玄丽彬纷
幼苗
共回答了16个问题采纳率:100% 举报
分析:所求正方形的边长即为AB的长,在等腰Rt△ACF、△CDE中,已知了CE、DE、CF的长均为10,根据等腰直角三角形的性质,即可求得AC、CD的长,由AB=AC+CD+BD即可得解.
如图;连接AB,则AB必过C、D
Rt△ACF中,AC=AF,CF=10;
则AC="5"
![](https://img.yulucn.com/upload/d/0d/d0d109f8fba4c939aab24c61e6be5202_thumb.jpg)
,AF="5"
![](https://img.yulucn.com/upload/d/0d/d0d109f8fba4c939aab24c61e6be5202_thumb.jpg)
;
同理可得BD="5"
![](https://img.yulucn.com/upload/d/0d/d0d109f8fba4c939aab24c61e6be5202_thumb.jpg)
;
Rt△CDE中,DE=CE=10,则CD="10"
![](https://img.yulucn.com/upload/d/0d/d0d109f8fba4c939aab24c61e6be5202_thumb.jpg)
;
所以AB="AC+CD+BD=20"
![](https://img.yulucn.com/upload/d/0d/d0d109f8fba4c939aab24c61e6be5202_thumb.jpg)
.
故答案为:20
![](https://img.yulucn.com/upload/d/0d/d0d109f8fba4c939aab24c61e6be5202_thumb.jpg)
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![](https://img.yulucn.com/upload/9/c5/9c500ebdf3a1b641e6508b179bee27bc_thumb.jpg)
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1年前
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