Derivation of Pythagorean Theorem

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Pythagorean Theorem
In any right triangle, the sum of the square of the two perpendicular sides is equal to the square of the longest side. For a right triangle with legs measures $ a $ and $ b $ and length of hypotenuse $ c $, the theorem can be expressed in the form
 

$ a^2 + b^2 = c^2 $

 

Proved by Pythagoras
 

Proof of Pythagorean Theorem by Pythagoras

 

Area of the large square = Area of four triangles + Area of small square
$ A_{total} = A_{four \, \, triangles} + A_{small \, \, square} $

$ (a + b)^2 = 4 \, \left( \frac{1}{2} ab \right) + c^2 $

$ a^2 + 2ab + b^2 = 2ab + c^2 $

$ a^2 + b^2 = c^2 $
 

Proved by Bhaskara
Bhaskara (1114 - 1185) was an Indian mathematician and astronomer.
 

Proof by Bhaskara

 

Area of the large square = Area of four triangles + Area of inner (smaller) square
$ A_{total} = A_{four \, \, triangles} + A_{small \, \, square} $

$ c^2 = 4 \, \left( \frac{1}{2} ab \right) + (b - a)^2 $

$ c^2 = 2ab + (b^2 - 2ab + a^2) $

$ c^2 = 2ab + b^2 - 2ab + a^2 $

$ c^2 = b^2 + a^2 $
 

Proved by U.S. Pres. James Garfield
 

Proof by Pres. James Garfield

 

Area of trapezoid = Area of 3 triangles
$ \frac{1}{2}(a + b)(a + b) = \frac{1}{2}ab + \frac{1}{2}c^2 + \frac{1}{2}ab $

$ (a + b)^2 = ab + c^2 + ab $

$ a^2 + 2ab + b^2 = 2ab + c^2 $

$ a^2 + b^2 = c^2 $
 

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