Geometric Proof of the Sum of Squares by Christina Martin Math 310, Fall 2012 a2a2 + b 2 = c 2 a2a2 a2a2 b2b2 b2b2.

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Presentation transcript:

Geometric Proof of the Sum of Squares by Christina Martin Math 310, Fall 2012 a2a2 + b 2 = c 2 a2a2 a2a2 b2b2 b2b2

b2b2 b2b2 a2a2 a2a2 -Begin with two squares of arbitrary size, a 2 and b 2. Stick them together to form a hexagon.

b2b2 b2b2 a2a2 a2a2 -Next, draw a line segment at the bottom of each square. That is, create one line with length a, and one line of length b. ab

b2b2 b2b2 -Then, switch the locations of the line segments: the b line is now at the base of the a 2 square, and the a line is now at the base of the b 2 square. a2a2 -Next, draw a line segment at the bottom of each square. That is, create one line with length a, and one line of length b.

b2b2 b2b2 a2a2 a2a2 -Draw a line from the top left corner of the a 2 square to the base where the b-line ends to form a triangle. -Create that triangle. -Rotate the triangle 90° counterclockwise. -Shift the triangle up to the top of the a 2 square.

a2a2 a2a2 -Repeat the process starting from the top of the b 2 square b2b2 -Draw a line from the top right corner of the b 2 square to the base where the a-line ends to form a triangle -Create that triangle. -Rotate the triangle 90° clockwise. -Shift the triangle up to the top of the b 2 square.

b2b2 b2b2 a2a2 a2a2 b2b2 Thus the final result is b2b2 b2b2 C2C2 a 2 +b 2 =c2c2 c c