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Math Help - Find the area in contact with water

  1. #1
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    Find the area in contact with water

    Diagram IV show the cross-section of the trough after it is filled with water. The cross-section has centre O and diameter 1.2m. From diagram IV, determine the area that is in contact with the water.

    Last edited by Punch; October 8th 2010 at 07:33 AM.
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  2. #2
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    Hello, Punch!

    Some information is missing.


    The diagram shows the cross-section of the trough partially filled with water.
    The cross-section has centre O and diameter 1.2 m.
    Determine the area of the side that is in contact with the water.


    Code:
      :     0.6     O     0.6     :
    A o ------------*-------------o B
      *          *  |  *          *
              *     |0.3  * 0.6
       *   *       E|        *   *
      C o-----------o-----------o D
          *         |0.3      *
             *      |     *
                  * o *
                    F

    The cross-section is a semicircle with radius:
    . . OA = OB = OC = OD = OF = 0.6

    The trough has water to a height of: . EF = 0.3 \quad\Rightarrow\quad OE = 0.3


    Right triangle OED has: . OE = 0.3,\;OD = 0.6

    . . \cos(\angle DOE) \:=\:\dfrac{0.3}{0.6} \:=\:0.5 \quad\Rightarrow\quad \angle DOE = 60^o = \dfrac{\pi}{3}

    Hence: . \angle COD \:=\:\dfrac{2\pi}{3}


    Length of arc: . s \:=\:r\theta
    We have: . \text{arc(CFD)} \;=\;(0.6)\left(\dfrac{2\pi}{3}\right) \:=\:0.4\pi


    \text{Therefore: }\;\text{Area of side} \;=\;(0.4\pi) \times \underbrace{\text{(length of trough)}}_{\text{not given}}

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  3. #3
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    Hi Soroban

    I found a way to solve this question!!

    I shall use your code to explain
    Code:
     :     0.6     O     0.6     :
    A o ------------*-------------o B
      *          *  |  *          *
              *     |0.3  * 0.6
       *   *       E|        *   *
      C o-----------o-----------o D
          *         |0.3      *
             *      |     *
                  * o *
                    F
    Find the area of triangle AOC,ODB and OCD.

    The sum of area for the three triangles will give the total area not in contact with water.

    Next, find the area of the whole semi circle.

    Substract the area of the three triangles from the area of the whole semi circle.

    U r done!!!!
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  4. #4
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    Quote Originally Posted by Punch View Post
    Hi Soroban

    I found a way to solve this question!! <=== NO.

    I shall use your code to explain
    Code:
     :     0.6     O     0.6     :
    A o ------------*-------------o B
      *          *  |  *          *
              *     |0.3  * 0.6
       *   *       E|        *   *
      C o-----------o-----------o D
          *         |0.3      *
             *      |     *
                  * o *
                    F
    Find the area of triangle AOC,ODB and OCD. <=== AOC and ODB are sectors and not triangles

    The sum of area for the three triangles will give the total area not in contact with water. <=== nearly (see above) but of the cross-section

    Next, find the area of the whole semi circle.

    Substract the area of the three triangles from the area of the whole semi circle.

    U r done!!!!
    According to your consideration you haven't understood your own question. Read Soroban's post again to see where you are mistaken.
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  5. #5
    Senior Member Educated's Avatar
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    I do not see what is confusing about this question or what information is missing so please tell me what I did wrong:

    Using Sorobans code:
    Code:
     :     0.6     O     0.6     :
    A o ------------*-------------o B
      *          *  |  *          *
              *     |0.3  * 0.6
       *   *       E|        *   *
      C o-----------o-----------o D
          *         |0.3      *
             *      |     *
                  * o *
                    F
    \angle COD = \frac {2 \pi}{3} = 120^{\circ}

    \text{area of sector COD} = \pi \cdot 0.6^2 \cdot \frac{120}{360} \approx 0.37699m^2

    \text{area of triangle COD} = \frac{1}{2} \cdot 0.6 \cdot 0.6 \cdot \sin 120^{\circ} \approx 0.15588m^2

    \text{Area in contact with water} = 0.37699m^2 - 0.15588m^2 = 0.2211m^2

    Did I do anything wrong?
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  6. #6
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    Quote Originally Posted by Educated View Post
    I do not see what is confusing about this question or what information is missing so please tell me what I did wrong:

    Using Sorobans code:
    Code:
     :     0.6     O     0.6     :
    A o ------------*-------------o B
      *          *  |  *          *
              *     |0.3  * 0.6
       *   *       E|        *   *
      C o-----------o-----------o D
          *         |0.3      *
             *      |     *
                  * o *
                    F
    \angle COD = \frac {2 \pi}{3} = 120^{\circ}

    \text{area of sector COD} = \pi \cdot 0.6^2 \cdot \frac{120}{360} \approx 0.37699m^2

    \text{area of triangle COD} = \frac{1}{2} \cdot 0.6 \cdot 0.6 \cdot \sin 120^{\circ} \approx 0.15588m^2

    \text{Area in contact with water} = 0.37699m^2 - 0.15588m^2 = 0.2211m^2

    Did I do anything wrong?
    1. You calculated the area of the waterfilled cross-section.

    2. According to the text of the question you should calculate the area of the trough (not it's cross-section) which is in contact with the water. (see attachment)
    Attached Thumbnails Attached Thumbnails Find the area in contact with water-watercontact.png  
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  7. #7
    Senior Member Educated's Avatar
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    Ah, I see. Sorry about that I misunderstood the question, and interpreted it completely wrong.

    Too many exam questions that ask for the area of the cross-sections.

    EDIT: Maybe it's a spherical trough?
    Last edited by Educated; October 9th 2010 at 01:44 AM.
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  8. #8
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    Quote Originally Posted by earboth View Post
    1. You calculated the area of the waterfilled cross-section.

    2. According to the text of the question you should calculate the area of the trough (not it's cross-section) which is in contact with the water. (see attachment)
    It was probably a mistake on my part in posting an "edited" question. The question is... determine the area of the side that is in contact with the water.

    Educated should be right
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