Posers and Puzzles

Posers and Puzzles

  1. Standard memberwolfgang59
    Mr. Wolf
    at home
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    14 Oct '07 14:57
    Here is an easy geometry puzzle for those that cannot do Ranjan's stinkers!

    What is the area of an isoceles triangle with a base of 12 and an apex of 120 degrees.

    Elegant and short proof required. 🙂
  2. Standard memberTheMaster37
    Kupikupopo!
    Out of my mind
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    25 Oct '02
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    20443
    14 Oct '07 16:38
    Originally posted by wolfgang59
    Here is an easy geometry puzzle for those that cannot do Ranjan's stinkers!

    What is the area of an isoceles triangle with a base of 12 and an apex of 120 degrees.

    Elegant and short proof required. 🙂
    Divide the triangle in two equal parts.

    Each of the halves is a 30-60-90 triangle with middle side 6.

    This means that the short side (wich is the height of the original triangle) of the halves is 6/sqrt(3).

    So the area of the original triangle is 1/2 * 12 * 6/sqrt(3) = 12 * sqrt(3)

    Elegant enough?
  3. Standard memberwolfgang59
    Mr. Wolf
    at home
    Joined
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    46124
    14 Oct '07 17:36
    Correct but you have assumed the reader knows ratios of the sides of a 30, 60, 90 triangle.

    I thought a neater idea was to spot that the described triangle is a third of an equilateral triangle of side 12.

    So using your shortcut we have area of equilateral triangle is
    0.5 * 12 * 6sqrt3 = 36sqrt3

    Therefore isoceles is 12sqrt3

    I said it was easy!!
  4. Standard memberTheMaster37
    Kupikupopo!
    Out of my mind
    Joined
    25 Oct '02
    Moves
    20443
    14 Oct '07 19:55
    Originally posted by wolfgang59
    Correct but you have assumed the reader knows ratios of the sides of a 30, 60, 90 triangle.

    I thought a neater idea was to spot that the described triangle is a third of an equilateral triangle of side 12.

    So using your shortcut we have area of equilateral triangle is
    0.5 * 12 * 6sqrt3 = 36sqrt3

    Therefore isoceles is 12sqrt3

    I said it was easy!!
    Ah, that is also a nice solution 🙂
  5. H. T. & E. hte
    Joined
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    3510
    16 Oct '07 15:062 edits
    Originally posted by wolfgang59
    Correct but you have assumed the reader knows ratios of the sides of a 30, 60, 90 triangle.

    I thought a neater idea was to spot that the described triangle is a third of an equilateral triangle of side 12.

    So using your shortcut we have area of equilateral triangle is
    0.5 * 12 * 6sqrt3 = 36sqrt3

    Therefore isoceles is 12sqrt3

    I said it was easy!!
    Really elegant. Also...it can be viewed as being equal to the area of an equilateral triangle of side 12/(sqrt(3)).
    The two halves of this equilateral triangle having been juxtaposed to form the given isosceles triangle with vertex angle 120 deg...
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