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2 ways to derive Pythagoras’ equation from scratch

The other day I discovered one more way to derive Pythagoras’ equation from scratch, completely by accident. I was deriving Pythagoras’ equation using the usual method, whilst navigating  a diagram similar to the one below, but without (B-A) measurements…

pythagoras' diagram

*Note (regarding diagram above): x+y = 90 degrees

The usual method goes like this…

The area of the largest square is:

{ \left( A+B \right)  }^{ 2 }

It is also:

4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }

Which means that:

{ \left( A+B \right)  }^{ 2 }=4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }\\ \\ { A }^{ 2 }+2AB+{ B }^{ 2 }=2AB+{ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }

Now, when I added the lengths (B-A) to my diagram, which are included in the diagram above, I discovered a new way to derive Pythagoras’ equation…

I did this by focusing on the area C^2. It turns out that:

4\cdot \frac { 1 }{ 2 } AB+{ \left( B-A \right)  }^{ 2 }={ C }^{ 2 }

And since:

{ \left( B-A \right)  }^{ 2 }\\ \\ ={ \left( A+B \right)  }^{ 2 }-4AB\\ \\ ={ A }^{ 2 }+2AB+{ B }^{ 2 }-4AB\\ \\ ={ B }^{ 2 }-2AB+{ A }^{ 2 }

I was able to say that:

4\cdot \frac { 1 }{ 2 } AB+\left\{ { B }^{ 2 }-2AB+{ A }^{ 2 } \right\} ={ C }^{ 2 }\\ \\ 2AB+{ B }^{ 2 }-2AB+{ A }^{ 2 }={ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }

Obviously, I was quite pleased. Have you discovered other ways in which to derive Pythagoras’ equation??


Related:

Video on how to come up with Pythagoras’s equation…

How To Come Up With Pythagoras’s Equation