a polygon with n number of sides will have how many digonals




d) n-3/2


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  1. hint: recall that 1+2+...+k = k(k+1)/2
    Now start counting your diagonals for an n-gon.

  2. c

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    posted by Candice
  3. c does not work if n is odd.
    Also, sorry about my hint. That does not apply here.

    There are n vertices.
    From each vertex, you can draw n-3 diagonals. (don't count your vertex or its two neighbors)
    So, there are n*(n-3) diagonals to draw, using each vertex as a base.
    But, now you have drawn each one twice, from both ends.
    So the answer is E: n(n-3)/2

  4. or, think of it this way:
    The number of possible lines between n points = C(n,2)
    but n of these are sides of the polygon, the rest are diagonals
    So, the number of diagonals
    = C(n,2) - n
    = n!((n-2)!2!) - n
    = n(n-1)/2 - 2n/2
    = (n^2 - n - 2n)/2
    = n(n-3)/2

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