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Math Help - Finding area of triangle

  1. #1
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    Smile Finding area of triangle

    I have a problem...I am given a triangle, with different heights, h_{1}, h_{2}, and h_{3}, green, blue, and red respectively.

    With these, I am to find the area of the triangle, but I'm not sure how if I'm not given a base at all. Can anyone help, please? Thank you!

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  2. #2
    Super Member malaygoel's Avatar
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    use Heros formula
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  3. #3
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    No, that doesn't work, I tried. Heron's formula involves using lengths of each side (a, b, c). My teacher said he would like a solution with heights alone, which is why I'm stumped.
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  4. #4
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    Quote Originally Posted by nathan02079 View Post
    No, that doesn't work, I tried. Heron's formula involves using lengths of each side (a, b, c). My teacher said he would like a solution with heights alone, which is why I'm stumped.
    This looks interesting.
    a,b,c are the known perpendicular lengths
    x,y,z are the UNKNOWN base or sides.

    a*x/2 = b*y/2 = c*z/2


    After a little thought:
    x1+x2 = x , base to a
    y1+y2 = y , base to b
    z1+z2 = z , base to c


     a^2 + (x1)^2 = (z1+z2)^2
     a^2 + (x2)^2 = (y1+y2)^2

     b^2 + (y1)^2 = (x1+x2)^2
     b^2 + (y2)^2 = (z1+z2)^2

     c^2 + (z1)^2 = (y1+y2)^2
     c^2 + (z2)^2 = (x1+x2)^2

    Six equations & six unknowns.
    I'll be back.
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  5. #5
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    Just to make sure I'm understanding it correctly...

    are you saying that the a, b, and c (the known perpendicular lengths) in your equation is equivelant to h_{1}, h_{2}, and h_{3}?
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  6. #6
    Super Member malaygoel's Avatar
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    Let us say that the area of triangle is \Delta

    and h_1,h_2,h_3 be the heights
    and a,b,c be the sides of the triangle respectively.

    Then,
    \Delta=\frac{1}{2}ah_{1}=\frac{1}{2}bh_{2}=\frac{1  }{2}ch_{3}

    which gives
    a=\frac{2\Delta}{h_{1}}

    b=\frac{2\Delta}{h_{2}}

    c=\frac{2\Delta}{h_{3}}

    and hence, semi-perimeter,s
    s=\frac{a+b+c}{2}

    s=\Delta(\frac{1}{h_{1}}+\frac{1}{h_{2}}+\frac{1}{  h_{1}})

    Now, substitute these in Heros formula
    Last edited by malaygoel; June 17th 2009 at 09:08 AM. Reason: spellings
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  7. #7
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    Quote Originally Posted by nathan02079 View Post
    Just to make sure I'm understanding it correctly...

    are you saying that the a, b, and c (the known perpendicular lengths) in your equation is equivelant to h_{1}, h_{2}, and h_{3}?
    YES.
    ----------------------------------------------------------

    After a waste of some scrap paper (and some time), the fog surrounding me evaporated. Someone has already solved this.

    "USE GOOGLE" (almost first respone pointed to the instant answer)


    Check here:
    http://mathforum.org/library/drmath/view/61893.html



    Synopsis:

    I =  \left ( \frac {1}{h_1} + \frac {1}{h_2} + \frac{1}{h_3} \right )

    I1 =  \left( \frac{-1}{h_1} + \frac{1}{h_2} + \frac{1}{h_3} \right )

    I2 =  \left ( \frac{1}{h_1} + \frac{-1}{h_2} + \frac{1}{h_3} \right )

    I3 =  \left ( \frac{1}{h_1} + \frac{1}{h_2} + \frac{-1}{h_3} \right )



    AREA_of_TRIANGLE =  \frac{1}{ \sqrt{ I \times I_1 \times I_2 \times I_3} }


    [reason for separation]
    Latex error
    Maximum Latex image size: 600x220
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