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Pythagoras Theorem: x2+y2 = z2. But Why ?

Updated on September 13, 2012

Pythagorean Theorem and its proof is considered as one of the first proofs of Mathematics.

The Theorem is stated as follows:

In any right angled triangle, the sum of squares of two sides having the right angle is equal to the square of its hypotenuse.

Please refer to the fig a which shows a right angled triangle with sides x and y having right angle and z as hypotenuse

So from fig.a and based on the above theorem

x2+y2 = z2.

Ever wondered how the above statement is true. Let's prove this geometrically.

  • Lets consider an arbitrary line of arbitrary length and consider an arbitrary point dividing the length in to x and y. Please refer to fig. b. So the length of the line will become x+y.
  • Now, lets draw a square having length x+y and please refer to fig.c
  • Now, the area of the square in fig.c is (x+y)2

(x+y)2 = x2+y2+2xy ----------- (i)

The geometrical proof for the above equation can be found in one of my hubs named Why (a+b)2 = a2+b2+2ab ?

  • Now, lets draw a square inside the square having length x+y taking the arbitrary points on the square as the corners and let the length of the square be z. Please refer to fig.d
  • Considering fig. d

Area of Outer Square of length (x+y) equals sum of areas of inner square of length z and area of 4 triangles.

Therefore

the area of inner square of

length z = z2 --------------------------(ii)

the area of triangle with

sides x ,y and z = (x*y)/2 -------------------(iii)

implies (x+y)2 = Area of inner square of length z + 4*(Area of triangles)

implies (x+y)2 = (ii) + (iii)

implies (x+y)2 = z2 + 4 * (x*y) / 2 --------------(iv)

Substituting (i) in (iv) (i.e above equation)

x2+y2+2 xy = z2 + 2 * (x*y) ---> x2+y2+2 xy = z2 + 2 xy

Cancelling 2xy from Left Hand Side and Right Hand Side from the above equation

Therefore we get

x2 + y2 = z2 which proves the Pythagoras Theorem







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      SUGHRANIL SAHA 23 months ago

      EXELENT

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