We missed Pythagoras Day

[Claude]

Still, for those who need Pythagoras formula explained, here is one sample of a concise and clear definition, courtesy Don Firth in the Sydney Morning Herald of today:

Don Firth, of Wooli, offers an alternative formula: "A tribal chief in America had taken three squaws as wives. With the first he used a tiger hide as their bed and she had one son. With the second a bear hide had been their bed and she produced twin sons. The third squaw had triplets, all boys. The chief explained that they had slept on a hippopotamus hide because he had heard that the squaw on the hippopotamus hide would equal the sons of the squaws on the other two hides." That's a wigwam wag's whim.

[Rudi]

That is brilliant.

Saving it for reference right now!

[AlanMiller]

I never cease to be amazed at the simplicity of its proof in this simple construction:



Alan

[BobH]

That Native American story made me !

Thanks for sharing that visual proof of the theorem, Alan. I have never seen it before but appreciate its simplicity and beauty.

[John Gray]

OK, I'll bite.
How is this single diagram related to the theorem of Pythagoras (other than having the usual letters a, b and c)?
The area of the surrounding blue square is (a+b)² or a²+ 2ab + b²
The area of the internal yellow square is c².
How can the comparative areas of the yellow square and the blue square be assessed?

[BobH]

John, I think one must intuit the relative sizes of the outer square divisions. Of course, with a sheet of paper one could fold it into thirds for the proof that b is twice the length of a.

[John Gray]

Bob - Pythagoras' theorem is general, not only for the specific case where b = 2a !

[AlanMiller]

How is this single diagram related to the theorem of Pythagoras (other than having the usual letters a, b and c)?
The area of the surrounding blue square is (a+b)² or a²+ 2ab + b²
The area of the internal yellow square is c².
How can the comparative areas of the yellow square and the blue square be assessed?

You've almost got it. The area of the surrounding blue square must equal the total area of all the shapes comprising it. Don't give up now!

Alan

[John Gray]

The other thing you need to know is that the area of each triangle is a*b/2, of course!

[HansV]

No, you don't. Be creative!

[John Gray]

That seemed fairly creative to me!

[HansV]

It isn't necessary to "know" that the area of each triangle is a*b/2. The only thing you need is the area of a rectangle with sides a and b is a*b, which is more or less the definition of area. By rearranging the shapes, you can use this.

[AlanMiller]

It isn't necessary to "know" that the area of each triangle is a*b/2. The only thing you need is the area of a rectangle with sides a and b is a*b, which is more or less the definition of area. By rearranging the shapes, you can use this.

Something like this:



Alan