题目内容
![](http://thumb.1010pic.com/pic3/upload/images/201303/22/6463931b.png)
10
10
.分析:过点E向BF作垂线,交BF于点C,求AB的长,根据勾股定理可得:AC2+BC2=AB2,据此解答即可.
解答:解:如图:过点E向BF作垂线,交BF于点C,
则:AB2=AC2+BC2
![](http://thumb.1010pic.com/pic3/upload/images/201303/75/e2a384fb.png)
则AB2=(7+1)2+(7-1)2,
AB=10;
答:AB长为10;
故答案为:10.
则:AB2=AC2+BC2
![](http://thumb.1010pic.com/pic3/upload/images/201303/75/e2a384fb.png)
则AB2=(7+1)2+(7-1)2,
AB=10;
答:AB长为10;
故答案为:10.
点评:此题属于中学习题,解答此题应根据勾股定理进行解答.
![](http://thumb2018.1010pic.com/images/loading.gif)
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