# Area of Projection

• Jan 16th 2010, 04:37 AM
reiward
Area of Projection

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.
• Jan 16th 2010, 05:53 AM
earboth
Quote:

Originally Posted by reiward

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:

$a=\frac12\cdot l^2 \cdot \sqrt{3}$
• Jan 16th 2010, 06:19 AM
reiward
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)
• Jan 16th 2010, 06:25 AM
earboth
Quote:

Originally Posted by reiward
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?
• Jan 17th 2010, 12:04 AM
reiward
Hmm, I have a question, is the shape of the projection triangle? Also, what kind of triangle?
• Jan 17th 2010, 01:08 AM
earboth
Quote:

Originally Posted by reiward
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.
• Jan 17th 2010, 08:50 PM
reiward
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)
• Jan 18th 2010, 06:32 AM
HallsofIvy
Quote:

Originally Posted by reiward
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.