Trapezoid

[MathIsEverything]

Hi! I've been struggling with this problem. I already spent 2 hours and i couldn't think of anything. Maybe you guys can help me.

A circle is drawn in the trapezium, the diameter of which is the shorter base of this trapezium having length k. This circle is also tangent to the longer base of the trapezium and intersects its diagonals at their centres. Determine the area of this trapezium(in terms of k)
Thank you

 

[The Highlander]

Hi! I've been struggling with this problem. I already spent 2 hours and i couldn't think of anything. Maybe you guys can help me.

A circle is drawn in the trapezium, the diameter of which is the shorter base of this trapezium having length k. This circle is also tangent to the longer base of the trapezium and intersects its diagonals at their centres. Determine the area of this trapezium(in terms of k)
Thank you

What kind of trapezium is it?
For example; the point of intersection of the diagonals of an isosceles trapezium lies midway between the parallel sides. Does that help?
Can you show us the original question, please? (Including any diagram(s) provided.)
Also, please show us what you have already tried including your sketch/drawing of the figure as you describe it.

 

[Dr.Peterson]

A circle is drawn in the trapezium, the diameter of which is the shorter base of this trapezium having length k. This circle is also tangent to the longer base of the trapezium and intersects its diagonals at their centres. Determine the area of this trapezium(in terms of k)
Thank you

Does "drawn in" mean "inscribed", so that it is tangent to all four sides? Or is it only tangent to the longer base? (I think it has to be the latter.)

And I assume it is the length of the diameter that is equal to the shorter base, because if the diameter itself were the base, then the circle would not be "in" the trapezium in any sense.

I think we really need a picture of what this means.

For example; the point of intersection of the diagonals of an isosceles trapezium lies midway between the parallel sides. Does that help?

I think you mean the midpoint of each diagonal. But that's true of any trapezium. Your statement is true of a parallelogram.

 

[The Highlander]

Does "drawn in" mean "inscribed", so that it is tangent to all four sides?I

Wouldn't that be a square?

Your statement is true of a parallelogram.

You're quite right, of course, I should have said: 'midway between the non-parallel sides.'
Not that it matters as I can't come up with any way to draw a trapezium that satisfies the stated conditions (at least as I interpret them).
We really do need picture(s)/more information.

 

[Dr.Peterson]

Wouldn't that be a square?

Here is an example of a circle inscribed in a (non-square) trapezium with one base equal to the diameter:

​

Of course, it's probably impossible for the circle to pass through those two midpoints. (I say "probably" only because I haven't proved it.)

I haven't tried yet to make a drawing that fits my alternative interpretation.

... but now I have:

​

This does seem to show that it's possible, if "in" just means "in the interior of".

 

[The Highlander]

This does seem to show that it's possible, if "in" just means "in the interior of".

Brilliant!
Is there enough information there to find the area of that?
(Beats me! lol.)

 

[Dr.Peterson]

Brilliant!
Is there enough information there to find the area of that?
(Beats me! lol.)

I haven't tried to solve it yet; I just moved things until I could see that such a figure can exist. The rest, of course, is up to @MathIsEverything (with us, perhaps, working in the background until we see some work to comment on).

 

[MathIsEverything]

The original question is on antoher language and i translated it. There is no attached picture. What I can deduce from the problem in the original language is that the shorter base IS the diameter, not that the diameter is the length of the base. So the trapezium would look like this. I thinks IT has to be isoscels trapezium because circle must cut diagonals at its centers. I am also curious if your interpretation of the question is solveable! In my interpretation I've already done the problem with using the knowledge that angle HBD is 15°.

 

[Dr.Peterson]

The original question is on another language and i translated it. There is no attached picture. What I can deduce from the problem in the original language is that the shorter base IS the diameter, not that the diameter is the length of the base. So the trapezium would look like this. I thinks IT has to be isosceles trapezium because circle must cut diagonals at its centers. I am also curious if your interpretation of the question is solvable! In my interpretation I've already done the problem with using the knowledge that angle HBD is 15°.

So the word "in", which I put a lot of weight on, was wrong? The circle is not in the trapezium, but is as I said:

... because if the diameter itself were the base, then the circle would not be "in" the trapezium in any sense.

I had seen some evidence (using my interpretation) that the area may be determined by k (regardless of angles), but had not been able to prove that. Now I can move on to your interpretation.

You're right about the angle (though the way you say it, "the knowledge that", makes it sound like that was given rather than deduced). In some ways this is a less interesting problem (because the figure has a fixed shape), but it has its challenges.

I'd be interested in seeing the original (so we can try translating it with Google or other help, and see what possibilities there are for the interpretation). In general, I prefer seeing both the original and a translation, so we can know how much to trust the translation, and not waste time solving the wrong problem.

 

[blamocur]

The original question is on antoher language and i translated it. There is no attached picture. What I can deduce from the problem in the original language is that the shorter base IS the diameter, not that the diameter is the length of the base. So the trapezium would look like this. I thinks IT has to be isoscels trapezium because circle must cut diagonals at its centers. I am also curious if your interpretation of the question is solveable! In my interpretation I've already done the problem with using the knowledge that angle HBD is 15°.

I get 15 degrees angle too. Looks like you have enough info to figure out AG, AB and the area.

 

[skeeter]

In my interpretation I've already done the problem with using the knowledge that angle HBD is 15°.

Is that so?

Trapezoid and circle

Hi! I've been struggling with this problem. I already spent 2 hours and i couldn't think of anything. Maybe you guys can help me. A circle is drawn in the trapezium, the diameter of which is the shorter base of this trapezium having length k. This circle is also tangent to the longer base of...

mathhelpforum.com

 

[MathIsEverything]

So the word "in", which I put a lot of weight on, was wrong? The circle is not in the trapezium, but is as I said:


I had seen some evidence (using my interpretation) that the area may be determined by k (regardless of angles), but had not been able to prove that. Now I can move on to your interpretation.

You're right about the angle (though the way you say it, "the knowledge that", makes it sound like that was given rather than deduced). In some ways this is a less interesting problem (because the figure has a fixed shape), but it has its challenges.

I'd be interested in seeing the original (so we can try translating it with Google or other help, and see what possibilities there are for the interpretation). In general, I prefer seeing both the original and a translation, so we can know how much to trust the translation, and not waste time solving the wrong problem.

So, this question is from competition i took part in. I spent 1 hour trying to figure out how to determine the area if circle IS in the trapezoid. After few days results came in and my friend who interpreted question as the circle may be outside (as on attached picture before) got full points using method with deducing angles.

Original question:

W trapezie narysowano okrąg, którego średnicą jest krótsza podstawa tego trapezu mająca długość k. Okrąg ten jest ponadto styczny do dłuższej podstawy trapezu i przecina jego przekątne w ich środkach. Wyznacz pole tego trapezu.

 

[The Highlander]

I get 15 degrees angle too. Looks like you have enough info to figure out AG, AB and the area.

I too have spent some considerable time vying with this problem but I’m still as confused as a baby in a topless bar! ???

The OP seems to have gone off quite happy with the ‘solution’ s/he has arrived at so maybe someone can help me out by providing details of how to arrive at the right answer based on the (second) approach outlined below.

The consensus appears to be that the correct interpretation of the problem is as shown in my (maybe not completely accurate) diagram here:-

​

Reference was made (along with a little dig at the OP, lol) to the solution(s) offered on the site linked below:-

Is that so?

Trapezoid and circle

Hi! I've been struggling with this problem. I already spent 2 hours and i couldn't think of anything. Maybe you guys can help me. A circle is drawn in the trapezium, the diameter of which is the shorter base of this trapezium having length k. This circle is also tangent to the longer base of...

mathhelpforum.com

One contributor there asserts that arc CF = 30° \(\displaystyle \because\angle\)FOC = 30° \(\displaystyle \Rightarrow\angle\)BDC = 15°

So, given that \(\displaystyle \angle\)FOC is 15° then I am quite happy that leads to AB = k\(\displaystyle (1+\sqrt{3})\) and, hence, the Area of the Trapezium works out at k²\(\displaystyle \left(\frac{1}{2}+\sqrt{\frac{3}{16}}\right)\).

However, a second contributor on that site offers an (ostensibly) different approach that comes up with the same answer but I just don’t see how it’s arrived at given the information provided.

Here is their post:-

​

How do you "compute DT" using just the values and the diagram provided???

Can anyone, please, explain (in detail) for me how the information/diagram provided in this post leads to the determination of half of the lower base of the Trapezium being k\(\displaystyle (1+\sqrt{3})\) ?

Maybe I’m just being dense or missing something obvious but any help would be much appreciated so I can stop obsessing over this damn puzzle.

 

[The Highlander]

Correction: (I've just noticed) the sentence below should read:

So, given that \(\displaystyle \angle\)BDC (not FOC) is 15° then I am quite happy that leads to AB = k\(\displaystyle (1+\sqrt{3})\) and, hence, the Area of the Trapezium works out at k²\(\displaystyle \left(\frac{1}{2}+\sqrt{\frac{3}{16}}\right)\).

 

[Dr.Peterson]

However, a second contributor on that site offers an (ostensibly) different approach that comes up with the same answer but I just don’t see how it’s arrived at given the information provided.

Here is their post:-

View attachment 34609​

How do you "compute DT" using just the values and the diagram provided???

Can anyone, please, explain (in detail) for me how the information/diagram provided in this post leads to the determination of half of the lower base of the Trapezium being k\(\displaystyle (1+\sqrt{3})\) ?

Maybe I’m just being dense or missing something obvious but any help would be much appreciated so I can stop obsessing over this damn puzzle.

I can't say what someone else thinks, but if I were them, I would do this:

 

[The Highlander]

I can't say what someone else thinks, but if I were them, I would do this:

View attachment 34610

Yup, that works for me, thanks. ?

I suppose I should have seen (even without calculation of the angle TDC) that:-

\(\displaystyle \mathsf{(ND)^2=k^2-\left(\frac{k}{2}\right)^2=k^2-\frac{k^2}{4}=\frac{3}{4}k^2}\)

\(\displaystyle \sf\Rightarrow ND=\sqrt{\frac{3}{4}}k=\frac{k}{2}\sqrt{3}\)

thus \(\displaystyle \sf TD=\frac{k}{2}+\frac{k}{2}\sqrt{3} \quad\&\quad AD\medspace(=2TD)=k(1+\sqrt{3})\)   QED. ?