=AB=BE,∠AEB=60°,

∵AB=1

2 BC,

∴AE=BE=1

2 BC,

∴AE=CE,故①正确;∴∠EAC=∠ACE=30°

∴∠BAC=90°,

∴S△ABC=1

2

AB?AC,故②错误;

∵BE=EC,

∴E为BC中点,O为AC中点,∴S△ABE=S△ACE=2 S△AOE,故③正确;∵四边形ABCD是平行四边形,∴AC=CO,

∵AE=CE,

∴EO⊥AC,

∵∠ACE=30°,

∴EO=1

2 EC,

∵EC=1

2 AB,

∴OE=1

4

BC,故④正确;

故正确的个数为3个,

故选:C.

【点睛】

此题考查平行四边形的性质,等边三角形的判定与性质.注意证得△ABE是等边三角形是解题关键.

16.如图,在平行四边形ABCD中,∠BAD的平分线交BC于点E,∠ABC的平分线交AD 于点F,若BF=12,AB=10,则AE的长为()

A.13 B.14 C.15 D.16

【答案】D

【解析】

【分析】

先证明四边形ABEF是平行四边形,再证明邻边相等即可得出四边形ABEF是菱形,得出AE

⊥BF,OA=OE,OB=OF=1

2

BF=6,由勾股定理求出OA,即可得出AE的长.

如图所示:

∵四边形ABCD 是平行四边形,

∴AD ∥BC ,

∴∠DAE=∠AEB ,

∵∠BAD 的平分线交BC 于点E ,

∴∠DAE=∠BAE ,

∴∠BAE=∠BEA ,

∴AB=BE ,同理可得AB=AF ,

∴AF=BE ,

∴四边形ABEF 是平行四边形,

∵AB=AF ,

∴四边形ABEF 是菱形,

∴AE ⊥BF ,OA=OE ,OB=OF=12BF=6, ∴OA=2222=106AB OB --=8,

∴AE=2OA=16.

故选D .

【点睛】

本题考查平行四边形的性质与判定、等腰三角形的判定、菱形的判定和性质、勾股定理等知识;熟练掌握平行四边形的性质,证明四边形ABEF 是菱形是解决问题的关键.

17.如图,在菱形ABCD 中,点A 在x 轴上,点B 的坐标轴为()4,1, 点D 的坐标为()0,1, 则菱形ABCD 的周长等于( )

A 5

B .3

C .45

D .20

【答案】C

【分析】

如下图,先求得点A 的坐标,然后根据点A 、D 的坐标刻碟AD 的长,进而得出菱形ABCD 的周长.

【详解】

如下图,连接AC 、BD ,交于点E

∵四边形ABCD 是菱形,∴DB ⊥AC ,且DE=EB

又∵B ()4,1,D ()0,1

∴E(2,1)

∴A(2,0)

∴AD=()()2220015-+-=

∴菱形ABCD 的周长为:45

故选:C

【点睛】

本题在直角坐标系中考查菱形的性质,解题关键是利用菱形的性质得出点A 的坐标,从而求得菱形周长.

18.如图点P 是矩形ABCD 的对角线AC 上一点,过点P 作//EF BC ,分别交AB 、CD 于点E 、F ,连接PB 、PD ,若1AE =,8PF =,则图中阴影部分的面积为( )

A .5

B .6

C .8

D .9

【答案】C

【解析】

【分析】 由矩形的性质可证明S △PEB =S △PFD ,即可求解.

作PM⊥AD于M,交BC于N.

则有四边形AEPM,四边形DFPM,四边形CFPN,四边形BEPN都是矩形,S△ADC=S△ABC,S△AMP=S△AEP,S△PBE=S△PBN,S△PFD=S△PDM,S△PFC=S△PCN,

∴S△DFP=S△PBE=1

2

×1×8=4,

∴S阴=4+4=8,

故选:C.

【点睛】

此题考查矩形的性质、三角形的面积,解题的关键是证明S△PEB=S△PFD.

19.如图,在□ABCD中,延长CD到E,使DE=CD,连接BE交AD于点F,交AC于点G.下列结论中:①DE=DF;②AG=GF;③AF=DF;④BG=GC;⑤BF=EF,其中正确的有()

A.1个B.2个C.3个D.4个

【答案】B

【解析】

【分析】

由AAS证明△ABF≌△DEF,得出对应边相等AF=DF,BF=EF,即可得出结论,对于①②④不一定正确.

【详解】

解:∵四边形ABCD是平行四边形,

∴AB∥CD,AB=CD,即AB∥CE,

∴∠ABF=∠E,

∵DE=CD,

∴AB=DE,

在△ABF和△DEF中,

∵===ABF E AFB DFE AB DE ∠∠??∠∠???

, ∴△ABF ≌△DEF (AAS ),

∴AF=DF ,BF=EF ;

可得③⑤正确,

故选:B .

【点睛】

此题考查平行四边形的性质、全等三角形的判定与性质、平行线的性质;熟练掌握平行四边形的性质,证明三角形全等是解题的关键.

20.如图,菱形ABCD 中,对角线BD 与AC 交于点O , BD =8cm ,AC =6cm ,过点O 作OH ⊥CB 于点H ,则OH 的长为( )

A .5cm

B .

52cm C .125cm D .245cm 【答案】C

【解析】

【分析】

根据菱形的对角线互相垂直平分求出OB 、OC ,再利用勾股定理列式求出BC ,然后根据△BOC 的面积列式计算即可得解.

【详解】

解:∵四边形ABCD 是菱形,

∴AC ⊥BD ,111163,842222

OC AC OB BD ==?===?= 在Rt △BOC 中,由勾股定理得,2222345BC OB OC ++=

∵OH ⊥BC ,

1122

BOC S OC OB CB OH ∴=?=?V ∴1143522

OH ??=?

12

5 OH

故选C.

【点睛】

本题考查了菱形的性质,勾股定理,三角形的面积,熟记性质是解题的关键,难点在于利用两种方法表示△BOC的面积列出方程.

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A swimming B to swim C swam D swim 8 Is this factory ______you visited last week? A that B where C the one D in which 9 The reason ____he was absent from the meeting was ____his car broke down on the way. A that; because B why; that C that; that D for; that 10 Is the river_____ through that town very large? A which flows B flows C that flowing D whose flows 11 The teacher told me that the students I wanted to see were seen___ football on the playground just now. A playing B to be playing C play D to play 12 The red rose is the only one _____I real like. A which B who C that D whom 13 All the apples _____ fell down were eaten by the pigs. A those B which C what D that 14 Don’t forget th e day ______you were received into the Youth League.

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2015-2016学年高三语法填空易错题汇编 (一)1.You may find_____ difficult to get to sleep if you have a problem or something else on your mind .This is _____ you need to relax. 2.You can also get up and read , but be sure to choose ____ book that is not too difficult._______, you may get so ________(interest) that you won’t go to sleep when you feel _________(sleep). 3.My 4-year –old son , Shane,__________(ask) for a puppy for _______(month), But his Daddy _________always say:‘‘No dogs! It will kill our rabbits.No dogs and that’s final.” 4.So I handed him a potato and _________(say) :‘‘Carry this ____it turned into a puppy. Keep it with you ______happens.” 5.The Third Ring Road runs through the center of the university campus,_________ (make) transportation very____________(convenience). 6.We both agree that life has________(base) been good to us. And we are very _______________(appreciate) of ______ we have been blessed with. 7.However, I do realize that sometimes life can get ____the way of our goals. 8.Almost all international conferences and _______(compete) ________(conduct) in English. 9.Contact people from all ______the world .Talk about your ideas and opinions with others in __________(discuss) groups on the Internet. Send _____ e-mail to __________(interest) people or learn about their life and culture. 10. Communicate _____people ______you go----English is spoken in more ______100 countries. https://www.360docs.net/doc/581605045.html,e your computer more ________(effect).Most computer __________(apply) are _____ English, so you will understand _____better if you know English. 12.For weeks we hadn’t had enough to eat and my pockets were full of food that I__________(take)from the table to carry home for my wife and______(hunger) children. 13.To them, among such things____ health, money, intelligence,_______(honest) and reputation, honesty is the only thing ______can be thrown away. 14.Inside the bag was a note, on which __________(write) these words: ‘‘This money is ______the thoughtful person who takes this stone away _____the road. Thank you.” 15.All of them complained _____the stone in the center of the road, but _____of them tried to move it away. 16.The _________(significant) of this is ______ we can defect if a person has depressive symptoms and severity of those symptoms _______asking them any questions. 17.Not really. My parents are fairly well_____, so I get money from my mother. 18.I share a flat with two _____boys. It’s not large but fairly tidy and the _______important point is _____the rent is quite low.

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中学英语易错集锦大全211道题(精华珍藏版) 1.Because he was ill yesterday, so he didn’t go to work. (×) Because he was ill yesterday, he didn’t go to work. (√) He was ill yesterday, so he didn’t go to work. (√) [析] 用though, but表示“虽然……,但是……”或用because, so 表示“因为……,所以……”时,though和but 及because和so 都只能择一而用,不能两者同时使用。 2.The Smiths have moved Beijing. (×) The Smiths have moved to Beijing. (√) [析] 不及物动词后接名词或代词作宾语时,要在动词之后加上适当的介词;但不及物动词后接home, here, there等副词作宾语时,动词之后不必加任何介词。 3.The box is too heavy for him to carry it. (×) The box is too heavy for him to carry. (√) [析] the box既是这句话的主语, 也是不定式to carry的逻辑宾语,若句末再加上it,就和the box重复了。 4.Each of the boys have a pen. (×) Each of the boys has a pen. (√) [析] 复数名词前有表个体的each of, one of, every,either of等词组修饰,或有表否定的neither of, none of 等词组修饰时,谓语动词要用单数形式。 5.例:那是你心软!我不就是一个例子吗? Neither he nor you is good at English. (×) Neither he nor you are good at English. (√) [析] either... or..., neither... nor..., not only..., but also... 等词组连接句子的两个主语时,

高三语法填空易错题汇编

高三语法填空易错题汇编 1. You may find_____ difficult to get to sleep if you have a problem or something else on your mind . 2. You may get so ____________〔interest〕 that you won’t go to sleep when you feel _________〔sleep〕. 3. I also had the chance of ______________〔stick〕 in a Beijing thunderstorm,bargaining in the market, and riding the high-speed train. 4. So I handed him a potato and _________〔say〕 :“Carry this ________it turned into a puppy. Keep it with you _______________happens.” 5. The Third Ring Road runs through the center of the university campus,_________ 〔make〕transportation very____________〔convenience〕. 6. We both agree that life has_____________〔base〕 been good to us. And we are very______________〔appreciate〕 of ___________ we have been blessed with. 7. However, I do realize that sometimes life can get ____the way of our goals. 〔conduct〕 in English. 9. Do you know the man ____________ 〔lie〕 under the apple tree? 10. The ___________〔conduct〕 was always on the watch for such doings. 〔apply〕 are _____ English, so you will understand __________better if you know English. 12.For weeks we hadn’t had enough to eat and my pockets were full of food that I_________ 〔take〕from the table to carry home for my wife and____________〔hunger〕 children. 13.To them, among such things____ health, money, intelligence,_______〔honest〕 and reputation, honesty is the only thing ______can be thrown away. 14.Inside the bag was a note, on which ________________〔write〕 these words: ‘‘The money is ______the thoughtful person who takes this stone away _____the road. Thank you.” 16. In the afternoon I would have culture classes _________ I would learn about Chinese calligraphy, Beijing opera,Chinese martial arts. 17.Not really. My parents are fairly well___________, so I get money from my mother. 18.Their reason is _________ photos of their beautiful home with the great mountains in the background could make people _____________〔see〕 them on social media jealous and sad. 19.Professor Paul cliff spent ten years ______〔look〕 at the personalities of rich people and _______ 〔find〕 that their behavior was different ______the behavior of ___________ 〔poverty〕people. 20.The poor give a_____________〔high〕 percentage of __________〔them〕 money _____________〔help〕 others than _____rich. 21.Miss Saigon is a popular musical that is set ______1975 during the Vietnam War and _________〔produce〕 by the _________〔create〕 of Les Miserables in 1989. 22.Despite the strong desire for parents to say yes to __________〔they〕 children’s wishes, research shows there’s a harmful side to_________〔seek〕 the newest thing others have. 23.Among the painters ________〔break〕 away from the_____________〔tradition style of

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