Help, Calculating height of Trapezium

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Homework Help Overview

The discussion revolves around calculating the height of a trapezium given its base length and area, with specific constraints on the angles of the trapezium's sides. The original poster seeks to maximize the height while adhering to an angle limit of 40 degrees.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the height, base, and top line of the trapezium, with some suggesting the need for additional information to solve for height. Questions arise regarding the implications of angle constraints on the trapezium's geometry.

Discussion Status

The conversation includes attempts to clarify the problem setup and the equations involved. Some participants express confusion about the units of area and the implications of the angle constraints, while others suggest alternative approaches to the problem. There is no explicit consensus on the next steps or the correctness of the equations presented.

Contextual Notes

Participants note potential issues with the stated area units and the need for clarity on the trapezium's properties. The original poster acknowledges missing information that could affect the problem-solving process.

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Homework Statement


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
 
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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"?
 
Well the slope/angles of the triangles (at either end) must be equal to or lower than 40 degrees. I don't need much help with that as of yet, once I am 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
 
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:
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:
Bump
(Sorry for this but i don't know what time it is over where everyone is (sleeping etc))
 
Anybody?
 
"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:
The area is m^2, but its not km^2.
 
  • #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.
 
  • #11
Bump. How are you supposed to do it when you got two variables in the equation, both h and the angle?
 
Last edited:
  • #12
Please delete thread, problem has been solved
 

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