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Thread: Pythagoras Question

  1. #1
    Member BobBali's Avatar
    May 2010

    Pythagoras Question

    Hi All, Happy New-Year to everyone.

    I've attached below a diagram of the question. My attempts at it have left me stumped...
    I have to find the length AB and length AB= x

    I started this off by finding length AC using pythagoras and got AC= 5

    Then I deduced that since AB= x, then the small length BC can be written as BC = x-5
    Also, I can note that length CD= a

    [TEX](x-5)^2 = a^2 + 1^2[\TEX]
    [TEX]a^2 = x^2 -10x - 24[\TEX]
    [TEX]a^2 = (x-6)(x-4)[\TEX]
    [TEX]a = \pm\sqrt{6} or \pm\sqrt{4}[\TEX]
    [TEX] a= 2.45 or 2[\TEX]

    So, if CD = 2.45 or 2, then using pthagoras again we can now > calculate
    [TEX]BC = \sqrt{2.45^2 + 1^2} = \sqrt{7}[\TEX]
    Which would result in length AB = 5 + \sqrt{7} = 7.65[\TEX]
    Using CD = 2 with the same above steps gives;
    [TEX]BC = \sqrt{5}[\TEX] and AB = 7.24

    However the answer is;
    [TEX]AB= 4\sqrt{2} or 5.657[\TEX]
    Where have i gone wrong? Thanks for the help.
    Attached Thumbnails Attached Thumbnails Pythagoras Question-triangle-q.png  
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  2. #2
    MHF Contributor ebaines's Avatar
    Jun 2008

    Re: Pythagoras Question

    I think the figure is causing some confusion. First, you have two pouints both lanbeled 'D', and second it's not clear what the 4 cm length applies to. I believe that a corrected version of this figure is per the attached - note that I renamed the right hand D' as D' and that the 4cm length applies to the distance from D to D'. Finding length AB is now a simple use of Pythagoras: its the hypotenuse given two two legs of a right riangle with height 3+1= 4 cm and width 4 cm. That gives you AB = sqrt(4^2 + 4^2) = 4 sqrt(2).

    The error you made are in here:

    $\displaystyle (x-5)^2 = a^2 + 1^2$ OK.
    $\displaystyle a^2 = x^2 -10x - 24$ This should be $\displaystyle a^2 = x^2 - 10x +24$
    $\displaystyle a^2 = (x-6)(x-4)$ OK
    $\displaystyle a = \pm\sqrt{6} or \pm\sqrt{4}$ How did you get this? It dies not come from the previous equation at all - what happened to the 'x' variable?

    BTW, the error in your LaTex expressions is that you have to use [/tex] to close it, not [\tex].
    Attached Thumbnails Attached Thumbnails Pythagoras Question-right_triagles.jpg  
    Thanks from topsquark
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  3. #3
    Forum Admin topsquark's Avatar
    Jan 2006
    Wellsville, NY

    Re: Pythagoras Question

    Here's a shortcut: Have you noticed that the two triangles are similar?

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