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Math Help - Area of Projection

  1. #1
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    Area of Projection

    Help please

    A flat roof has an inclination of 30 degrees to the floor of a building. The roof is square, 20m long (being parallel to the floor, 20m wide. Find the area of the projection of roof upon the floor of the building.
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  2. #2
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    Quote Originally Posted by reiward View Post
    Help please

    A flat roof has an inclination of 30 degrees to the floor of a building. The roof is square, 20m long (being parallel to the floor, 20m wide. Find the area of the projection of roof upon the floor of the building.
    1. Draw a rough sketch (see attachment)

    2. In the right triangle the width of the floor is the adjacent leg of the angle of 30.

    3. Let l denote the length of the square side. Then the area of the projektion of the roof upon the floor is:

    a = l \cdot l\cdot \cos(30^\circ)

    With \cos(30^\circ) = \frac12 \cdot \sqrt{3} the area becomes:

    a=\frac12\cdot l^2 \cdot \sqrt{3}
    Attached Thumbnails Attached Thumbnails Area of Projection-vert_projektion.png  
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    I dont get it. What is the value of L? I think its harder for me to visualize the image because it has been thought to us differently. Something like the one I attached. Can you explain it with my illustration? (illustration is bad, can someone fix it with the same idea, I just dont know where to put 30 degrees)
    Attached Thumbnails Attached Thumbnails Area of Projection-untitled.bmp  
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  4. #4
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    Quote Originally Posted by reiward View Post
    I dont get it. What is the value of L? I think its harder for me to visualize the image because it has been thought to us differently. Something like the one I attached. Can you explain it with my illustration? (illustration is bad, can someone fix it with the same idea, I just dont know where to put 30 degrees)
    1. With your example l = 20 m.

    2. I've some difficulties to recognize the angles and the orientation of the 2 planes. Which lines are invisible because they are hidden by an area in the foreground?
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    Hmm, I have a question, is the shape of the projection triangle? Also, what kind of triangle?
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  6. #6
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    Quote Originally Posted by reiward View Post
    Hmm, I have a question, is the shape of the projection triangle? Also, what kind of triangle?
    I've added some labels to make clear how I have understood your question. Hope that helps a little bit further.
    Attached Thumbnails Attached Thumbnails Area of Projection-vert_projektion.png  
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  7. #7
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    Hmm, I still dont understand it. Uhmm can you illustrate it in how we did it in class(see attachment).

    (The one attached has a roof 50m long and 20m wide)
    Attached Thumbnails Attached Thumbnails Area of Projection-sample.jpg  
    Last edited by reiward; January 18th 2010 at 12:56 AM.
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    Quote Originally Posted by reiward View Post
    Hmm, I still dont understand it. Uhmm can you illustrate it in how we did it in class(see attachment).

    (The one attached has a roof 50m long and 20m wide)
    That's essentially the same picture earboth showed (except that his is larger and clearer!). The point is that the sloping surface makes an angle of 30 degrees with any horizontal line (such as the floor) and so makes a right triangle with hypotenuse of 20 m and angle 30 degrees.

    The base leg, the projection onto the floor and the "near side" in the triangle, has length 20 cos(30). The other length is parallel to the floor and so its projection onto the floor is 20. the projection onto the floor is a rectangle with length 20 cos(30) and width 20 so its area is (20)(20 cos(30)), exactly earboth's formula.
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