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Volume of pyramid

  1. Mar 19, 2011 #1
    1. The problem statement, all variables and given/known data
    The volume of a pyramid can be evaluated by using the equation V=1/3*A*h where h is
    the height of the pyramid and A is the area of the base of the pyramid. The problem is
    to design such a triangular pyramid where the volume is V = 72 cm3 and the height is
    h = 12 cm. The base of the pyramid is a rectangular isosceles triangle.


    2. Relevant equations
    a) Determine the area of the base of the pyramid. Pay attention to the unit of your answer.
    b) Draw a picture of the base triangle of the pyramid in a proper scale
    3. The attempt at a solution
    a)
    V=1/3*A*h
    A=72/(12/3)=18cm2

    b) A = 18 = b*y
    b=y
    so: 18=2*b => b = 9cm
     

    Attached Files:

  2. jcsd
  3. Mar 19, 2011 #2

    LCKurtz

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    Your picture didn't come through, but I'm guessing your formula A = 18 = b*y is wrong. Remember the area of a triangle is (1/2)base * height. And even if it were correct, b = y wouldn't give you 18 = 2b, it would be 18 = b2. Two things to fix.
     
  4. Mar 20, 2011 #3

    HallsofIvy

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    Well, I can see your picture and it isn't even of a triangle!

    Did you simply misread the problem?
     
  5. Mar 20, 2011 #4
    Yes i did :redface: my mistake, i thought it's a pyramid with 4 faces..

    but then we have the area of a triangle that equal 18m2
    18=(b*y)/2
    To be able to draw that triangle we need to find the length of just one side (assuming the sides are equal) and the problem is..we don't have that
     

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  6. Mar 22, 2011 #5
    some please give me a hint
     
  7. Mar 22, 2011 #6

    LCKurtz

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    You still haven't drawn a picture of your pyramid. When you do I think you will find what you need in my earlier post. (reply #2).
     
  8. Mar 22, 2011 #7
    i'm not good in drawing but i have an image of it in my mind..and i can't see how i will find the length of the side of that triangle
     
  9. Mar 22, 2011 #8

    LCKurtz

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    Earlier you had that V = (1/3)Ah and you are given that V = 72 and h = 12.

    So A = 3V/h = 3*72/12 = 18.

    You had that a long time ago. So the area of your isosceles "rectangular" triangle is 18. I presume you know what isosceles means and I presume that "rectangular" triangle means what is usually called a right triangle. And as I pointed out in post #2, the area of a triangle is (1/2)*base* height. What exactly are you stuck on, given you have all this information?
     
  10. Mar 22, 2011 #9
    how you know that the triangle is isosceles ? isn't suppose to be equilateral ?
     
  11. Mar 22, 2011 #10

    LCKurtz

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    Didn't you read your own statement of the problem?
     
  12. Mar 22, 2011 #11
    ok i must be blind :/
    but still we can't find the dimension of that triangle..it's about the triangle and we need the height of the triangle, not the pyramid
     
  13. Mar 22, 2011 #12

    LCKurtz

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    I asked you before if "rectangular" triangle means right triangle. You didn't answer that, but I assume it does. You have an isosceles right triangle and you know its area is 18. Draw a picture of it. You should be able to figure out its dimensions from the picture.
     
  14. Mar 23, 2011 #13
    here it is...
    i guess that the base of this triangle = it's height.
     

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  15. Mar 23, 2011 #14

    HallsofIvy

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    Yes, and taking "s" to be the length of one of the legs, the area of the base is
    [tex]\frac{1}{2}s^2= 18[/tex].
     
  16. Mar 23, 2011 #15
    ahh ok..now i get it..
    s2 = 36
    s = 6 cm

    and then we find the hypotenuse = approximately 8.5 cm
     

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