I was having a discussion on how to double a square which has an area of 1, as discussed in Platos mino dialogue. The method that is proven to work is to bisect the square into 2 traingles, and use the hypatnuse as the side of the new square which has exactly an area of 2. I wanted to see if this solution is also correct. Can anyone tell me the length of the Hypatenuse in this picture?
http://www.chamberband.com/ali/images/square.jpg
The length/width/area of the origional black square is 1. Thus each side is 1. Thus the bisected triangle at the north of the square have a base length of .5 (lets say .5 meters, or whatever metric value you want to give it). So...This is a 45 degree triangle bisected into 2 90 degree triangles. Can anyone give me the length of the hypatnuse of the triangles? I'm tryign to see if this method also doubles the square or not. Thank you in advance.
http://www.chamberband.com/ali/images/square.jpg
The length/width/area of the origional black square is 1. Thus each side is 1. Thus the bisected triangle at the north of the square have a base length of .5 (lets say .5 meters, or whatever metric value you want to give it). So...This is a 45 degree triangle bisected into 2 90 degree triangles. Can anyone give me the length of the hypatnuse of the triangles? I'm tryign to see if this method also doubles the square or not. Thank you in advance.
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