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(1)甲种剪法所得的正方形面积更大,理由见解析(2)
![](https://img.yulucn.com/upload/d/0c/d0c957063fbd4812a45426466961e010_thumb.jpg)
,
![](https://img.yulucn.com/upload/1/43/143f878ece7f0f57acae4d69ad22ceb1_thumb.jpg)
(3)
(1)解法1:如图甲,由题意,得AE=DE=EC,即EC=1,S
正方形CFDE =1
2 =1
如图乙,设MN=x,则由题意,得AM=MQ=PN=NB=MN=x,
∴
![](https://img.yulucn.com/upload/6/bf/6bf3ac97465beee95a826aad4512a5f2_thumb.jpg)
,
解得
∴
又∵
∴甲种剪法所得的正方形面积更大.
说明:图甲可另解为:由题意得点D、E、F分别为AB、AC、BC的中点,S
正方形OFDE =1.
解法2:如图甲,由题意得AE=DE=EC,即EC=1,
如图乙,设MN=x,则由题意得AM=MQ=QP=PN=NB=MN=x,
则
![](https://img.yulucn.com/upload/3/4e/34e03390d009cb0887e163cee1a4c16d_thumb.jpg)
,
解得
![](https://img.yulucn.com/upload/d/1c/d1ca9e7c73052a7b29b248c5a118608b_thumb.jpg)
,
又∵
![](https://img.yulucn.com/upload/f/b4/fb4dee24b589c33d1592b022072f599b_thumb.jpg)
,即EC>MN.
∴甲种剪法所得的正方形面积更大.
(2)
![](https://img.yulucn.com/upload/d/0c/d0c957063fbd4812a45426466961e010_thumb.jpg)
,
![](https://img.yulucn.com/upload/1/43/143f878ece7f0f57acae4d69ad22ceb1_thumb.jpg)
.
(3)解法1:探索规律可知:
剩余三角形面积和为2﹣(S
1 +S
2 +…+S
10 )=2﹣(1+
![](https://img.yulucn.com/upload/9/a7/9a77375254e6fa6dd9c159086b411139_thumb.jpg)
+…+
![](https://img.yulucn.com/upload/6/99/699909101a8d2e5b47b6ce5731da605d_thumb.jpg)
)=
解法2:由题意可知,
第一次剪取后剩余三角形面积和为2﹣S
1 =1=S
1 第二次剪取后剩余三角形面积和为
![](https://img.yulucn.com/upload/c/41/c416770199566231ae4de79c6ac24016_thumb.jpg)
,
第三次剪取后剩余三角形面积和为
![](https://img.yulucn.com/upload/6/7b/67ba55a39755a9f27e5601d12f4ebe69_thumb.jpg)
,
…
第十次剪取后剩余三角形面积和为
![](https://img.yulucn.com/upload/a/99/a99319f7af1998bf7754a2aaecda106c_thumb.jpg)
.
(1)分别求出甲、乙两种剪法所得的正方形面积,进行比较即可;
(2)按图1中甲种剪法,可知后一个三角形的面积是前一个三角形的面积的
![](https://img.yulucn.com/upload/c/ca/ccab4d0edfc429683f3c57a5cd260612_thumb.jpg)
,依此可知结果;
(3)探索规律可知:
![](https://img.yulucn.com/upload/9/98/9988c54829ff4812273231eaf9aa1be8_thumb.jpg)
,依此规律可得第10次剪取后,余下的所有小三角形的面积之和.
1年前
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