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【329563】2.5.2 矩形的判定2

时间:2025-02-02 18:23:59 作者: 字数:4937字

2.5.2 矩形的判定



1.矩形具有而一般平行四边形不具有的性质是( )

A.对角相等 B.对边相等 C.对角线相等 D.对角线互相垂直


2.下列叙述中能判定四边形是矩形的个数是( )

对角线互相平分的四边形;②对角线相等的四边形;③对角线相等的平行四边形;④对角线互相平分且相等的四边形.

A1 B2 C3 D4


3.下列命题中,正确的是( )

A.有一个角是直角的四边形是矩形  <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> B.三个角是直角的多边形是矩形

C.两条对角线互相垂直且相等的四边形是矩形 D.有三个角是直角的四边形是矩形


4.如图1所示,矩形ABCD中的两条对角线相交于点O,∠AOD=120°AB=4cm,则矩形的对角线的长为_____

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>  <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>



1 2

5.若四边形ABCD的对角线ACBD相等,且互相平分于点O,则四边形ABCD_____形,若∠AOB=60°,那么ABAC=______


6.如图2所示,已知矩形ABCD周长为24cm,对角线交于点OOE⊥DC于点EOF⊥AD于点FOF-OE=2cm,则AB=______BC=______


7 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> .如图所示,ABCD的四个内 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> 角的平分线分别相交于EFGH,试说明四边形EFGH是矩形.








8.如图所示,△ABC中,CECF分别平分∠ACB和它的邻补角∠ACDAE⊥CEEAF⊥CFF,直线EF分别交ABACMN两点,则四边形AECF是矩形吗?为什么?

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>






9 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> .(一题多解题)如图所示,△AB <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> C为等腰三角形,AB=ACCD⊥ <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> ABDPBC上的一点,过P点分别作PE⊥ABPF⊥CA,垂足分别为EF,则有PE+PF=CD,你能说明为什么吗?





1  <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> 0.如图所示,△ABC中,AB=ACADBC边上的高,AE是∠CAF的平分线且∠CAF是△ABC的一个外角,且DE∥BA,四边形ADCE是矩形吗?为什么?













11.如图所示是一个书架,你能用一根绳子检查一下书架的侧边是否和上下底垂直吗?为什么?


 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>





12.已知AC为矩形ABCD的对角线,则下图中∠1与∠2一定不相等的是( )

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>  <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>  <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>  <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>

1 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>  <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> 3.正方形通过剪切可以拼成三角形.方法如图1所示,仿照图1上用图示的方法,解答下面问题:如图2,对直角三角形,设计一种方案,将它分成若干块,再拼成一个与原三角形等面积的矩形.

1 2





1 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> 4.(展开与折叠题)已知如图所示,折叠矩形纸片ABCD,先折出折痕(对角线)BD,再过点 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> D折叠,使AD落在折痕BD上,得另一折痕DG,若AB=2BC=1,求AG的长度.







参考答案

1C 2B 3D 48cm 5.矩;12 68cm4cm

7.解:∠HAB+∠HBA=90°,所以∠H=90°.同理可求得∠HEF=∠F=∠FGH=90°

所以四边形EFGH是矩形.

8.解:四边形AECF是矩形.∠ECF= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> (∠ACB+∠ACD=90°.∠AEC=∠AFC=90°

点拨:本题是通过证四边形中三个角为直角得出结论.还可以通过证其为平行四边形,再证有一个角为直角得出结论.

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>


9.解法一:能.如图1所示,过P点作PH⊥DC,垂足为H

四边形PHDE是矩形.所以PE=DHPH∥BD.所以∠HPC=∠B <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>1

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> 因为AB=AC,所以∠B=∠ACB.所以∠HPC=∠FCP

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> 因为PC=CP,∠PHC=∠CFP=90°,所以△PHC≌△CFP.所以PF=HC

所以DH+HC=PE+PF,即DC=PE+PF <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>

2

解法二:能.延长EP,过C点作CH⊥EP,垂足为H,如图2所示,

四边形HEDC是矩形.所以EH=PE+PH=DCCH∥AB.所以∠HCP=∠B

PHC≌△PFC,所以PH=PF,所以PE+PF=DC

10.解:是矩形;理由:∠CAE=∠ACB,所以AE∥BC.又DE∥BA,所以四边形ABDE是平行四边形,所以AE= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> BD,所以AE=DC.又因为AE∥DC,所以四边形ADCE是平行四边形.又因为∠ADC=90°,所以四边形ADCE是矩形.

11.解:能;首先用绳子量一下书架的两组对边,再用绳子量一下书架的对角线,若对角线相等,则书架的侧边和上下底垂直,否则不垂直.12D

13.解:本题有多种拼法,下面提供几种供参考:

方法一:如图(1),方法二:如图(2

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>




14.解:如图所示,过点GGE⊥BD于点E, 则AG=EGAD=ED.在Rt△ABD中,由勾股定理,得BD= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> ,所以BE=BD-DE=BD-AD= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> -1BG=AB-AG=2-AG,设AG=EG=x,则BG=2-x.在Rt△BEG中,由勾股定理,得BG2=EG2+BE2,即(2-x2= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> -12+x2

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>x= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a> ,即AG= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/900/" title="矩形" class="c1" target="_blank">矩形</a>