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Centroid of non-right triangle

  1. Apr 15, 2013 #1
    1. The problem statement, all variables and given/known data

    See attachment

    2. Relevant equations



    3. The attempt at a solution

    If the y-coordinate of the center of mass is given by (1/A)*∫y dA, how come the solution uses bh/2 as the area? This triangle isn't right angled so that area formula should not hold. What am I missing?
     

    Attached Files:

  2. jcsd
  3. Apr 15, 2013 #2
    The area of any triangle is given by bh/2, not just a right-angled one.
     
  4. Apr 15, 2013 #3
    What he said.

    A right triangle is basically like half of a square, cut down in a diagonal. The area of a square is b*h (or side^2). So a right triangle's area should be bh/2.

    In the case of non-right triangles, this still holds true. Cut a parallelogram diagonally and you get two triangles (just like the square, except without 90 degree angles). The area of a parallelogram is b*h, just like the square's area. Therefore the triangle's area is still bh/2.
     
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