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anonymous

  • one year ago

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?

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  1. jdoe0001
    • one year ago
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    got pic?

  2. anonymous
    • one year ago
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    no, there is no picture @jdoe0001

  3. anonymous
    • one year ago
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    but, how do I find the side lengths?

  4. anonymous
    • one year ago
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    So, how would I do that for a 5-gon inscribed in the circle?

  5. dan815
    • one year ago
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    are u saying theyre both 60 or together they are 60?

  6. anonymous
    • one year ago
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    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

  7. dan815
    • one year ago
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    oh hey ok im here

  8. dan815
    • one year ago
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    do u want me to solve mahimas question or yours jhanny

  9. dan815
    • one year ago
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    mahmia this is what u do okay first see the length of one of the sides of tthis n gon and multibly by n

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

  11. anonymous
    • one year ago
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    and do you mean multiply the length of the side by n?

  12. Jhannybean
    • one year ago
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    Sorry, forget my explanation. I was missing some key concepts in it.

  13. dan815
    • one year ago
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    |dw:1438992661973:dw|

  14. dan815
    • one year ago
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    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

  15. dan815
    • one year ago
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    now u can use the cos law to find the length of each side

  16. dan815
    • one year ago
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    or write a vector subtraction magnitude equation

  17. dan815
    • one year ago
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    |dw:1438992837654:dw|

  18. anonymous
    • one year ago
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    could help me solve one of them so that i completely understand?

  19. anonymous
    • one year ago
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    how would i use the law of cos to find the length of each side?

  20. dan815
    • one year ago
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    |dw:1438992861468:dw|

  21. dan815
    • one year ago
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    that is the length of each side so the periment is n*L

  22. dan815
    • one year ago
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    where L is a function of n defined up there

  23. anonymous
    • one year ago
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    |dw:1439175975995:dw|

  24. dan815
    • one year ago
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    that formujla was from cos law

  25. dan815
    • one year ago
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    |dw:1438993061593:dw|

  26. anonymous
    • one year ago
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    @surjithayer would use that equation for all of the n-gons?

  27. Jhannybean
    • one year ago
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    But dan, what is the actual formula?

  28. dan815
    • one year ago
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    that is

  29. anonymous
    • one year ago
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    @surjithayer would i substitutes the number of sides for n in that equation?

  30. Jhannybean
    • one year ago
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    Well, you could also use \(P(n) = 2rn\sin\left(\frac{180}{n}\right)\)

  31. Jhannybean
    • one year ago
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    Where r is the radius of the circle and n is the number of sides

  32. anonymous
    • one year ago
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    how did you come up with that formula? @Jhannybean

  33. dan815
    • one year ago
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    |dw:1438993370012:dw|

  34. dan815
    • one year ago
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    |dw:1438993380437:dw|

  35. anonymous
    • one year ago
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    oh, that makes sense @dan815

  36. Jhannybean
    • one year ago
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    I finally got it.

  37. Jhannybean
    • one year ago
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    Somewhat.

  38. Jhannybean
    • one year ago
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    So if we have a pentagon, right, let's draw that out. |dw:1438994108310:dw|

  39. Jhannybean
    • one year ago
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    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. \(c^2=a^2+b^2-2ab\cos(C)\)

  40. anonymous
    • one year ago
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    okay

  41. Jhannybean
    • one year ago
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    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|

  42. anonymous
    • one year ago
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    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

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

  44. Jhannybean
    • one year ago
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    No, when r=1 you do not get 0.

  45. anonymous
    • one year ago
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    oh, order of operations, that's right i forgot

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

  47. anonymous
    • one year ago
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    okay, thank you so much for your help, that makes so much sense

  48. Jhannybean
    • one year ago
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    Great :D Glad i could help!

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