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Math Help - need help.

  1. #1
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    need help.

    A long rope is pulled out between two opposite shores of a lake. It's pulled so tight that it's perfectly straight.
    Because the earth is spherical most of the rope is under water.
    The length of the portion of rope that is under water is 70 km long.
    How many meters below the surface is the rope at its deepest point?
    The earths radius is assumed to be 6370 km.
    Attached Thumbnails Attached Thumbnails need help.-lake.gif  
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  2. #2
    Super Member

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    Hello, Bobby!

    A long rope is pulled out between two opposite shores of a lake.
    It's pulled so tight that it's perfectly straight.
    Because the earth is spherical most of the rope is under water.
    The length of the portion of rope that is under water is 70 km long.
    How many meters below the surface is the rope at its deepest point?
    The earths radius is assumed to be 6370 km.
    Code:
                     C
                   * * *
               *     :x    *
            *        :   35   *
        A * - - - - -+- - - - - * B
            \       D:        /
              \      :R-x   /
              R \    :    / R
                  \  :  /
                    \:/
                     *
                     O

    The center of the earth is O. . OA = OB = OC = R (radius of the earth).

    The 70-km rope is AB. .We see that: AD = DB = 35.

    Let x = CD be the distance the rope is underwater at its center.
    Then DO = R - x.


    From right triangle ODB:\;\;DO^2 + DB^2\:=\:OB^2

    So we have: . (R - x)^2 + 35^2\:=\:R^2\quad\Rightarrow\quad x^2 - 2Rx + 1225\;=\;0

    Quadratic Formula: . x\;=\;\frac{2R \pm\sqrt{4R^2 - 4900}}{2}\;=\;R \pm \sqrt{R^2 - 1225}


    Since R = 6370, we have: . x \;= \;6370 \pm \sqrt{6370^2 - 1225}

    Therefore: . x\;=\;0.096154575\text{ km}\;\approx\;96.2\text{ meters.}

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  3. #3
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
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    Quote Originally Posted by Soroban
    Hello, Bobby!

    Code:
                     C
                   * * *
               *     :x    *
            *        :   35   *
        A * - - - - -+- - - - - * B
            \       D:        /
              \      :R-x   /
              R \    :    / R
                  \  :  /
                    \:/
                     *
                     O

    The center of the earth is O. . OA = OB = OC = R (radius of the earth).

    The 70-km rope is AB. .We see that: AD = DB = 35.

    Let x = CD be the distance the rope is underwater at its center.
    Then DO = R - x.


    From right triangle ODB:\;\;DO^2 + DB^2\:=\:OB^2

    So we have: . (R - x)^2 + 35^2\:=\:R^2\quad\Rightarrow\quad x^2 - 2Rx + 1225\;=\;0

    Quadratic Formula: . x\;=\;\frac{2R \pm\sqrt{4R^2 - 4900}}{2}\;=\;R \pm \sqrt{R^2 - 1225}


    Since R = 6370, we have: . x \;= \;6370 \pm \sqrt{6370^2 - 1225}

    Therefore: . x\;=\;0.096154575\text{ km}\;\approx\;96.2\text{ meters.}

    Hello Soroban.
    We could do it in this way also.
    Extend CO to meet the circle at E
    Now,
    AB and CE are two chords of a cicle intresecting at D
    Hence, AD.DB = CD.DE
    35.35 = x(DO + OE)
    1225 = x(2R - x)

    KeepSmiling

    Malay
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  4. #4
    Super Member

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    Very nice, Malay!

    An excellent application of that intersecting-chords theorem
    . . . which I obviously forgot.
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