# 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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Is this homework? If so, please add the [homework] tag. –  Jonathan Christensen Jan 16 '13 at 23:35
No, just a query –  Daniel K Jan 16 '13 at 23:37
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
@DanielK Did you mean to use $\pi = 2.14$? –  anorton Jan 16 '13 at 23:50

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.

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Great answer finally actually avoiding the $\pi$ :) –  Juris Jan 17 '13 at 0:35

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 –  anorton 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.

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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

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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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