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Math Help - X in relation to A on a triangle

  1. #1
    Newbie Critter314's Avatar
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    X in relation to A on a triangle



    I tried using pythagoras but I end up with equation with a lot of radicands. And to remove radicands I have to multiply each side on itself and end up with 6 defree equation.
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  2. #2
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    This is a very confusing question.
    Are you saying that all the 'red' segments are congruent?
    Are you saying the base is 20 units?
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  3. #3
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    Base is 20. A is number. X is number and yes all 3 red lines are same lenght.
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    Code:
    A
    
    
    x
    
          a
    D
    
          x
    
    C    10     B
    Use right half of your triangle; change your A to a.
    Extend vertical red line to base; label as shown in my diagram.
    So we have: a = AB, x = AD = BD, BC = 10 (half the given 20).

    So CD = sqrt(x^2 - 100)

    So a^2 = [x + sqrt(x^2 - 100)]^2 + 100

    A bit of simplifying will lead to: a^2 - 2x^2 = (2x)sqrt(x^2 - 100)
    Square both sides:
    a^4 - 4a^2 x^2 + 4x^4 = 4x^2(x^2 - 100)
    Simplify:
    4a^2 x^2 - 400x^2 - a^4 = 0
    x^2(4a^2 - 400) = a^4

    Take over...

    EDIT:
    above simplifies to: x = a^2 / [2sqrt(a^2 - 100]
    Last edited by Wilmer; May 19th 2011 at 06:48 AM. Reason: Soroban's post!!
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  5. #5
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    Mistaken post
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    Hello, Critter314!

    Base is 20. .A is a number. .X is a number.
    And yes, all 3 red lines are same length.

    If all this is true, the triangle is equilateral.

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  7. #7
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    Disagree Soroban; do you mean isosceles?
    Code:
    D
    
    
    c
    
           d
    B
    
    a      c
    
    C     b      A
    All you need is start with any right triangle ABC; a = BC, b = AC, c = AB.

    Next, extend CB to CD where BD = c; let d = AD

    And cases with integer solutions exist:
    as example, a = 7, b = 24, c = 25;
    triangle ACD: 24-32-40, thus isosceles triangle: 40-40-48

    AND c = d^2 / [2sqrt(d^2 - b^2)]
    With the original problem, b = 10, AD = A and AB = x:
    x = A^2 / [2sqrt(A^2 - 10^2)]
    Last edited by Wilmer; May 19th 2011 at 07:11 AM.
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  8. #8
    Newbie Critter314's Avatar
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    A bit of simplifying will lead to: a^2 - 2x^2 = (2x)sqrt(x^2 - 100)
    What bit of simplifying?
    I tried to simplify and I get a^2-2x^2=2x\cdot(x^2-100)^2-100

    EDIT: Never mind I was wrong.
    Last edited by Critter314; May 24th 2011 at 12:05 AM.
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