Finding circumference without using $\pi$
If the area of a circle is $254.34\ldots\text{ cm}^2$ it has a diameter of $18\text{ cm}$, is it possible to find the circumference without using or making the irrational constant Pi ($\pi=3,1415926535\ldots)$ in any way at all, and if it is possible, how?
To find the area of $254.34$ I used $\pi$ as $3.14$ in shorthand. The formula you give should allow me to find the Circumference as $56.52\text{ cm}$ (as this also uses $\pi$ as 3.14)
Sorry! There was a mistake in the question!
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$\begingroup$Consider a $n$ sided regular polygon as shown in the figure below.
Let the diameter of the polygon as shown be $d$ and the side of the polygon be $a$. Then the area of the triangle as shown in the figure is $$T = \dfrac12 \times a \times \dfrac{d}2 = \dfrac{ad}4$$The perimeter of the polygon is $$P = na,$$ while the area of the polygon is $$A = nT = \dfrac{nad}4.$$ Hence, we get that $$A = \dfrac{Pd}4$$Letting the number of sides $n$ tend to infinity, the polygon "tends" to a circle and we get that$$\text{Area of the circle} = \dfrac{\text{Circumference }\times \text{ diameter}}4$$or to put it the other way around$$\text{Circumference} = \dfrac{4 \times \text{Area of the circle}}{\text{ diameter}}$$As you see from this, there is no need for $\pi$ anywhere.
$\endgroup$ 2 $\begingroup$Recall: $$A = \pi \left(\frac{D}{2}\right)^2$$ $$C = \pi D$$ Where $C$ is circumference, $A$ is area, and $D$ is diameter.
Thus: $$\frac{A}{C} = \frac{\pi \left(\frac{D}{2}\right)^2}{\pi D}$$ The $\pi$'s cancel out, as well as one of the D's: $$\frac{A}{C} = \frac{D}{4}$$
Solving for C: $$C = \frac{4A}{D}$$
$\endgroup$ 5 $\begingroup$Let's write down all our formulas:
$$A = \pi r^2 = 254.34$$ $$C = 2\pi r$$ $$d = 2r = 18$$
where $A$ is area, $r$ is radius, $C$ is circumference, and $d$ is diameter. We're trying to find the expression in the second; if we look at the first formula it's almost there, but it's missing a factor of 2 (easy to solve) and there's an extra factor of $r$. But if we divide the expression in the first formula by the expression in the third formula, we get
$$\frac{254.34}{18} = A/d = \frac{\pi r}{2}$$
Almost there! Now we just need to multiply by 4:
$$2 \pi r = 4\frac{\pi r}{2} = 4A/d = 4*\frac{254.43}{18} = 56.52.$$
$\endgroup$ 3 $\begingroup$Hint: $$\frac{4\times\text{Area}}{\text{Diameter}}=\frac{4\pi r^2}{2r}=2\pi r$$
$\endgroup$ 5 $\begingroup$See Approximation of circle by n-sided polygons in interative Java Apllets for Circle Area Approximation. For Exemple here.
Since $A=\pi r^2$ then $A/r^2=\pi$. But $C=2\pi r$ and so $C=2r A/r^2=2A/r=4A/d$, where $d$ is the diameter.
$\endgroup$ 2 $\begingroup$I'm surprised no-one has yet given this answer, so: $$C=\tau r=\frac{\tau}{2}d$$ So, no, you don't need $\pi$ ;)
$\endgroup$ 2 $\begingroup$The relation between area of a circle and its circunference can be seen by dissecting it, according to your tastes, as an onion, or as a pizza.
$\endgroup$ $\begingroup$I think this answer might help also.
Suppose you know a priori that the only way to get what you want either by algorithms that approximate the area A of the circle area polígnos with n sides.
If you want an algorithm which does not mention the constant $\pi$ irrational then you may have a dozen algorithms that provide the area $A>0$ of the circle by approximation of areas $A_{n_k}$ polignos of $n_k$ sides. Here $k$ indicates the $k$-th step of the algorithm. In general all these algorithms are the algorithm as a coneceitual as described below.
Conceptual algorithm
Initialization: Set $k = 0$ and one (or more if you want) poligno $P_{n_k}$ with $n_k$ sides. A number $\epsilon> 0$, Area $A>0$ of the circle.
Step Interactive: some effective procedure that gets a poligno $P_{n_{k +1}}$ by polignos from the previous step that makes the sequence $ A_k $ of areas of polignos $P_{n_{k}}$ converges.
Stopping criterion: say something like $| A - A_{n_{k +1}} | <\epsilon$
If we do not have a priori the area $A>0$ of the circle then such an conceptual algorithm is possible but a little more complex.
The answer to your question is then yes and not:
Yes, by cause the conceptual algorithm not mentioned the irrational constant $\pi$.
Not, by cause, for all above type algorithms work by providing an indirect calculation for $\pi $. To see this, just remember that the area of circle is $$A = \pi\cdot r^2.$$ And by any above type algorithms $$\lim_{k\to\infty} A_{n_k} = A.$$ Then $$ \pi = \frac{1}{r^2}\cdot\lim_{k\to\infty} A_{n_k}. $$
I hope that helped.
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