题目内容
如图,一张三角形纸片沿DE对折,点B与点A重合,若AB=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022162438659164177/SYS201310221624386591641006_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022162438659164177/SYS201310221624386591641006_ST/images1.png)
【答案】分析:首先由折叠的性质,即可求得BD的长,然后由∠B=30°,在Rt△BDE中,利用∠B的正切,即可求得折痕DE的长.
解答:解:根据折叠的性质可得:BD=AD=
AB=
×2
=
,∠BDE=90°,
∵∠B=30°,
∴在Rt△BDE中,tan∠B=tan30°=
=
=
,
∴DE=1.
∴故答案为:1.
点评:此题考查了折叠的性质,直角三角形中的三角函数的知识.此题难度不大,解题的关键是掌握数形结合思想的应用,注意由折叠的性质,找到等量关系.
解答:解:根据折叠的性质可得:BD=AD=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022162438659164177/SYS201310221624386591641006_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022162438659164177/SYS201310221624386591641006_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022162438659164177/SYS201310221624386591641006_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022162438659164177/SYS201310221624386591641006_DA/3.png)
∵∠B=30°,
∴在Rt△BDE中,tan∠B=tan30°=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022162438659164177/SYS201310221624386591641006_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022162438659164177/SYS201310221624386591641006_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022162438659164177/SYS201310221624386591641006_DA/6.png)
∴DE=1.
∴故答案为:1.
点评:此题考查了折叠的性质,直角三角形中的三角函数的知识.此题难度不大,解题的关键是掌握数形结合思想的应用,注意由折叠的性质,找到等量关系.
![](http://thumb2018.1010pic.com/images/loading.gif)
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