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I was asked by a colleague yesterday about how the formula for the arc length of a circle is derived. I wanted to give them a correct answer, so I said I'd get back to them once I'd thought about it properly. Would the following be a correct derivation?

We know that the circumference of a circle is $2\pi r$ (for some fixed radius $r$) and as such we can write down the following relation $$ \frac{arclength}{circumference} = \frac{s}{2\pi r} =t$$ where $0\leq t \leq 1$. Now, we also know that $t=0$ when $\theta =0$ and $t=1$ when $\theta =2\pi$; furthermore, we know that (for a circle) $0\leq\theta\leq 2\pi$ and therefore $\theta =2\pi t\Rightarrow t=\frac{\theta}{2\pi}$. Hence we conclude that $$\frac{s}{2\pi r} =\frac{\theta}{2\pi}\Rightarrow s=r\theta$$ In a similar fashion we can also deduce the area of an arc sector of a circle $$\frac{Area\,\,of\,\,arc-sector}{Area\,\,of\,\,circle} =\frac{A}{\pi r^{2}} =\frac{\theta}{2\pi} \Rightarrow A=\frac{1}{2}r^{2}\theta$$ Where, in both cases, $0\leq\theta\leq 2\pi$ is measured in radians.

Would this be a correct description? Thanks for your time.

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  • $\begingroup$ Here's Euclid's proof. $\endgroup$ – David Mitra Dec 17 '14 at 11:46
  • $\begingroup$ @DavidMitra Thanks for the link. Would what I put be acceptable as well? (Quite keen now to make sure that my understanding is correct). $\endgroup$ – Will Dec 17 '14 at 12:05
  • $\begingroup$ It seems you're assuming what you need to prove: that in a given circle, the ratio of the lengths of the arcs subtended by two angles is the same as the ratio of the angles. $\endgroup$ – David Mitra Dec 17 '14 at 12:09
  • $\begingroup$ See this post, though. $\endgroup$ – David Mitra Dec 17 '14 at 12:11
  • $\begingroup$ I guess that's true, although I was trying to get around it (perhaps a bit cheekily) by equating the ratio of arc length to circumference to some parameter $t$ that varies between 0 and 1 and then trying to argue how this relates to the angle?! $\endgroup$ – Will Dec 17 '14 at 13:20

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