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Math Help - Diagonals in an Octagon

  1. #1
    Senior Member DivideBy0's Avatar
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    Diagonals in an Octagon

    In a regular octagon, what is the ratio of the length of its longest diagonal to the length of its shortest diagonal?

    I would easily be able to do this if it were talking about a hexagon, because its so easy to construct equilateral triangles in one, but in an octagon, all the angles and lengths are messed up and .
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  2. #2
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    Quote Originally Posted by DivideBy0 View Post
    In a regular octagon, what is the ratio of the length of its longest diagonal to the length of its shortest diagonal?

    I would easily be able to do this if it were talking about a hexagon, because its so easy to construct equilateral triangles in one, but in an octagon, all the angles and lengths are messed up and .
    Hello,

    I've attached an image of an octagon and its diagonals.

    let the longest diagonal be D = 2r

    then the shortest diagonal is d = √(2r) = r√(2)

    the ratio is: q = 2r/r√(2) = √(2)

    EB
    Attached Thumbnails Attached Thumbnails Diagonals in an Octagon-octagon_diagon.gif  
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  3. #3
    Senior Member DivideBy0's Avatar
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    Doh! Thanks, I tried starting by giving the octagon a side length of 1 and working my way inwards :P

    Say, if you tried with my approach, how would you get the length of the longer diagonal?
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  4. #4
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    Quote Originally Posted by DivideBy0 View Post
    Doh! Thanks, I tried starting by giving the octagon a side length of 1 and working my way inwards :P

    Say, if you tried with my approach, how would you get the length of the longer diagonal?
    Hello,

    in this case you have to use a little bit trigonometrie:

    let s be the length of one side of the octagon.

    Then you deal with 8 isosceles triangle with the base s, the two legs r and the angle at the vertex (that is the center of the circle) is 45.
    Now divide one isoscele triangle into 2 right triangles:
    The hypotenuse is r and one leg is s, the angle opposite of this leg is 22.5.

    Now you can calculate r:
    Code:
     s                         s
    ----- = sin(22.5) ==> r = ----------
      r                         sin(22.5)
    The longest diagonal D = 2r

    Let d be the shortest diagonal. Then d is the height in one of the isoscele triangles:
    Now you can calculate d:
    Code:
     d                        d
    ----- = sin(45) ==> r = ----------
      r                        sin(45)
    By the way: This last equation is exactly the same as the one I've posted in my previous post.

    EB
    Attached Thumbnails Attached Thumbnails Diagonals in an Octagon-octagon_raius.gif  
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