题目内容
如图,已知矩形ABCD,E为AD上一点,F为CD上一点,若将矩形沿BE折叠,点A恰与点F重合,且△DEF为等腰三角形,DE=1,求矩形ABCD的面积
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20120718/201207182202225344744.png)
解:∵四边形ABCD是矩形,
∴∠A=∠D=∠C=90°,AD=BC,
根据折叠的性质可得:AE=FE,∠EFB=∠A=90°,
∵△DEF为等腰三角形,
∴∠DEF=∠DFE=45°,
∴DE=1,DF=1,EF=
,
∴AE=EF=
,
∴AD=AE+DE=
+1,
∴BC=
+1,
∵∠EFD+∠BFC=90°,
∴∠BFC=45°,
∴∠FBC=45°,∠BFC=∠FBC,
∴FC=BC=
+1,
∴CD=DF+FC=1+
+1=
+2,
∴矩形ABCD的面积为:SCDAB=(
+2)(
+1)=4+3
。
∴∠A=∠D=∠C=90°,AD=BC,
根据折叠的性质可得:AE=FE,∠EFB=∠A=90°,
∵△DEF为等腰三角形,
∴∠DEF=∠DFE=45°,
∴DE=1,DF=1,EF=
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20120718/20120718220222577709.png)
∴AE=EF=
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20120718/20120718220222619709.png)
∴AD=AE+DE=
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20120718/20120718220222662694.png)
∴BC=
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20120718/20120718220222714694.png)
∵∠EFD+∠BFC=90°,
∴∠BFC=45°,
∴∠FBC=45°,∠BFC=∠FBC,
∴FC=BC=
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20120718/20120718220222776694.png)
∴CD=DF+FC=1+
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20120718/20120718220222819694.png)
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20120718/20120718220222882694.png)
∴矩形ABCD的面积为:SCDAB=(
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20120718/20120718220222941694.png)
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20120718/20120718220222987694.png)
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20120718/20120718220223068694.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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