Pythagoras was a Greek philosopher and mathematician who lived in the sixth century B.C.E. He acquired a following in his time, and there was a cult built up around him in which he was worshiped as a demi god1. Today, he is most well known for his contribution to Euclidian geometry: the Pythagorean Theorem. Today's challenge is to prove the Pythagorean Theorem.
Problem:
Given any right triangle ABC, prove A2 + B2 = C2 using only simple geometric formulas. (No trigonometric functions allowed).
Solution:
This problem seeped into my head as I was trying to fall asleep some nights ago, and I never could solve it lying in bed in the dark. It wasn't until I sat down with a pencil and paper and studied the problem from several different angles (during the second and third quarters of the Superbowl) that I came up with a solution.
The first thing I did was to redraw the diagram like this:
That way, I could visualize the quantities, A2, B2 and C2 as the areas of the squares made from the sides of the triangles.
Next, I started to think about this:
Maybe we can find a relationship between a square with the side lengths of A + B and the grey square with lengths C.
The final thing you need to know is that the area of a triangle is equal to half the base times the height: The answer becomes clear when you draw in the rest of the triangles. Look at the red square.The lengths of each of its sides are equal to A + B, so it has the area (A + B)2.
If you look at the triangles shaded in pink......you should see that each has a height of A and a base of B, so their areas are ½AB. Therefore, the area of the grey square is equal to the area of the red square from above minus the areas of the four triangles.
Wednesday, March 19, 2008
Squares of Triangles
Solved by J Function at 11:10 AM
Labels: geometry, Pythagorean Theorem
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