all the sq are divided into right angle triangle. if the sq are of side are 3,4,5 , it will comes equal to the area of sq root 50 after measurement of each triangle . central triangle. figure cannot be drawn on same sheet ihave roof area equal to the three sq.

TRANGLE IN WHICH SIN=4/5^2=16/25 AND COS=3/5^2=9/25 SO EQUAL TO 25/25=1 THEREFORE 25=25 AFTER WHICH LEFT DISOLVE INTO 4^2+3^2=25 .IF ASSUME SIDE OF MIDDLE TRIANGLE ARE 3,4.,HYPOTINEOUS=5. PLEASE COMMENT.

I am not sure, when I searched online I found this as the reason.
While the proof looks like many other proofs by dissection and rearrangement, this one is not complete in that the construction does not go through where one of the legs of the given triangle is essentially less than the other
This is the website:http://www.cut-the-knot.org/pythagoras/FalseProofs.shtml
refer to proof #3

The main concern is that the picture doesn't deal with all possible cases. In this picture, we see that the small square is broken up into 3 vertical strips, and we can see how to rearrange them. The central strip juts nicely into the brown and blue triangles, with the green triangles placed below.

However, there is no reason why there are only 3 vertical strips. As the larger base grows larger, the smaller square could take many many vertical strips, after which it is unclear how to glue everything together. What is the role of the brown and blue triangles, and where do they go?

This is akin to drawing a special picture for a geometry question, say by setting the triangle to be equilateral or isosceles, and finding the answer for the specific case. We know that the answer is only valid in that scenario, and not in the more general case.

Yeah, this is how I was taught in school actually (well, everyone already knew it, but the teacher thought it was a cool thing to help explain why it works).

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TopNewestCan it be generalized for all possible pairs of squares?

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my answer was 0.2 pleace comment differnce 0.217& 0.2

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all the sq are divided into right angle triangle. if the sq are of side are 3,4,5 , it will comes equal to the area of sq root 50 after measurement of each triangle . central triangle. figure cannot be drawn on same sheet ihave roof area equal to the three sq.

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solve the theorem in the middle of square 3/5^2+4/5^2=25 make sq of 25=9+16=25

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TRANGLE IN WHICH SIN=4/5^2=16/25 AND COS=3/5^2=9/25 SO EQUAL TO 25/25=1 THEREFORE 25=25 AFTER WHICH LEFT DISOLVE INTO 4^2+3^2=25 .IF ASSUME SIDE OF MIDDLE TRIANGLE ARE 3,4.,HYPOTINEOUS=5. PLEASE COMMENT.

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I am not sure, when I searched online I found this as the reason. While the proof looks like many other proofs by dissection and rearrangement, this one is not complete in that the construction does not go through where one of the legs of the given triangle is essentially less than the other This is the website:http://www.cut-the-knot.org/pythagoras/FalseProofs.shtml refer to proof #3

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The main concern is that the picture doesn't deal with all possible cases. In this picture, we see that the small square is broken up into 3 vertical strips, and we can see how to rearrange them. The central strip juts nicely into the brown and blue triangles, with the green triangles placed below.

However, there is no reason why there are only 3 vertical strips. As the larger base grows larger, the smaller square could take many many vertical strips, after which it is unclear how to glue everything together. What is the role of the brown and blue triangles, and where do they go?

This is akin to drawing a special picture for a geometry question, say by setting the triangle to be equilateral or isosceles, and finding the answer for the specific case. We know that the answer is only valid in that scenario, and not in the more general case.

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Thanks

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I give up. Can you explain why it is not a valid proof?

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Yeah, this is how I was taught in school actually (well, everyone already knew it, but the teacher thought it was a cool thing to help explain why it works).

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