c²… c² run

a² + b² = c², and that is a fact! At least I think it is. That’s what I learned in grade 8 anyway, so it’s probably true, especially since my grade 8 teacher taught the theorem to our class via song and interpretive dance. People rarely tell lies through interpretive dance, so I had no reason way back in grade 8 to doubt the validity and universality of the Pythagorean theorem.

In my old age, however, I am crusty and skeptical and disinclined to take Pythagoras’ (and Mr. Baumgartner’s) word as established truth. I recently decided to embark on a quest to figure out whether there is any evidence in support of the Pythagorean theorem. What follows, in brief, is what I found…

Is there any evidence to support the Pythagorean theorem?

Yes. Boatloads. There are hundreds of published mathematical proofs of the Pythagorean theorem. Check out this excellent page for 92 of them.

What exactly does the Pythagorean theorem state?

The theorem states that for any right triangle, the squared length of the hypotenuse is equal to the sum of the squared lengths of the other two sides, commonly written as: a² + b² = c².

This equality can be thought of in terms of areas of squares build on each side of a right triangle, as illustrated in the figure below. Because the area of a square is simply the squared length of any of its sides, the Pythagorean theorem essentially states that the area of the square on the hypotenuse (yellow square) is equal to the sum of the areas of the squares on the other two sides (blue and pink squares).

Prove it...

Here is one proof that involves simple geometry and just a splash of algebra:

Start with 4 identical copies of a right triangle.

Rotate the triangles to 0°,  90°, 180°, and 270°, respectively.

Form these triangles into a square such that each side of the large square is the hypotenuse (C) of the original triangles. Note that the area of this large square will be C².

The large square has a square hole inside of it. The sides of this small square are equal to A - B.

The area of the large square (C²) must equal the summed areas of the constituent shapes – the 4 triangles and the small inner square.

The area of each triangle (½base • height) is ½AB, and the area of the small inner square is (A-B)².

So we have:

C²  =  4(½AB) + (A - B)²

C²  =  2AB + (A² - 2AB + B²)

C²  =  A² + B²

Hurray!

Was Scarecrow correct?

No. In The Wizard of Oz (1939), Scarecrow demonstrates his ‘knowledge’ by claiming “The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side. Oh joy! Rapture! I’ve got a brain!”

Not even close, Scarecrow. You can watch this epic math fail here via YouTube.