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Parallelograms projection on a plain

  1. Oct 14, 2011 #1
    1. The problem statement, all variables and given/known data

    Parallelograms 5cm side is located on a plain alpha and its 3cm side forms an angle with the plain equal to 30 degrees. Find the area for the parallelograms projection on the plain alpha, if an angle in the original parallelogram is 60 degrees.

    2. Relevant equations

    all kinds

    3. The attempt at a solution

    I think Im visualizing something wrong. Because what I did gave me an answer or 11,25 cm2, but the answer is supposed to be 10,6cm2 according to the textbook.

    Here's what I did:

    Find the length of the 3cm side's projection on the plain: d=3cm*cos30=3*sqrt3/2 cm
    Directly find the area of the projection with 2 sides of the projection and the angle between them: S=5*d*sin60=5*3*sqrt3*sqrt3/2*2=11,25 cm2

    Where did I go wrong?

    Thanks in advance,
    fawk3s
     
    Last edited: Oct 14, 2011
  2. jcsd
  3. Oct 14, 2011 #2
    I got the same answer as you. *Shrug*
     
  4. Oct 14, 2011 #3
    I just happened to figure it out. I have to go out at the moment, but incase you are interested in the solution dacruick, I can post it up later on.
     
  5. Oct 14, 2011 #4
    yes I'd like to see :)
     
  6. Oct 14, 2011 #5
    [PLAIN]http://img267.imageshack.us/img267/1040/mathlol.png [Broken]

    The important thing to realize is that β is not equal to 60o.
    Think of it this way: the side a of the parallelogram which is not on the plain alpha, is shifted by d relative to the side a which IS on the plain. As is its projection, no matter what the angle between the parallelogram and the plain.

    d=b*cos60o=3*cos60o=1,5 cm

    Since we know that the angle between b and the plain is 30o (which I accidently forgot to add in the drawing), we can find b's projection. Let it be b'.

    b'=cos30o*b=3*sqrt(3)/2

    We can now find β, because cosβ=d/b'. So β=54,74o (approx.)
    Now we can find the area

    S=a*b*sinβ=10,606601...
     
    Last edited by a moderator: May 5, 2017
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