#1**+2 **

(a)

The " vertex angle of measure " is \(\frac{360^{\circ}}n\) , and angle 1 is half of this angle. So...

m∠1 = \(\frac12*\frac{360^{\circ}}n=\frac{360^{\circ}}{2n}=\frac{180^{\circ}}{n}\)

(b)

sin (an angle) = (side opposite) / (hypotenuse)

sin (angle 1) = x / r

sin ( \(\frac{180^{\circ}}{n}\) ) = x/r

r * sin ( \(\frac{180^{\circ}}{n}\) ) = x

x = r * sin ( \(\frac{180^{\circ}}{n}\) )

(c)

Two x's fit on each side...and if there are n sides, then

perimeter = 2x * n

perimeter = 2 * r * sin ( \(\frac{180^{\circ}}{n}\) ) * n

(d)

The more sides you add, the closer the shape gets to becoming a circle. Think about a pentagon (5 sides) compared to a decagon (10 sides). The decagon is a lot more like a circle than the pentagon. So...the bigger "n" gets, the more circle-like the shape is, and the perimeter gets bigger and bigger the more sides you add.

Look: \(5\sin(\frac{180}{5})\approx 2.94 \\~\\ 10\sin(\frac{180}{10})\approx 3.09\)

(e)

Sorry Kakarot...I don't really think I know this one!

hectictar
May 25, 2017

#1**+2 **

Best Answer

(a)

The " vertex angle of measure " is \(\frac{360^{\circ}}n\) , and angle 1 is half of this angle. So...

m∠1 = \(\frac12*\frac{360^{\circ}}n=\frac{360^{\circ}}{2n}=\frac{180^{\circ}}{n}\)

(b)

sin (an angle) = (side opposite) / (hypotenuse)

sin (angle 1) = x / r

sin ( \(\frac{180^{\circ}}{n}\) ) = x/r

r * sin ( \(\frac{180^{\circ}}{n}\) ) = x

x = r * sin ( \(\frac{180^{\circ}}{n}\) )

(c)

Two x's fit on each side...and if there are n sides, then

perimeter = 2x * n

perimeter = 2 * r * sin ( \(\frac{180^{\circ}}{n}\) ) * n

(d)

The more sides you add, the closer the shape gets to becoming a circle. Think about a pentagon (5 sides) compared to a decagon (10 sides). The decagon is a lot more like a circle than the pentagon. So...the bigger "n" gets, the more circle-like the shape is, and the perimeter gets bigger and bigger the more sides you add.

Look: \(5\sin(\frac{180}{5})\approx 2.94 \\~\\ 10\sin(\frac{180}{10})\approx 3.09\)

(e)

Sorry Kakarot...I don't really think I know this one!

hectictar
May 25, 2017

#2**+1 **

HMMMMMMM !!

I just looked at part e some more...

The formula for the "perimeter" of a circle is

c = 2 * r * pi

The formula for the perimeter of a n-gon is

p = 2 * r * sin(\(\frac{180^{\circ}}{n}\)) * n

With really big values of n , sin( \( \frac{180^{\circ}}{n} \) ) * n must be close to pi !!

So...I tried that

with n = 1,000: sin( \(\frac{180^{\circ}}{1000}\) ) * 1000 ≈ 3.141587

and

with n = 50,000: sin(\(\frac{180^{\circ}}{50,000}\) ) * 50,000 ≈ 3.14159265

I thought that the limit as n approaches infinity would make it exactly pi,

but WolframAlpha doesn't seem to say that.. here

But if I put pi instead of 180º , it works...? here

hectictar
May 25, 2017

#4**+2 **

Hectictar's surmise that n * sin (180 /n) approaches "pi" as n approaches infinity is exactly correct...see the graph here :

https://www.desmos.com/calculator/7pjtxl5jl1

I've used "x" instead of "n" but the idea is the same

So...putting this fact together with the answer she gave in part C, the perimeter of an n-gon for some large value of n approaches 2 * r * [ pi ] = 2 *pi*r = the circumference of a circle with a given radius of "r"

CPhill
May 25, 2017