If we divide a circle with radius [math]r[/math] into [math]n[/math] sectors of equal size, we can rearrange them in the pattern shown at the right. At first, this does not seem to be very helpful, but notice what happens as we increase the number of sectors.
What shape does this figure resemble as [math]n\rightarrow\infty[/math]?
What does the height, [math]h[/math], approach as [math]n\rightarrow\infty[/math]??
As [math]n\rightarrow\infty[/math], the height becomes closer and closer to the height of one of the red sectors, which is just the radius of the circle.[br][math]h\rightarrow r[/math]
As [math]n\rightarrow\infty[/math], what does the base, [math]b[/math], approach?
The arcs add up to half of the circumference of the circle: [math]\frac{C}{2}=\frac{2\pi r}{2}=\pi r[/math][br]As [math]n\rightarrow\infty[/math], they path formed by the arcs gets closer and closer to forming a straight line, which makes the base approach [math]\pi r[/math].[br][math]b\rightarrow\pi r[/math]
As [math]n\rightarrow\infty[/math], what does the area of the figure approach?
The larger [math]n[/math] is, the more closely the figure will resemble a parallelogram, which has area [math]A=bh[/math].[br]The more sectors we use, the straighter the top and bottom of the parallelogram will become, and thus the base will be closer and closer to [math]\pi r[/math], while the height will be closer and closer to [math]r[/math].[br][math]A\rightarrow bh\rightarrow\left(\pi r\right)\left(r\right)=\pi r^2[/math]
In the diagram notice a few relationships:[br]Looking at the right triangle we have [math]\cos\theta=\frac{a}{r}[/math], which means the apothem has length:[br][math]a=r\cos\theta[/math] [br][br]If [math]s[/math] is the side length of the regular polygon, then we also have [math]\sin\theta=\frac{\frac{s}{2}}{r}=\frac{s}{2r}[/math], so we can express the side length as:[br][math]s=2r\sin\theta[/math][br][br]Therefore, the perimeter is:[br] [math]P=ns=2rn\sin\theta[/math][br][br]Regarding the angle [math]\theta[/math], we have two copies of the central angle [math]\theta[/math] corresponding to each of the [math]n[/math] sides of the polygon, and all of these add up to [math]360^\circ[/math], so [math]\theta=\frac{360^\circ}{2n}=\frac{180^\circ}{n}[/math].
As [math]n\rightarrow\infty[/math], what does the length of the apothem approach?
Visually, it is clear that the apothem will approach the length of the radius.[br][math]a\rightarrow r[/math][br][br]Also, since [math]\theta=\frac{180^\circ}{n}[/math], as [math]n\rightarrow\infty[/math], we have [math]\theta\rightarrow0^\circ[/math], but [math]\cos0^\circ=1[/math], which means that:[br] [math]a=r\cos\left(\frac{180^\circ}{n}\right)\rightarrow r\left(1\right)=r[/math][br]
As [math]n\rightarrow\infty[/math], what does the perimeter approach?
Visually, we can see that the perimeter will approach the circumference of the circle:[br][math]P\rightarrow C=2\pi r[/math][br][br]To show this using [math]P=2rn\sin\left(\frac{180^\circ}{n}\right)[/math] is a lot more complicated since there is also an [math]n[/math] in the coefficient. Calculus is required to show that [math]n\sin\left(\frac{180^\circ}{n}\right)\rightarrow\pi[/math]
Therefore, as [math]n\rightarrow\infty[/math], what does the area approach?
Combining the two previous results we have:[br][math]a\rightarrow r[/math] and [math]p\rightarrow2\pi r[/math][br][br]So [math]A=\frac{1}{2}ap\rightarrow\frac{1}{2}\left(r\right)\left(2\pi r\right)=\pi r^2[/math]