Can someone please help me?
A regular n-gon is inscribed in the unit circle. What is the perimeter for each n below?
a. 3
b. 5
c. 6
d. 10
e. 57
f. 542
g. n
The perimeter in part f should be close to what number? How close is it?

- anonymous

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- jdoe0001

got pic?

- anonymous

no, there is no picture @jdoe0001

- anonymous

but, how do I find the side lengths?

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## More answers

- anonymous

So, how would I do that for a 5-gon inscribed in the circle?

- dan815

are u saying theyre both 60 or together they are 60?

- anonymous

is there some sort of formula to solve this? because to find the perimeter of a 542-gon inscribed in the circle, using a formula to figure it out is what makes sense

- dan815

oh hey ok im here

- dan815

do u want me to solve mahimas question or yours jhanny

- dan815

mahmia this is what u do okay first see the length of one of the sides of tthis n gon and multibly by n

- anonymous

sorry, i don't understand. What do you mean by "see the length of one of the sides"

- anonymous

and do you mean multiply the length of the side by n?

- Jhannybean

Sorry, forget my explanation. I was missing some key concepts in it.

- dan815

|dw:1438992661973:dw|

- dan815

see the vectors from the center have 360/3 separation for 3 god
360/4 for 4 gon
360/5 for 5 gon and so on

- dan815

now u can use the cos law to find the length of each side

- dan815

or write a vector subtraction magnitude equation

- dan815

|dw:1438992837654:dw|

- anonymous

could help me solve one of them so that i completely understand?

- anonymous

how would i use the law of cos to find the length of each side?

- dan815

|dw:1438992861468:dw|

- dan815

that is the length of each side so the periment is n*L

- dan815

where L is a function of n defined up there

- anonymous

|dw:1439175975995:dw|

- dan815

that formujla was from cos law

- dan815

|dw:1438993061593:dw|

- anonymous

@surjithayer would use that equation for all of the n-gons?

- Jhannybean

But dan, what is the actual formula?

- dan815

that is

- anonymous

@surjithayer would i substitutes the number of sides for n in that equation?

- Jhannybean

Well, you could also use \(P(n) = 2rn\sin\left(\frac{180}{n}\right)\)

- Jhannybean

Where r is the radius of the circle and n is the number of sides

- anonymous

how did you come up with that formula? @Jhannybean

- dan815

|dw:1438993370012:dw|

- dan815

|dw:1438993380437:dw|

- anonymous

oh, that makes sense @dan815

- Jhannybean

I finally got it.

- Jhannybean

Somewhat.

- Jhannybean

So if we have a pentagon, right, let's draw that out. |dw:1438994108310:dw|

- Jhannybean

Now we know that each of these triangles would be an isosceles triangle, where 2 sides of the triangle are considered the `radii` of the circle, whereas the opposite side touching the circle is what we have to find.
So now, \(a=b=\text{radius (r)}\)
Therefore, we can apply the law of cosines. \(c^2=a^2+b^2-2ab\cos(C)\)

- anonymous

okay

- Jhannybean

And just like @dan815 demonstrated, but with \(l=c\)
\(c^2=r^2+r^2-2r^2\cos(C)\)
Now we're missing one thing, and that is \(\angle C\)
\(\angle C = \dfrac{360^\circ}{n}\) - replace this back in formula, and you get
\(c^2 = r^2+r^2 -2r^2\cos\left(\dfrac{360^\circ}{n}\right)\)
That will give give you the side length of the pentagon, c |dw:1438994818712:dw|

- anonymous

but since the radius is 1, when you substitute it into the formula wouldn't you get 0 times the cos of 360/n? making the answer 0, which wouldn't make sense

- Jhannybean

Solving for c, we have: \[c= \sqrt{2r^2-2r^2\cos\left(\frac{360^\circ}{n}\right)}\iff c=r\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}\]

- Jhannybean

No, when r=1 you do not get 0.

- anonymous

oh, order of operations, that's right i forgot

- Jhannybean

And lastly, all we have to do is multiply by the number of sides \[P= n\cdot c ~~\sf\text{where} ~ c=r\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}\]

- anonymous

okay, thank you so much for your help, that makes so much sense

- Jhannybean

Great :D Glad i could help!

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