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Presented By: Alexander Sarkissian

Presented By: Alexander Sarkissian. Pythagoras’s Theorem First we will start with the common right triangle. Side a= Adjacent side Side b= Opposite

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Page 1: Presented By: Alexander Sarkissian. Pythagoras’s Theorem  First we will start with the common right triangle.  Side a= Adjacent side  Side b= Opposite

Presented By:

Alexander Sarkissian

Page 2: Presented By: Alexander Sarkissian. Pythagoras’s Theorem  First we will start with the common right triangle.  Side a= Adjacent side  Side b= Opposite

Pythagoras’s Theorem

First we will start with the common right triangle.

Side a= Adjacent side

Side b= Opposite side

Side c= Hypotenuse

Page 3: Presented By: Alexander Sarkissian. Pythagoras’s Theorem  First we will start with the common right triangle.  Side a= Adjacent side  Side b= Opposite

We start out with four right triangles.Three of these have been rotated 90o, 180o, and 270o.

Each has an area of ab/2

Page 4: Presented By: Alexander Sarkissian. Pythagoras’s Theorem  First we will start with the common right triangle.  Side a= Adjacent side  Side b= Opposite

Pythagoras’s Theorem

Let's put them together without additional rotations so that they form a square with side c.

Page 5: Presented By: Alexander Sarkissian. Pythagoras’s Theorem  First we will start with the common right triangle.  Side a= Adjacent side  Side b= Opposite

Pythagoras’s Theorem

The square has a square hole with the side (a-b). Summing up its area (a-b)2 and 2ab, the area of the four triangles is (4·ab/2).

Page 6: Presented By: Alexander Sarkissian. Pythagoras’s Theorem  First we will start with the common right triangle.  Side a= Adjacent side  Side b= Opposite

Formula c2 = (a-b)2+2ab = a2-2ab+b2+2ab=

a2+b2 Thus we simplify the equation and

end up with a²+b²=c²

Page 7: Presented By: Alexander Sarkissian. Pythagoras’s Theorem  First we will start with the common right triangle.  Side a= Adjacent side  Side b= Opposite

The End

Sources: http://www.cut-the-knot.org/pythagoras/index.shtml