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Topic: Area of circle to its circumference is a constant -approach to Pi
Replies: 7   Last Post: Mar 19, 2013 1:33 PM

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 Eur Ing Panagiotis Stefanides Posts: 567 Registered: 12/3/04
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
&gt;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 ]