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Help, Calculating height of Trapezium

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data
    I know the base of the trapezium is 20km long. I also know the area of it has to be 1x10^5 m^3. I was just wondering how would you be able to calculate the height of the trapezium without knowing the length of the top line, i want to do it so the height is maximised at angles lower than 40 degrees.

    I have managed to get to the equation of:
    h=2A/a+b where a=base, b=top line

    Can somebody please guide me on how i can get to want i need? Thanks
     
  2. jcsd
  3. Sep 22, 2009 #2

    LCKurtz

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    You mean h = 2A/(a+b), the parentheses are needed. You need one of h or b to calculate the other.

    Can you explain more precisely what you mean by "the height is maximised at angles lower than 40 degrees"?
     
  4. Sep 22, 2009 #3
    Well the slope/angles of the triangles (at either end) must be equal to or lower than 40 degrees. I dont need much help with that as of yet, once im able to find the height with the known angle that should be fine because ill be able to do it with other angles then. So basically, the bottom length must be 20km long, and the slope of the triangles must not be greater than 40 degrees. Thanks
     
  5. Sep 22, 2009 #4

    LCKurtz

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    I somehow think you haven't stated the whole problem. If you assume it is an isosceles trapezoid with angle φ at the bottom, from the geometry you can get the equation:

    A = ah - h2cot(φ)

    This determines implicitly a relation between h and φ. Put in your numbers for A and a and find the max of h remembering that:

    0 ≤ φ ≤ 2π/9
     
    Last edited: Sep 23, 2009
  6. Sep 22, 2009 #5
    Yeh i forgot to put that up, i got to the equation you showed but didnt know if it was correct. The thing is do i go from there, sub in the values for A and a, as well as the angle and use the quadratic formula to determine the height, because i try this nd get VERY big values. How can i do this because shouldn't i only get one value for h for the equation to meet the desired area and angle range? Thanks
     
    Last edited: Sep 23, 2009
  7. Sep 23, 2009 #6
    Bump
    (Sorry for this but i dont know what time it is over where everyone is (sleeping etc))
     
  8. Sep 27, 2009 #7
    Anybody?
     
  9. Sep 27, 2009 #8

    LCKurtz

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    "20km long. I also know the area of it has to be 1x10^5 m^3."

    Just out of curiosity, what about those units? Area would be in units2, not units3. So you should have written m2. But did you really mean m2 and not km2? Have you tried drawing it to scale?
     
    Last edited: Sep 27, 2009
  10. Sep 27, 2009 #9
    The area is m^2, but its not km^2.
     
  11. Sep 27, 2009 #10
    Just use a different value for the area and bottom length as that might make it easier to explain then ill be able to do the rest by myself with my values.
     
  12. Sep 29, 2009 #11
    Bump. How are you supposed to do it when you got two variables in the equation, both h and the angle?
     
    Last edited: Sep 30, 2009
  13. Oct 2, 2009 #12
    Please delete thread, problem has been solved
     
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