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Math Help - Finding a solution for non-linear simultaneous equations

  1. #1
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    Manchester, England
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    Finding a solution for non-linear simultaneous equations

    Hi

    I am not sure if the title accurately reflects my issue, apologies if this is not the case.

    I am an academic engineer, I have encountered a problem where i would like to equate a circle to a triangle to simplfy a problem..... My criteria is to transform a circle to a 120, 30, 30 triangle with equivalent area and perimeter, when i quickly derived the simultaeous equations i get the following two expressions

    y1 = b+2*h
    y2=(b*h)/2

    where:
    y1=the circles perimeter
    y2=the circles area
    b= base of triangle
    h= height of triangle

    I thought i would ask as i do not want to waste time if it is not possible to solve and satisfy the 120, 30, 30 condition......... Are there any methods that can be used to tackle this issue?
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  2. #2
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    Tejas
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    Re: Finding a solution for non-linear simultaneous equations

    if b,h are given, y_1 and y_2 may simply be calculated. so i assume that y_1, y_2 are given, and you wish to find b and h.

    write: b = y_1 - 2h = \frac{2y_2}{h}

    from y_1 - 2h = \frac{2y_2}{h} we get:

    2h^2 - y_1h - 2y_2 = 0, which is a quadratic in h.

    the discriminant is y_1^2 - 8y_2

    however, y_1,y_2 aren't "independent" they both depend on the radius of the circle, r:

    y_1 = 2\pi r
    y_2 = \pi r^2

    so y_1^2 - 8y_2 = (2\pi r)^2 - 8\pi r^2 = -4\pi r^2 < 0, therefore: no real solutions for h.

    however, in reviewing your original equations, it appears that your formula for either the area or the perimeter of the isoceles triangle is incorrect, if h is the height, then the perimeter length should be:

    b + 2(\sqrt{h^2 + \frac{b^2}{4}}).

    you might want to review that, and confirm (the calcuations get a lot messier if this is true).
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  3. #3
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    Manchester, England
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    Re: Finding a solution for non-linear simultaneous equations

    Thanks, I did make a couple of mistakes actually, sorry about that, the equations should be

    y1 = b+2*(sqrt(b^2+h^2))
    y2=(b*h)/2

    where:
    y1=the circle z's perimeter (GIVEN)
    y2=the circle z's area (GIVEN)
    b= base of triangle
    h= height of triangle


    I want to calculate values for b, and h for a 120, 30, 30 triangle, my main question is there a solution? i.e. can you take a circle and construct a 120, 30, 30 triangle with an identicle Area and perimeter?
    Last edited by malby; March 6th 2012 at 04:32 AM.
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