anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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jdoe0001
  • jdoe0001
got pic?
anonymous
  • anonymous
no, there is no picture @jdoe0001
anonymous
  • anonymous
but, how do I find the side lengths?

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More answers

anonymous
  • anonymous
So, how would I do that for a 5-gon inscribed in the circle?
dan815
  • dan815
are u saying theyre both 60 or together they are 60?
anonymous
  • 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
  • dan815
oh hey ok im here
dan815
  • dan815
do u want me to solve mahimas question or yours jhanny
dan815
  • 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
  • anonymous
sorry, i don't understand. What do you mean by "see the length of one of the sides"
anonymous
  • anonymous
and do you mean multiply the length of the side by n?
Jhannybean
  • Jhannybean
Sorry, forget my explanation. I was missing some key concepts in it.
dan815
  • dan815
|dw:1438992661973:dw|
dan815
  • 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
  • dan815
now u can use the cos law to find the length of each side
dan815
  • dan815
or write a vector subtraction magnitude equation
dan815
  • dan815
|dw:1438992837654:dw|
anonymous
  • anonymous
could help me solve one of them so that i completely understand?
anonymous
  • anonymous
how would i use the law of cos to find the length of each side?
dan815
  • dan815
|dw:1438992861468:dw|
dan815
  • dan815
that is the length of each side so the periment is n*L
dan815
  • dan815
where L is a function of n defined up there
anonymous
  • anonymous
|dw:1439175975995:dw|
dan815
  • dan815
that formujla was from cos law
dan815
  • dan815
|dw:1438993061593:dw|
anonymous
  • anonymous
@surjithayer would use that equation for all of the n-gons?
Jhannybean
  • Jhannybean
But dan, what is the actual formula?
dan815
  • dan815
that is
anonymous
  • anonymous
@surjithayer would i substitutes the number of sides for n in that equation?
Jhannybean
  • Jhannybean
Well, you could also use \(P(n) = 2rn\sin\left(\frac{180}{n}\right)\)
Jhannybean
  • Jhannybean
Where r is the radius of the circle and n is the number of sides
anonymous
  • anonymous
how did you come up with that formula? @Jhannybean
dan815
  • dan815
|dw:1438993370012:dw|
dan815
  • dan815
|dw:1438993380437:dw|
anonymous
  • anonymous
oh, that makes sense @dan815
Jhannybean
  • Jhannybean
I finally got it.
Jhannybean
  • Jhannybean
Somewhat.
Jhannybean
  • Jhannybean
So if we have a pentagon, right, let's draw that out. |dw:1438994108310:dw|
Jhannybean
  • 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
  • anonymous
okay
Jhannybean
  • 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
  • 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
  • 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
  • Jhannybean
No, when r=1 you do not get 0.
anonymous
  • anonymous
oh, order of operations, that's right i forgot
Jhannybean
  • 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
  • anonymous
okay, thank you so much for your help, that makes so much sense
Jhannybean
  • Jhannybean
Great :D Glad i could help!

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