Krdytkyu

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.

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:
$$C = frac{4A}{D}$$

Let's write down all our formulas:

$$A = pi r^2 = 254.34$$
$$C = 2pi r$$
$$d = 2r = 18$$

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:

$$2 pi r = 4frac{pi r}{2} = 4A/d = 4*frac{254.43}{18} = 56.52.$$

Hint: $$frac{4timestext{Area}}{text{Diameter}}=frac{4pi r^2}{2r}=2pi r$$

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 $C=2pi r$ and so $C=2r A/r^2=2A/r=4A/d$, where $d$ is the diameter.

I'm surprised no-one has yet given this answer, so:
$$C=tau r=frac{tau}{2}d$$
So, no, you don't need $pi$ ;)

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.

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

1. 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.




2. 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.




3. 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 $$A = picdot r^2.$$ And by any above type algorithms $$lim_{ktoinfty} A_{n_k} = A.$$ Then
  $$
  pi = frac{1}{r^2}cdotlim_{ktoinfty} A_{n_k}.
  $$

I hope that helped.