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Math Help - Triangle geometry problem

  1. #1
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    Triangle geometry problem

    Please can somebody help me with the following problem. I am trying to find x in terms of a,b and c (as per the attached diagram). I came up with the equation below, but this has turned out quite complicated to resolve. Can anyone think of a simpler solution?

    Many thanks.

    x^2 = b^2 + (c - (ac/(x+a)))^2
    Attached Thumbnails Attached Thumbnails Triangle geometry problem-pic.jpg  
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  2. #2
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    Are these right-angle triangles?
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  3. #3
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    Yes, they're right angle triangles.
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  4. #4
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    May be easier by setting up an "example case"; say extending the 3-4-5 right triangle:
    Code:
    B
    
    6   10
    
    E   8    D
    
    
    9        9     15
    
    
    C   8    F     12       A
    I'll complicate your life by changing your variables to different letters!

    Start with right triangle ABC; a = BC = 15, b = AC = 20, c = AB = 25
    Draw line ED such that similar right triangle DBE is formed; e = BE = 6, d = DE = 8, f = DB = 10
    Similar right triangle ADF also is formed; DF = a-e = 9, AF = b-d = 12, AD = c-f (your x) = 15

    Givens are BC = a = 15, DB = f = 10 and AF = b-d = 12
    PROBLEM: what is (c-f) in terms of a, f and (b-d) ?

    I'll try this out later...
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  5. #5
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    Quote Originally Posted by EdCoventry View Post
    x^2 = b^2 + (c - (ac/(x+a)))^2
    Well, tried and tried...but MESSY it is!!
    That equation of yours (quite correct) can be simplified to:
    x^2 = b^2 + [cx / (a + x)]^2
    However, I could not find a way to end up with anything simpler than:
    x^4 + 2a x^3 + (a^2 - b^2 - c^2) x^2 - 2ab^2 x - a^2 b^2 = 0

    Using my example in previous post (and switching to match your variables),
    we have a=10, b=12 and c=15; and x=15.

    Long division by x - 15 gives:
    x^3 + 35x^2 + 256x + 960; so we have:
    (x - 15)(x^3 + 35x^2 + 256x + 960) = 0 ; yes, of little use!

    Funny how something "appearing so simple" (similar right triangles et al...)
    turns out like this; a bit like requiring Heron's area of triangle formula for
    a triangle that can be divided in 2 right triangles by a height line...

    Enough to make a Saint use all the French Quebec swear words
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