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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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Do you want to avoid using value of $\pi$ or the concept of $\pi$ and all the formulas that contain $\pi$? – Juris Jan 16 '13 at 23:38
The area of a circle needs to be in square linear units; square cm instead of cm. The area cannot be 279.46 cm. – robjohn Jan 16 '13 at 23:40
Not to nitpick, but all the answers use $\pi$ in some way... Just for fun, consider using the method here. you know the area of the "rectangle" there and one side length. You can then compute the other side length and thus find the circumference. – David Mitra Jan 16 '13 at 23:48
@DanielK Did you mean to use $\pi = 2.14$? – apnorton Jan 16 '13 at 23:50
Your data are inconsistent. A radius of 9 cm gives an area of $81 \pi \approx 254.47$ and a circumference of $\approx 56.549$ – Ross Millikan Jan 16 '13 at 23:50

Consider a $n$ sided regular polygon as shown in the figure below. enter image description here

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.

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Great answer finally actually avoiding the $\pi$ :) – Juris Jan 17 '13 at 0:35
But P is not constant and depends on a which changes with every new value of n. In the end we do not actually know the area of the circle. – fleablood Dec 15 '15 at 0:37

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

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Tricky answer :) but if we just have D how we can calculate A without using pi? – Vins Jan 17 '13 at 0:29
vins: it will be awfully hard to calculate A without using pi since A is precisely pi*d! – Steven Stadnicki Jan 17 '13 at 0:34
Not difficult. There are two easy ways to approximate it, for the methods I say it's much better to do it with some program. One would be to use Montecarlo (2 random variables, x and y, with a module < R) and the other to integrate by parts $ y = \sqrt {r^2 - x^2} $ from 0 to r and multiply for 4 – Francisco Presencia Jan 17 '13 at 0:40
@StevenStadnicki It wouldn't be oh so hard... just define a new constant, say $\tau = \frac{C}{R}$ Now you can find the area in terms of $\tau$! :P – apnorton Jan 17 '13 at 2:18

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:

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

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@Sigur that would result in an an area smaller than the true area, not larger. His given area and diameter imply $\pi \approx 3.45$. – Jonathan Christensen Jan 16 '13 at 23:45
Wow, 4 equivalent answers but noone (including me) had noticed that the circle is impossible. Great caution there :) – Juris Jan 16 '13 at 23:45
Some times we see people using such strange values to approximate $\pi$. I'll not be surprised with this value, unfortunately. – Sigur Jan 16 '13 at 23:47

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

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Sorry this didn't give me the right answer – Daniel K Jan 16 '13 at 23:38
@DanielK, what is your right answer? – Sigur Jan 16 '13 at 23:41
@DanielK Perhaps it didn't give you the right answer because your circle is impossible? – Jonathan Christensen Jan 16 '13 at 23:44
@DanielK: assuming that your area is $279.46\text{ cm}^2$ and the diameter is $18\text{ cm}$, we get the circumference to be $$ \frac{4\times279.46\text{ cm}^2}{18\text{ cm}}=62.1\text{ cm} $$ However, a circle with a diameter of $18\text{ cm}$ would have an area of $254.469\text{ cm}^2$ – robjohn Jan 16 '13 at 23:46
@DanielK I don´t understand, why are you asking if you already know the "right answer"? – dwarandae Jan 17 '13 at 1:19

See Approximation of circle by n-sided polygons in interative Java Apllets for Circle Area Approximation. For Exemple here.

enter image description here

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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.

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So you're saying (Area x 2) divided by Diameter? Because if you are this also didn't give me the answer I was looking for – Daniel K Jan 16 '13 at 23:41
@DanielK: It looks like you are a factor 2 off, but your data are not correct, either. – Ross Millikan Jan 16 '13 at 23:51

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

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Did you mean to write C instead of A? – Martin Jan 17 '13 at 12:35
@Martin You are right of course. Thanks for spotting this. – Christian Jan 21 '13 at 14:01

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.

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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 $$A = \pi\cdot r^2.$$ 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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