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# I can't do math...

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Kakarot_15  May 25, 2017

#1
+4749
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(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
Sort:

#1
+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
#2
+4749
+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
#3
+149
+1

So what should I do for E?

Kakarot_15  May 25, 2017
#4
+77081
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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
edited by CPhill  May 25, 2017
#5
+149
+2

Thank you both for helping

Kakarot_15  May 26, 2017

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