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Geometric Proofs for Pythagoras’ Theorem

Four triangles with measures of A, B and C and C being the hypotenuse. Find the area of the triangles. First we need to get a visual of the triangles and label their sides. Go ahead and start on that and see if it’s the same as mine. Now arrange them to form a square with the hypotenuse of each triangle being used as a side of the square. Would you agree that we now have a big square of side C and a little square of the side A minus B? And, we also have four right triangles inside the big square therefore the area of the triangles plus the area of the little square should equal the area of the big square. Let’s see. There are four triangles each with base A and a height of B so their areas total four times the quantity A times B over two. The area of the small square is A minus B squared and the area of the large square is C squared so we have four times AB over two plus the quantity of A minus B squared all equal to C squared. This gives us two AB plus A minus B times A minus B equals C squared. Two AB plus the quantity of A squared minus two AB plus B squared equals C squared bringing us back to A squared plus B squared equals C squared. What did you think about that?

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