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Math Help - Chords and Arcs

  1. #1
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    Chords and Arcs

    Help please!

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  2. #2
    MHF Contributor alexmahone's Avatar
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    Volume of the cylindrical tank=bh
    =\pi*(\frac{59}{2})^2*470
    =1284314.95 in^3

    Volume of milk= \frac{20}{59}*1284314.95 in^3<br />
                   =435361 in^3

    231 in^3=1 gal
    So, 435361 in^3=1885 gal
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  3. #3
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    Hello, juvenilepunk!

    This requires some trig.


    Look at the circular cross-section . . .

    Code:
                  * * *
              *           *
            *               *
           *                 *
    
          *         O         *
          *         *         *
          *      *  |θ * 29.5 *
              *  9.5|     *
          A*- - - - * - - - -*B
            *:::::::C:::::::*
              *:::::::::::*
                  * * *
                    D

    The circle has radius OA = OB = 29.5 inches.
    The depth of the milk is 20 inches, so OC = 9.5 inches.

    We want the area of segment ADB
    . . which is: .(area of sector OADB) - (area of triangle OAB).


    In right triangle OCB\!:\;\cos\theta \:=\:\frac{9.5}{29.5}\:=\:0.322033898
    . . Hence: . \theta \:\approx\:71.214^o
    And: . \angle AOB \:=\:2\theta \:\approx\:142.43^o

    \text{Area of sector }OADB \:=\:\frac{142.43^o}{360^o}\times \pi(29.5^2) \:=\:1081.665251 in².
    \text{Area of }\Delta AOB \:=\:\frac{1}{2}(29.5^2)\sin142.43^o \:=\:265.30887 in².

    Hence: . \text{Area of segment }ADC \:=\:1081.665251 - 265.39887 \;=\;816.356381 in².


    The volume of the milk is: . 816.356381 \times 470 \;=\;383687.4991 in³.

    Then: . V \;=\;\frac{383687.4991\text{ in}^3}{1} \times \frac{1\text{ gal}}{231\text{ in}^3} \;=\;1660.984844 \;\approx\;1661\text{ gallons}

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  4. #4
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    AlexMahone's answer is incorrect be 20/59 of the diameter would be 20/59 of the volume only if the volume were a linear function of the diameter- and it isn't.
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  5. #5
    MHF Contributor alexmahone's Avatar
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    ^Sorry.

    Soroban is right.
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