题目内容
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193402650720798/SYS201311011934026507207015_ST/images0.png)
【答案】分析:先求出矩形的对角线AC,根据中位线定理可得出EF,继而可得出△AEF的周长.
解答:解:在Rt△ABC中,AC=
=10cm,
∵点E、F分别是AO、AD的中点,
∴EF是△AOD的中位线,EF=
OD=
BD=
AC=
,AF=
AD=
BC=4cm,AE=
AO=
AC=
,
∴△AEF的周长=AE+AF+EF=9cm.
故答案为:9.
点评:本题考查了三角形的中位线定理、勾股定理及矩形的性质,解答本题需要我们熟练掌握三角形中位线的判定与性质.
解答:解:在Rt△ABC中,AC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193402650720798/SYS201311011934026507207015_DA/0.png)
∵点E、F分别是AO、AD的中点,
∴EF是△AOD的中位线,EF=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193402650720798/SYS201311011934026507207015_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193402650720798/SYS201311011934026507207015_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193402650720798/SYS201311011934026507207015_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193402650720798/SYS201311011934026507207015_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193402650720798/SYS201311011934026507207015_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193402650720798/SYS201311011934026507207015_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193402650720798/SYS201311011934026507207015_DA/7.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193402650720798/SYS201311011934026507207015_DA/8.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193402650720798/SYS201311011934026507207015_DA/9.png)
∴△AEF的周长=AE+AF+EF=9cm.
故答案为:9.
点评:本题考查了三角形的中位线定理、勾股定理及矩形的性质,解答本题需要我们熟练掌握三角形中位线的判定与性质.
![](http://thumb2018.1010pic.com/images/loading.gif)
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