
Area of circle to its circumference is a constant approach to Pi
Posted:
Sep 12, 2004 12:49 PM


The area of a circle to its circumference is:
{[pi*d^2]/4}/pi*d = d/4 , where d is circle's diameter.
If we inscribe an orthogonal triangle ABC with AB=diameter d ,as its hypotenuse, and construct the two orthogonal sides AC and CD so that the ratio of AB/AC =d/4 ,and AC/CB = d/4 ,
[We are allowed to do this **** REF BELOW NOTE ]
we get the value of AC=4[AB]/d=4d/d=4 , and the value of CB=4[AC]/d=16/d >From this orthogonal triangle we have from Pythagoras Theorem:
AB^2=AC^2+CB^2 ,or
d^2=4^2+[16/d]^2 ,or
d^4  [4^2]*d^2  16^2 =0 or d^4  [4^2]*d^2  4^4 =0 Which gives positive and real root for d:
d=4*[SQRT{1+SQRT5}/2] = 4T =
4*[SQUARE ROOT OF THE GOLDEN SECTION]
T = Tan[THETA] =AC/CB=4/[16/d]=4/[16/4T]=T So: AB=4T AC=4 CB=4/T 
If we call 16/d=Pi =[CB] so d=16/Pi =[AB]
area of circle= Pi[16/Pi]^2/4 =64/Pi circumference =Pi[16/Pi]=16 Their ratio =4/Pi , [equals d/4 =[16/Pi]/4= 4/Pi ] Then : [AB]/[AC]=4/Pi , or [AC]=[AB]*Pi/4 =[16/Pi]*Pi/4= 4
Thus we have the orthogonal triangle inscribed in circle diameter d=hypotenuse=16/Pi with big vertical side =4 and small horizontal= Pi. Using Pythagoras:
[16/Pi]^2 = 4^2 + Pi^2 or
Pi^4 +4^2*Pi^2  4^4 =0
This gives as positive real root: Pi=4T =4*[SQRT{1+SQRT5}/2]
NOTE**** [We are allowed to do this BECAUSE WE HAVE A KNOW SIMILAR ORTHOGONAL TRIANGLE having sides T^3 hypotenuse , T^2 bigger orthogonal , and T^1 smaller orthogonal. having ratio T ,DETWEEN EACH OTHER and T=SQRT[GOLDEN SECTION] If T^3=D CIRCLE'S DIAMETER and at the same time the HYPOTENUSE of an inscribed orthogonal triangle with orthogonal sides: T^2 ,and T^1 then the circle's area={Pi*[T^3]^2}/4 Its circumference =Pi*[T^3] ratio of area to circumference=T^3]/4 = D/4 ]
Ref:<a href="http://www.stefanides.gr/piquad.htm">http://www.stefanides.gr/piquad.htm</a> <a href="http://www.stefanides.gr/quadcirc.htm">http://www.stefanides.gr/quadcirc.htm</a> <a href="http://www.stefanides.gr/theo_circle.htm">http://www.stefanides.gr/theo_circle.htm</a> <a href="http://www.stefanides.gr/quad.htm">http://www.stefanides.gr/quad.htm</a>
Regards, Panagiotis Stefanides <a href="http://www.stefanides.gr">http://www.stefanides.gr</a>

