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Kakarot_15  May 25, 2017

Best Answer 

 #1
avatar+4749 
+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
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5+0 Answers

 #1
avatar+4749 
+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
avatar+4749 
+1

HMMMMMMM !!  surprise

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
 #3
avatar+149 
+1

So what should I do for E?

Kakarot_15  May 25, 2017
 #4
avatar+77081 
+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"

 

 

 

cool cool cool

CPhill  May 25, 2017
edited by CPhill  May 25, 2017
 #5
avatar+149 
+2

Thank you both for helping

Kakarot_15  May 26, 2017

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