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共回答了17个问题采纳率:94.1% 举报
(Ⅰ)如图(1),折叠后点B与点A重合,连接AC,
则△ACD≌△BCD,
设点C的坐标为(0,m)(m>0),
则BC=OB-OC=4-m,
于是AC=BC=4-m,
在Rt△AOC中,由勾股定理,得AC
2 =OC
2 +OA
2 ,
即(4-m)
2 =m
2 +2
2 ,解得m=
![](https://img.yulucn.com/upload/8/88/888894702f06316a4b344e50106b2839_thumb.jpg)
,
∴点C的坐标为
![](https://img.yulucn.com/upload/c/41/c415aa29a57f4d96066c539924c785d7_thumb.jpg)
;
(Ⅱ)如图(2),折叠后点B落在OA边上的点为B′连接B′C,B′D,
则△B′CD≌△BCD,
由题设OB′=x,OC=y,
则B′C=BC=OB-OC=4-y,
在Rt△B′OC中,由勾股定理,
得B′C
2 =OC
2 +OB′
2 ,
∴(4-y)
2 =y
2 +x
2 ,
即
![](https://img.yulucn.com/upload/2/ff/2ff3810f6eec9880a204847ba9280261_thumb.jpg)
,
由点B′在边OA上,有0≤x≤2,
∴解析式
![](https://img.yulucn.com/upload/2/12/2128c71c8eeda88b8b0710535ecf656e_thumb.jpg)
(0≤x≤2)为所求,
∵当0≤x≤2时,y随x的增大而减小,
∴y的取值范围为
![](https://img.yulucn.com/upload/9/d9/9d9be0fffcb6d79b9ac9ad588ec13324_thumb.jpg)
;
(Ⅲ)如图(3),折叠后点B落在OA边上的点为B′,连接B′C,B′D,B′D∥OB,
则∠OCB′=∠CB′D,
又∵∠CBD=∠CB′D,
∴∠CB′=∠CBD,
∴CB′∥BA,
∴Rt△COB′∽Rt△BOA,
有
![](https://img.yulucn.com/upload/b/8b/b8bb5bcd6237b80f70ca86e7d5f2b5ea_thumb.jpg)
,
得OC=20B′,
在Rt△B′OC中,设OB′=x
0 (x
0 >0),则OC=2x
0 ,
由(Ⅱ)的结论,得2x
0 =
![](https://img.yulucn.com/upload/d/27/d2715f4d1538a77de64dca17cd8f46a7_thumb.jpg)
,
解得x
0 =
![](https://img.yulucn.com/upload/1/c0/1c047468612dbe8cdb9b3547f6916225_thumb.jpg)
,
∵x
0 >0,
∴x
0 =
![](https://img.yulucn.com/upload/1/c0/1c047468612dbe8cdb9b3547f6916225_thumb.jpg)
,
∴点C的坐标为
![](https://img.yulucn.com/upload/d/68/d6831f500f109f0b5805884f2ba8e214_thumb.jpg)
。
1年前
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