Advanced Area of a Trapezoid

[philthy]

I am back!!! I was so impressed with the information provided to me for the triangle that I think I am ready to attempt the trapezoid I mentioned in the original post of my first thread.

For this problem I will have three inputs: AB, DC, & h (see attached file) and d=b (this will always be equal and will not be an input-it is what it is). Using these inputs, I need to figure out the area of a 1” border all around (just like the triangle example in the above posts). I truly appreciate any help you can provide. Thank you!

 

[Deleted member 4993]

Well, if b=d and AB is parallel to CD, then
all 4 angles shown are equal to each other,
and what you have is the bottom portion
of an isosceles triangle.

Agree?
So give it a shot using knowledge from your 1st post.

No!!

\(\displaystyle \displaystyle{\alpha = \beta \\ and \\ \delta = \gamma}\)

 

[Ishuda]

I am back!!! I was so impressed with the information provided to me for the triangle that I think I am ready to attempt the trapezoid I mentioned in the original post of my first thread.

For this problem I will have three inputs: AB, DC, & h (see attached file) and d=b (this will always be equal and will not be an input-it is what it is). Using these inputs, I need to figure out the area of a 1” border all around (just like the triangle example in the above posts). I truly appreciate any help you can provide. Thank you!

You really do need one other piece of information and that is the "orientation" of the trapezoid. The information might be presented in several ways. One would be to say that it was an Isosceles trapezoid, that is for the given angles, \(\displaystyle \alpha=\beta\, and\, \delta=\gamma\). Another would be to say that DC was centered on AB and there are other possible statements.

Given that, obviously the outside trapezoid would have a bottom line one inch below and parallel to AC and the top line one inch above and parallel to DC. For the two side lines, go back and look at the other solution. The same type of work applies.

 

[philthy]

You really do need one other piece of information and that is the "orientation" of the trapezoid. The information might be presented in several ways. One would be to say that it was an Isosceles trapezoid, that is for the given angles, \(\displaystyle \alpha=\beta\, and\, \delta=\gamma\). Another would be to say that DC was centered on AB and there are other possible statements.

Given that, obviously the outside trapezoid would have a bottom line one inch below and parallel to AC and the top line one inch above and parallel to DC. For the two side lines, go back and look at the other solution. The same type of work applies.

Thanks for the replays everyone. I am out of town right now but I will give this a shot when I get back on Sunday.

@Ishuda - Did you mean that the bottom will be one inch below and paralell to AB? Also, this will always be an "isosceles trapezoid".

 

[Ishuda]

Thanks for the replays everyone. I am out of town right now but I will give this a shot when I get back on Sunday.

@Ishuda - Did you mean that the bottom will be one inch below and paralell to AB? Also, this will always be an "isosceles trapezoid".

Well of course I meant AB, it's this dang keyboard that makes the misteak, not me.

 

[Ishuda]

Join me in the corner...keyboard did same to me...

be left there