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Math Help - Algebra problem

  1. #1
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    Exclamation Algebra problem

    A friend recently emailed me this one (apparently it's from a game he plays)

    AB=81, AD=180, AE=108, BE=135,CD=84, CE=156, BC=square root of ?

    I've been poring over this for a bit, but I'm obviously out of practice.

    Anyone care to have a bash?
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  2. #2
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    Question

    Quote Originally Posted by Davey Jones View Post
    A friend recently emailed me this one (apparently it's from a game he plays)

    AB=81, AD=180, AE=108, BE=135,CD=84, CE=156, BC=square root of ?

    I've been poring over this for a bit, but I'm obviously out of practice.

    Anyone care to have a bash?
    is that addition or multiplication?

    AB= A*B
    AB=A+B
    Last edited by princess_21; February 3rd 2009 at 09:16 PM.
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  3. #3
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    \frac{AE}{BE} = \frac{108}{135} = \frac{A}{B}

    \frac{A}{B} \times AB = \frac{108}{135} \times 81 = \frac{324}{25} = A^2

    So A = \frac{18}{\sqrt{5}}

    AD = 180, so D = \frac{180}{A} = 10\sqrt{5}

    CD = 84, so C = \frac{84}{D} = \frac{84}{10\sqrt{5}} = \frac{42}{5\sqrt{5}}

    AB = 81, so B = \frac{81}{A} = \frac{81\sqrt{5}}{18} = \frac{9\sqrt{5}}{2}

    \therefore BC = \frac{9\sqrt{5}}{2} \times \frac{42}{5\sqrt{5}} = \frac{189}{5}
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  4. #4
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    Quote Originally Posted by nzmathman View Post
    \frac{AE}{BE} = \frac{108}{135} = \frac{A}{B}

    \frac{A}{B} \times AB = \frac{108}{135} \times 81 = \frac{324}{25} = A^2

    So A = \frac{18}{\sqrt{5}}

    AD = 180, so D = \frac{180}{A} = 10\sqrt{5}

    CD = 84, so C = \frac{84}{D} = \frac{84}{10\sqrt{5}} = \frac{42}{5\sqrt{5}}

    AB = 81, so B = \frac{81}{A} = \frac{81\sqrt{5}}{18} = \frac{9\sqrt{5}}{2}

    \therefore BC = \frac{9\sqrt{5}}{2} \times \frac{42}{5\sqrt{5}} = \frac{189}{5}

    why did u not squared \frac{189}{5} because it is said that BC=is the squared of ?

    i did my own calculations i think i'm lost.

    ABCD=ABCD

    BC= x

    81*84=180*\sqrt{x}

    6804=180\sqrt{x}

    \frac{6804}{180}=\sqrt{x}

    (\frac{189}{5})^{2}=(\sqrt{x})^{2}

    x=\frac{35721}{25}
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  5. #5
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    Hello, Davey Jones!

    Is there a typo?
    The equations are inconsistent . . .


    A friend recently emailed me this one (apparently it's from a game he plays)

    . . \begin{array}{cccc}(1)&AB&=&81 \\ (2)&AD&=&180 \\ (3)& AE&=&108 \\ <br />
(4) & BE&=&135 \\ (5)&CD&=&84 \\ (6)& CE&=&156\\ \\ & BC&=& \sqrt{??} \end{array}

    . . \begin{array}{cccc}\text{Divide (5) by (2):} & \dfrac{CD}{AD} \:=\:\dfrac{84}{180} &\Rightarrow& \dfrac{C}{A} \:=\:\dfrac{7}{15} \\ \\[-2mm]<br />
\text{Divide (6) by (3):} & \dfrac{CE}{AE} \:=\:\dfrac{156}{108} &\Rightarrow& \dfrac{C}{A}\:=\:\dfrac{13}{9} \end{array} . ??

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  6. #6
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    Quote Originally Posted by Davey Jones View Post
    A friend recently emailed me this one (apparently it's from a game he plays)

    AB=81, AD=180, AE=108, BE=135,CD=84, CE=156, BC=square root of ?

    I've been poring over this for a bit, but I'm obviously out of practice.

    Anyone care to have a bash?
    The way that I interpret this question, A, B, C, D, E are meant to be points in a plane, and AB, ..., are the distances between these points.

    If that is the case, then ABE is a right-angled triangle, because 81^2+108^2=135^2. However, the remaining information is not sufficient to fix the positions of C and D. You need an extra piece of information, such as the distance DE, before you can pin down the distance BC.
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  7. #7
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    There was a typo sorry!

    CD = 60
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  8. #8
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    Even CD=60, the equations are stil not consistent
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