The Non Transcendental, Exact Value of π and the Squaring of the Circle
Posted: Mon Aug 18, 2025 9:29 am
We take a square and a circle of the same perimeter π and write the equation
4b = π
We calculate the areas, using 4b for π, for the area of the circle.. Respectively we obtain :
As = b^2
Ac = b
Now we use these values as the legs of a right triangle to apply Pythagoras and find out the value of b or π/4.
(b)^2 + (b^2)^2 = 1
b^2 + b^4 – 1 = 0
b = 0.7861513
4b = π = 3.1446
The reason to take the Hypotenuse = 1 is simple. The value of b and b^2 are calculated for Diameter = 1.
So, the only semicircle where we can inscribe this right triangle to preserve the values of b and b^2 is one of Diameter = 1
4b = π
We calculate the areas, using 4b for π, for the area of the circle.. Respectively we obtain :
As = b^2
Ac = b
Now we use these values as the legs of a right triangle to apply Pythagoras and find out the value of b or π/4.
(b)^2 + (b^2)^2 = 1
b^2 + b^4 – 1 = 0
b = 0.7861513
4b = π = 3.1446
The reason to take the Hypotenuse = 1 is simple. The value of b and b^2 are calculated for Diameter = 1.
So, the only semicircle where we can inscribe this right triangle to preserve the values of b and b^2 is one of Diameter = 1