An Easter Conundrum

Discussion in 'Math' started by studiot, Apr 8, 2009.

  1. studiot

    Thread Starter AAC Fanatic!

    Nov 9, 2007
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    Most folks have no trouble doing a 3-4-5 triangle, however there are other interesting ones some even without right angles.

    None of the sides in the obtuse triangle in the sketch are whole numbers, yet each forms a whole number when squared. So the squares on each side have whole number areas.

    What is the area of the triangle itself?

    Is it also a whole number?
     
  2. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
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    28 using Heron's formula.

    John
     
  3. jasperthecat

    Member

    Mar 26, 2009
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    From your diagram

    the lengths of the sides are Sqrt(41), Sqrt(97) and 14

    using

    area = sqrt(s*(s-a)*(s-b)*(s-c))

    where a,b,c are sides and s = semi perimeter

    I'd be surprised if the area is a whole number
     
  4. jasperthecat

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    Mar 26, 2009
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    My surprise for the day
     
  5. Ratch

    New Member

    Mar 20, 2007
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    To the Ineffable All,

    A simpler way is to drop a perpendicular line from the top vertice to the base. Then solving x^2+y^2 = 41, (14-x)^2+y^2 = 97 , where x is the distance from the left vertice to the perpendicular line and y is the altitude of the triangle. Solving, we easily get y=4 . So (1/2)*(14)*4 = 28

    Ratch
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    The answer I obtained was 28.

    hgmjr
     
    Last edited: Apr 8, 2009
  7. zgozvrm

    Member

    Oct 24, 2009
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    Whoops!

    If you meant for the sides of the triangle to NOT be whole numbers in order to make the problem more difficult, you failed ... you have the square on the bottom of the triangle as having an area of 196 sq. m. The square root of 196 is 14 which IS a whole number!
     
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