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Area Of Paralellogram

  1. Aug 5, 2009 #1
    What are the formulas to find out the area of a parallelogram? Yep, that's as much detail I can put in.
     
  2. jcsd
  3. Aug 5, 2009 #2

    CRGreathouse

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    Height times width. In this picture, b * h:
    parallelogram_1.gif
     
  4. Aug 5, 2009 #3
    A = b*h

    A - Area
    b - length of base
    h - height
     
  5. Aug 5, 2009 #4
    Thanks. Is that it? wow its pretty simple.
     
  6. Aug 5, 2009 #5
    Can you derive this formula though? Try breaking up the parallelogram into shapes that you know the area of (i.e. rectangle and right triangles), and then add up those areas to see if you get base*height for your area equation. This should be good practice for you.
     
  7. Aug 5, 2009 #6
    Actually, the parallelogram is from a graph which means I should know the coordinates. Oh and I just forgot! I have to solve it by algebraic means.
     
  8. Aug 5, 2009 #7
    Last edited: Aug 5, 2009
  9. Aug 6, 2009 #8
    So its ad-bc then eh? By looking at the diagram on wiki. Can a,b,c & d be x and y values? Or am I completely off track?
     
  10. Aug 6, 2009 #9
    oh no wait! do they represent vectors?
     
  11. Aug 6, 2009 #10
    They are vectors.
     
  12. Aug 6, 2009 #11
    If you know the lengths of two adjacent sides of the parallelogram and the value of any of its angles. Then the area of the parallelogram is ab*sin(theta), where a and b are the adjacent sides and theta is the angle.
     
  13. Aug 7, 2009 #12
    thanks, I just hope I get it right.
     
  14. Aug 7, 2009 #13
    hang on, why don't I just post the question from the assignment.
    "Find the area enclosed by the four tangents by geometrical and algebraic means."
     
  15. Aug 8, 2009 #14

    rock.freak667

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    what 4 tangents now?
     
  16. Aug 8, 2009 #15
    from the linear graph i made.
     
  17. Aug 8, 2009 #16

    rock.freak667

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    then you can find the coordinates of each corner of the parallelogram and use any of the above methods posted.
     
  18. Aug 8, 2009 #17
    And that's using geometrical and algebraic methods yeah? thanks.
     
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