Finding height of trapezoid with area, angles and one base only

[morkfromork]

Hi all


This is a very tricky problem that has bothered me for a long time, and now I thought my brain might not be so brilliant as yours.

I am trying to construct an equation to find the height of a trapezoid of which I know the area and the angles but only one base.

So for example, if I have a trapezoid that has a base of 12, angles 45, 135, 90 and 90, and an area of 40, how would I find out its height?



Of course this can be done by some guessing as these are all easy rounded numbers, but I am curious about what the equation would be.
I know it must be possible, but I cannot find any clue online on how to do it.

I hope someone can help me crack this one!

 

[Deleted member 4993]

Hi all


This is a very tricky problem that has bothered me for a long time, and now I thought my brain might not be so brilliant as yours.

I am trying to construct an equation to find the height of a trapezoid of which I know the area and the angles but only one base.

So for example, if I have a trapezoid that has a base of 12, angles 45, 135, 90 and 90, and an area of 40, how would I find out its height?

View attachment 23836

Of course this can be done by some guessing as these are all easy rounded numbers, but I am curious about what the equation would be.
I know it must be possible, but I cannot find any clue online on how to do it.

I hope someone can help me crack this one!

Can you find the "expression" for

the length of the shorter "parallel" side as a function of 'h'​

​

- noting that m<BAD = 45^o?​

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[morkfromork]

I apologize, first I should confess that this is not a school assignment but a personal interest and that it has been a long time since I went to school.

The example presented is my own with easy to do angles and so in this convenient case
if we call the base b₁ = 12 and then m<BAD = 45^o
b₂ = b₁-h

But knowing neither h or b₂ I am stuck.

I could put together this for finding out the area
area = (2b₁-h)/2 * h

But then I forgot how to transpose it to isolate the h... I think that is my real embarrasing problem in this instance.

Of course, by simple addition using the trapezoid area equation increasing h by 1 at a time we easily get to the answer that h = 4. But I am curious about the equation that would do that, for any kind of angles.

 

[Deleted member 4993]

I apologize, first I should confess that this is not a school assignment but a personal interest and that it has been a long time since I went to school.

The example presented is my own with easy to do angles and so in this convenient case
if we call the base b₁ = 12 and then m<BAD = 45^o
b₂ = b₁-h

But knowing neither h or b₂ I am stuck.

I could put together this for finding out the area
area = (2b₁-h)/2 * h

But then I forgot how to transpose it to isolate the h... I think that is my real embarrasing problem in this instance.

Of course, by simple addition using the trapezoid area equation increasing h by 1 at a time we easily get to the answer that h = 4. But I am curious about the equation that would do that, for any kind of angles.

and an area of 40

area = (2b1-h)/2 * h

40 = 12*h - h^2/2

h^2 - 24*h + 80 = 0 .............................................[edited]

This is a quadratic equation. You should be able to solve for 'h' from here.

If you forgot - do a google search with key words "solution of quadratic equation" and see what you can learn (re-learn) from those.

If you are still stuck, please come back to discuss.

 

[morkfromork]

Thank you so much!

I think I got around some of it. So the general equation for the trapezoid problem would be h^2 - 2*b*h + 2*area = 0, I assume.

With your help, I looked into solving it with the quadratic formula and realize that it gives two roots, h = 4 and h = 20.

But naturally, in the trapezoid only one could be true. Plotting it I found that the larger result makes the trapezoid intersect itself to make two triangles (a positive and a negative?) that when subtracted from eachother leaves the trapezoid.



I wonder, does it mean that the answer in the trapezoid problem is always the lesser result of the quadratic formula? In the quadratic formula, should I always only subtract the discriminant, or is the larger result always an imaginary number of sorts? Is there a common rule to which is the right answer? I am sorry for asking so much, but this is a new world to me, and it is very exciting.

 

[Deleted member 4993]

Thank you so much!

I think I got around some of it. So the general equation for the trapezoid problem would be h^2 - 2*b*h + 2*area = 0, I assume.

With your help, I looked into solving it with the quadratic formula and realize that it gives two roots, h = 4 and h = 20.

But naturally, in the trapezoid only one could be true. Plotting it I found that the larger result makes the trapezoid intersect itself to make two triangles (a positive and a negative?) that when subtracted from eachother leaves the trapezoid.

View attachment 23843

I wonder, does it mean that the answer in the trapezoid problem is always the lesser result of the quadratic formula? In the quadratic formula, should I always only subtract the discriminant, or is the larger result always an imaginary number of sorts? Is there a common rule to which is the right answer? I am sorry for asking so much, but this is a new world to me, and it is very exciting.

remember one of the side-length = 12 - h

h = 20 will give you length < 0 - impossible!!

 

[morkfromork]

Of course, so the right result would be the one that is > 0 and < b (the side-length)

I will try to work with other arbitrary angles to build further on this.

Thank you so much!