Trapezoid calculation: Need help clarifying my understanding

[Leaderofrouge]

Hi all,


My question is: Should a mathematical statement that describes a right angle trapezoid be generalisable to other trapezoid.

Below is the context.


I am trying to derive a generalised equation to calculate for an intermediate length of a badminton shuttlecock which takes the form of a trapezoid as shown below:




The length in question is ef.


I wasn't too sure on how to get started initially and went on to do a quick search over the Internet where I found some suggestions about using the law of sines which initially seemed to be working out (see below):



(Since my question about the understanding and not the calculation itself, I won't be including the step-by-step working unless someone wants to see it for a particular reason.)

Using the above equation, I was indeed able to rearrange and solve for c; so, everything seems logical in my mind at the time and did not think too much about it.


Then, a couple of days ago I came back to the equation and tried to write a generalised statement but got stumped, as it does not seem to work:



As can be seen, an extra "a-c" has appeared. I understand that I have only been working with half the trapezoid to begin with, but I did not think the general equation should change --- i.e., I would the ratio of the two "a-c" to be doubled/halved depending on the perspective but that is not what I am doing here (at least I don't think I am).

So, I am wondering if a right angle trapezoid is some sort of special case and cannot be generalised?


Thanks in advance and much would appreciated if someone could point out where my misunderstanding is occurring from.

Cheers!


P.S. I would also be interested if anyone else has another suggestion on how to approach the problem.

 

[Dr.Peterson]

As can be seen, an extra "a-c" has appeared. I understand that I have only been working with half the trapezoid to begin with, but I did not think the general equation should change --- i.e., I would the ratio of the two "a-c" to be doubled/halved depending on the perspective but that is not what I am doing here (at least I don't think I am).

So, I am wondering if a right angle trapezoid is some sort of special case and cannot be generalised?

First, I wouldn't use the law of sines, or even trigonometry at all; similar triangles are all you need. But sines are a way to talk about proportions, so it's not wrong.

But to work with the isosceles trapezoid, I would just cut it in half and work with one of the two right-angled trapezoids. If you do work with the entire trapezoid, then, yes, you do need to take into account the doubling (which is caused by having two sloped sides). I would also use the fact that the figure is symmetrical, so the angles on each side are the same.

 

[Dr.Peterson]

Then, a couple of days ago I came back to the equation and tried to write a generalised statement but got stumped, as it does not seem to work:

As can be seen, an extra "a-c" has appeared. I understand that I have only been working with half the trapezoid to begin with, but I did not think the general equation should change --- i.e., I would the ratio of the two "a-c" to be doubled/halved depending on the perspective but that is not what I am doing here (at least I don't think I am).

Since I didn't look at the details of what you wrote, I ought to point out at least one specific error:

​

The difference a-c includes both of those distances on the bottom, so each is half of a-c.

Now, one way to continue, using your diagram but using similar triangles rather than trig, is to use triangles ABC and DBE as labeled here:

​

You need to find DE, given (I think) AC, BD, and BE. That will be a simple proportion. Then you can use that to answer your actual question.

 

[Leaderofrouge]

Thanks for the clarification and it makes a lot of sense.

I saw it as "2(a-c)" originally, because I started working on the problem as a right angle trapezoid. So, when I put them back together I thought "since I now have 2 right angle trapezoids" I will need to multiple by two and confused myself there.

I give your suggestion a go and see how everything works out.


Just out of curiosity, while we know my approach was not the simplest/most appropriate way to solving the problem at hand---like you said it's not wrong---was the method capable of solving the problem? If so, how would you have approached it?

Was the reason I got stuck just due to me not having understood a-c correctly?


Thanks again.

 

[Dr.Peterson]

Just out of curiosity, while we know my approach was not the simplest/most appropriate way to solving the problem at hand---like you said it's not wrong---was the method capable of solving the problem? If so, how would you have approached it?

If I were to use trig at all, I would not use the law of sines in the isosceles triangle you show (which doesn't involve the vertical height that I presume you want to use). I would just use the definition of the sine or cosine in the right triangles in the original diagram. I'm curious about the site you used; I suspect you may have misinterpreted it.