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