## 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?

1. anonymous

got pic?

2. anonymous

no, there is no picture @jdoe0001

3. anonymous

but, how do I find the side lengths?

4. anonymous

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

5. dan815

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

6. 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

7. dan815

oh hey ok im here

8. dan815

do u want me to solve mahimas question or yours jhanny

9. 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

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. dan815

|dw:1438992661973:dw|

14. 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

15. dan815

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

16. dan815

or write a vector subtraction magnitude equation

17. dan815

|dw:1438992837654:dw|

18. anonymous

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

19. anonymous

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

20. dan815

|dw:1438992861468:dw|

21. dan815

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

22. dan815

where L is a function of n defined up there

23. anonymous

|dw:1439175975995:dw|

24. dan815

that formujla was from cos law

25. dan815

|dw:1438993061593:dw|

26. anonymous

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

27. anonymous

But dan, what is the actual formula?

28. dan815

that is

29. anonymous

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

30. anonymous

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

31. anonymous

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

32. anonymous

how did you come up with that formula? @Jhannybean

33. dan815

|dw:1438993370012:dw|

34. dan815

|dw:1438993380437:dw|

35. anonymous

oh, that makes sense @dan815

36. anonymous

I finally got it.

37. anonymous

Somewhat.

38. anonymous

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

39. anonymous

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

okay

41. anonymous

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

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. anonymous

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. anonymous

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

45. anonymous

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

46. anonymous

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

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

48. anonymous

Great :D Glad i could help!