My Math Forum Square in a right triangle.

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 April 11th, 2019, 03:17 AM #11 Senior Member     Joined: Feb 2010 Posts: 711 Thanks: 147 My response was incorrect. I now agree with Skipjack. I forgot to add the side of the square when calculating the hypotenuse.
 April 11th, 2019, 05:51 AM #12 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Damn! I erroneously used the longer leg instead of the hypotenuse when doing one of my "similar triangle a-go-go!". Agree with Skip. Turns out to be quite an easy problem.
 April 11th, 2019, 11:01 AM #13 Senior Member   Joined: Dec 2015 From: somewhere Posts: 636 Thanks: 91 Piece of cake.
 April 11th, 2019, 11:45 AM #14 Senior Member     Joined: Jan 2012 Posts: 140 Thanks: 2 Thanks. Got the result from the concept of similarity of triangles.
April 11th, 2019, 01:16 PM   #15
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Quote:
 Originally Posted by happy21 Thanks. Got the result from the concept of similarity of triangles.
This works as a formula:
length of sides of square = abc / (a^2 +b^2 + a*b)
(a = shorter leg, b = other leg, c = hypotenuse)

Only primitive case that has integer results (up to a = 999):
a = 111, b = 148, c = 185 : square sides = 60

April 11th, 2019, 11:27 PM   #16
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Quote:
 Originally Posted by Denis This works as a formula: length of sides of square = abc / (a^2 +b^2 + a*b) (a = shorter leg, b = other leg, c = hypotenuse) Only primitive case that has integer results (up to a = 999): a = 111, b = 148, c = 185 : square sides = 60
The above formula was derived from that concept of similarity only...right?

April 12th, 2019, 04:58 AM   #17
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Quote:
 Originally Posted by happy21 The above formula was derived from that concept of similarity only...right?
Of course...

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