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Math Help - Weather Surface of a conical roof

  1. #1
    Dug
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    Weather Surface of a conical roof

    A cylindrical tower 30ft in diameter has a conical roof the length of whose eaves is 2ft. An element of the roof is inclined 45 degrees to the horizontal. Find the weather surface (see figure).


    Answer at the back of the book: 61.061 square yards.

    what i did was,
    i got the radius (2+30)/2
    i used cosine function to get the slant height,
    i used the radius to get the circumference,
    then i got the lateral area of the cone by using LA=1/2(circumference)(slant height)
    the answer is on the unit square feet. so i divided the answer by 3 so that i can get square yards. my answer is very far from the answer on the book.
    i don't know what i did wrong. or is the book wrong? couldn't be.
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  2. #2
    Member HappyJoe's Avatar
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    The length of the eaves is 2 ft. Wouldn't this mean that 2 ft of the roof is beyond the walls? In that case, the radius wouldn't be (2+30)/2, but x + 30/2, where x is the horizontal distance from the tower to the tip of the roof (which would be sqrt(2)).

    To go from unit square feet to unit square yards, you need to divide by 9, not 3. This is because of the squares.
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  3. #3
    Dug
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    Quote Originally Posted by HappyJoe View Post
    The length of the eaves is 2 ft. Wouldn't this mean that 2 ft of the roof is beyond the walls? In that case, the radius wouldn't be (2+30)/2, but x + 30/2, where x is the horizontal distance from the tower to the tip of the roof (which would be sqrt(2)).

    To go from unit square feet to unit square yards, you need to divide by 9, not 3. This is because of the squares.
    i didn't get that part. wouldn't be the length of the diameter of the cone be 30ft(diameter of the tower) + 2 (eaves on each side of the tower) = 32ft?
    i can't visualize what you're talking about sorry and thanks btw.
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  4. #4
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    Hi dug,
    The lateral area of a rt circular cone = pi r l where r= radius of base and l=slant height. You have two cones and from givens you must find the r and l for the larger one. Make a sketch Showing how the two cones overlap. Use the properties of rt isosceles triangles to find the r and l




    bjh
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  5. #5
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    Hello, Dug!

    You're wrong about the eaves.


    A cylindrical tower 30 ft in diameter has a conical roof.
    The length of whose eaves is 2 ft.
    An element of the roof is inclined 45 degrees to the horizontal.
    Find the weather surface.

    Answer at the back of the book: 61.061 square yards.
    Code:
                      A
                      o
                    * | *       _
                  *   |   *  15√2
                *     |15   *
              *       |       *
            *         |         *  B
        C o - - - - - + - - - - - o   2
        * |   15      M      15   | *
      *   |                     E * - o D
          |                       |
          |                       |

    The diameter of the tower is 30 feet: . CB = 30,\;CM = MB = 15

    \text{Angle }ABC = 45^o \quad\Rightarrow\quad AB = 15\sqrt{2}

    The eaves are 2 feet: . B\!D = 2

    Note that: ED = \sqrt{2}


    \text{The radius of the circular base of the cone is: }\:MB + ED \:=\:15 + \sqrt{2}

    \text{The circumference of the circular base is: }\:2\pi(15 + \sqrt{2})


    Flatten the conical roof and we have a major sector of a circle.

    Code:
                  * * *
           P  *           *  Q
            o               o
           *  *           *  *
                *       * R
          *       * O *       *
          *         *         *
          *         @         *
          
           *                 *
            *               *
              *           *
                  * * *
                         _
              s = 2π(15+√2)

    \text{We have the formula: }\:s \:=\:R\theta

    . . \text{where }\:s \:=\:2\pi(15 + \sqrt{2}),\;\;R \:=\:15\sqrt{2} + 2 \:=\:\sqrt{2}(15 + \sqrt{2})


    \text{Then: }\:2\pi(15+\sqrt{2})\:=\:\sqrt{2}(15+\sqrt{2}) \theta

    . . \theta \:=\:\frac{2\pi(15+\sqrt{2})}{\sqrt{2}(15+\sqrt{2}  )} \;=\;\pi\sqrt{2}


    \text{The area of the sector is: }\:A \:=\:\tfrac{1}{2}R^2\theta

    . . \text{where: }\,R \:=\:\sqrt{2}(15+ \sqrt{2}),\;\theta \:=\:\pi\sqrt{2}


    \text{Hence: }\;A \;=\; \tfrac{1}{2}\bigg[\sqrt{2}(15 + \sqrt{2})\bigg]^2(\pi\sqrt{2}) \;=\;1197.029986\text{ ft}^2

    . . \text{Therefore: }\;A \;=\;\frac{1197.029986}{9} \;\approx\;133\text{ yd}^2

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  6. #6
    Dug
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    Quote Originally Posted by Soroban View Post
    Hello, Dug!

    You're wrong about the eaves.



    Code:
                      A
                      o
                    * | *       _
                  *   |   *  15√2
                *     |15   *
              *       |       *
            *         |         *  B
        C o - - - - - + - - - - - o   2
        * |   15      M      15   | *
      *   |                     E * - o D
          |                       |
          |                       |

    The diameter of the tower is 30 feet: . CB = 30,\;CM = MB = 15

    \text{Angle }ABC = 45^o \quad\Rightarrow\quad AB = 15\sqrt{2}

    The eaves are 2 feet: . B\!D = 2

    Note that: ED = \sqrt{2}


    \text{The radius of the circular base of the cone is: }\:MB + ED \:=\:15 + \sqrt{2}

    \text{The circumference of the circular base is: }\:2\pi(15 + \sqrt{2})


    Flatten the conical roof and we have a major sector of a circle.

    Code:
                  * * *
           P  *           *  Q
            o               o
           *  *           *  *
                *       * R
          *       * O *       *
          *         *         *
          *         @         *
          
           *                 *
            *               *
              *           *
                  * * *
                         _
              s = 2π(15+√2)

    \text{We have the formula: }\:s \:=\:R\theta

    . . \text{where }\:s \:=\:2\pi(15 + \sqrt{2}),\;\;R \:=\:15\sqrt{2} + 2 \:=\:\sqrt{2}(15 + \sqrt{2})


    \text{Then: }\:2\pi(15+\sqrt{2})\:=\:\sqrt{2}(15+\sqrt{2}) \theta

    . . \theta \:=\:\frac{2\pi(15+\sqrt{2})}{\sqrt{2}(15+\sqrt{2}  )} \;=\;\pi\sqrt{2}


    \text{The area of the sector is: }\:A \:=\:\tfrac{1}{2}R^2\theta

    . . \text{where: }\,R \:=\:\sqrt{2}(15+ \sqrt{2}),\;\theta \:=\:\pi\sqrt{2}


    \text{Hence: }\;A \;=\; \tfrac{1}{2}\bigg[\sqrt{2}(15 + \sqrt{2})\bigg]^2(\pi\sqrt{2}) \;=\;1197.029986\text{ ft}^2

    . . \text{Therefore: }\;A \;=\;\frac{1197.029986}{9} \;\approx\;133\text{ yd}^2


    The answer at the back of the book is 61.061 square yards.
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  7. #7
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    Hi Dug,
    I also calculate lateral area is 133 sq yds
    s =pi r l=pi (15+rad2)(15rad2+2)=1197 sq ft = 133 sq yds


    bjh
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  8. #8
    Dug
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    Thank you. I also double checked by calculating. The answer is 133yds.
    The book must be wrong then. (typo error?)
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