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Math Help - Rectangle breakdown

  1. #1
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    Rectangle breakdown

    Code:
    A                                   E           B
    
    
    
    
                                                    F
    
    
    
    
    
    
    D                           G                   C
    Rectangle ABCD.
    Givens: AD = 180, DG = GF = 160, angle GFE = 90 degrees, angle AFE = angle BFE.
    Find lengths of CF and CG.

    Been playing around with that one for a while; can't wrap it up!
    It's evident that angle AFG = angle CFG and triangle BEF is similar to triangle CGF.

    Was able to bruteforce solution: CF = 96 and CG = 128.

    Can anyone see a way to solve this directly?
    Thanks muchly.
    Last edited by Wilmer; December 7th 2013 at 11:00 AM.
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  2. #2
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    Re: Rectangle breakdown

    Hey Wilmer.

    As an exercise the first thing you should do in these kinds of situations is draw a diagram with every piece of information involved and then convert those to expressions and/or equations. I think it's fair to ask you to do that first and then show us where you went wrong or had trouble.

    By creating a graph with all the angles and lengths along with a line by line set of mathematical expressions/equalities, it will not only help you but be easier for us as well.

    Please understand that sometimes when people are tired, having these things already prepared in advance will help us and show that you have put in effort. This is not directed at you per se, but it is a general statement that applies to all posters and I am not taking any shots at you in any way.
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  3. #3
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    Re: Rectangle breakdown

    I know all that stuff Chiro. I'm a volunteer helper here, and at other sites.
    I guess (since I solved it by brute force) that I could put it up as a challenge.

    Anyway, thanks for replying.

    Anybody else interested?
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  4. #4
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    Re: Rectangle breakdown (SOLVED)

    OK, just realised that there is no unique solution (another information required in order to be unique).
    I was getting nowhere due to using INTEGERS.
    Line GF can "rotate" (within a limit) thus creating multiple solutions.
    By limit, I means cannot be positionned such that line FE has "point E" left of point A,
    even if that is a valid solution, since not prohibited in the problem statement.

    So ya'll can quit working on it(!), many thanks, particularly to my friend Chiro
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  5. #5
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    Re: Rectangle breakdown

    Finally got an equation!

    a = DC, b = AD, x = CF

    x^2 - aSQRT(a^2 - x^2) + xSQRT(b^2 + x^2) + bx -a^2 = 0
    or:
    x^2 + x[b + SQRT(b^2 + x^2)] - a[a + SQRT(a^2 - x^2)] = 0

    If a = 160 and b = 180 then x = 96

    solve x^2+x*(180+sqrt(180^2+x^2))-160*(160+sqrt(160^2-x^2))=0 for x - Wolfram|Alpha

    Thank you, Sir Wolfram
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