I should have made this post a long time ago, because it’s a bone of contention I’ve always had with trapezoids. Or…not with trapezoids–I like trapezoids–but a bone of contention I have with the *definition* of trapezoids. In my humble opinion, it’s a major problem with Geometry as it’s currently taught. Here’s the usual definition of a trapezoid (taken from our school’s Geometry text book, by Holt Rinehart and Winston):

“A quadrilateral with

one and only onepair of parallel sides.”

I’ve emphasized the words “**one and only one**,” which is what I want to comment about in this post. (Here’s another source and another source and another source that say it that way, too.) Sometimes it’s also said, “a quadrilateral with *exactly* one pair of parallel sides.”

**I’ve prepared a simple GeoGebra applet and posted it here**. It allows you to play with the trapezoid, moving its vertices and edges. As you drag it around, at all times, one pair of sides will be parallel. But wait, it’s not always a trapezoid, is it? According to the Geometry book, there’s one moment, as you’re dragging it around, that it stops being a trapezoid and for that one second is *exclusively *a parallelogram. Here’s the moment I’m talking about:

That’s right, using the Geometry textbook’s definition of a trapezoid, if *both* pairs of opposite sides of the quadrilateral happen to be parallel, it’s not a trapezoid anymore. At this point, the mathematical reader should be crying, “*Foul! How did we ever let this happen? This definition of a trapezoid is so inelegant!!*” And I couldn’t agree more.

**We don’t do this with the definition of any other quadrilateral. Why do it with a trapezoid?** If I were to make another little applet that lets you drag around a rectangle, would we say “it’s not a rectangle” at the moment you make the four sides equal? No! That would be absurd.

The definition of a trapezoid, in my opinion (and thankfully in the opinion of some others) should read:

“A quadrilateral with

at leastone pair of parallel sides.”

And the hierarchical diagram should look like this one, I found online (taken from a mathematically enlightened author):

Here’s a nice paragraph from the wikipedia entry on trapezoid:

There is also some disagreement on the allowed number of parallel sides in a trapezoid. At issue is whether parallelograms, which have two pairs of parallel sides, should be counted as trapezoids. Some authors

^{[2]}define a trapezoid as a quadrilateral havingexactlyone pair of parallel sides, thereby excluding parallelograms. Other authors^{[3]}define a trapezoid as a quadrilateral withat leastone pair of parallel sides, making the parallelogram a special type of trapezoid (along with the rhombus, the rectangle and the square). The latter definition is consistent with its uses in higher mathematics such as calculus. The former definition would make such concepts as the trapezoidal approximation to a definite integral be ill-defined.

This site and this site also get it right. So there’s hope for the Geometry community and for teachers everywhere. But please, let’s work hard to eradicate the “exclusive” definition from ALL the textbooks. It’s hideous.

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Love it! Let’s keep fighting this battle!

You are right – it is hideous. I think that the “only” is just due to poor writing and lack of understanding on the part of some textbook writers. You might want to consider tearing that section out of your textbook (or maybe chuck the whole thing).

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Thanks for leaving comments, guys!

@Dan: I don’t think the “only” is just due to poor writing. It’s very intentional. I claim it’s due to poor *thinking*!

I just posted some additional thoughts today. Check out the post here:

https://mrchasemath.wordpress.com/2011/02/18/why-i-hate-the-definition-of-trapezoids-again/

Just a question then – what do you do with the following two theorems regarding isosceles trapezoids when the trapezoid becomes a parallelogram?

Base Angles Theorem for Isosceles Trapezoids – the base angles of an isosceles trapezoid are congruent – not true for basic parallelograms (only rectangles).

Theorem (I don’t think it has a specific name) – the diagonals of an isosceles trapezoid are congruent – not true for basic parallelograms (only rectangles).

Do you then say that these theorems only apply for non-parallelogram isosceles trapezoids or that isosceles trapezoids can’t be parallelograms?

Seems to be an issue. If you have a solution for it, I would love to hear it because this is the one thing that seems to be a problem for the difference in definitions.

Andy, thanks for the comment. I don’t think there’s a problem here though. The statement I made was “a parallelogram is a trapezoid.” It’s not necessarily true that “a parallelogram is an isosceles trapezoid,” though it may happen to be true in the case of a rectangle (the ‘good’ diagrams I highlighted illustrate this). If a parallelogram is an isosceles trapezoid, then these theorems apply. The wording of the theorems does not have to be changed at all.

It’s the same problem as the following (which is really no problem at all): a rhombus is a special case of a parallelogram. A rectangle is always a parallelogram, and it may also be a rhombus in the case of a square. But theorems about a rhombus don’t have to also hold for rectangles. For example, “the diagonals of a rhombus bisect each other.” Shouldn’t you cry foul here, too? You might say, but wait, this theorem doesn’t apply to rectangles! But that’s because, as we just said, it’s not usually the case that a rectangle IS a rhombus. If a rectangle happens to be a rhombus, then the bisecting theorem applies.

In reality, you have no problem with the definition of a parallelogram even though it allows for the issue above. Likewise, you should have no problem with the definition of a trapezoid.

Not sure if that helps. I can say more if need be!

Well, it seems to me that you short-change (or completely change) the definition of an isosceles trapezoid then if its not necessarily true that ” a parallelogram is an isosceles trapezoid.”

The usual definition of an isosceles trapezoid is one where the legs of the trapezoid are congruent. This should hold true for any parallelogram. One set of sides is parallel (actually both sets are) – the “bases” of the trapezoid, and the other set of sides is congruent (actually both sets are) – the “legs” of the trapezoid. Now, all parallelograms would have to be isosceles trapezoids – quadrilaterals where one set (or possibly 2 sets by the inclusive definition) of opposite sides are parallel and the other set of sides (which could possibly be parallel) are congruent.

I don’t see how your answer answers that problem.

Andy, you’re totally right. I’m sorry for ignoring the nuance of your post. You DO in fact have to redo the definition of an isosceles triangle if your definition is as you state.

Someone else brought this up a while ago and I hadn’t thought about it before then. I proceeded to do some research and found that the definition of an isosceles triangle is also hotly contested. Here’s a link to that conversation.

In my reply I link to some sources around the web that define ‘isosceles triangle’ in ways that are inconsistent with the inclusive definition of a trapezoid (like your definition of the isosceles triangle), and other sources that define it in ways that

areconsistent with the inclusive definition (based on symmetry or base angles).I appreciate that response. I have enjoyed the conversation overall.

Honestly, I don’t see how this will ever be resolved. I think it will be one of those things that in the long run, mathematicians agree to disagree on, but I think the thought process that goes into it can be beneficial for both sides.

Like I posted (on the other conversation I believe), I think that the inclusive definition of a trapezoid, the inclusive definition of an isosceles trapezoid (based on its usual definition of congruent sides rather than congruent angles or symmetry), and the three theorems (base angles and its converse and the congruent diagonals theorem) can’t co-exist very well. It seems that one of those will have to give somewhere in my opinion.

Which one giving to make things more fluid or more “beautiful” mathematically is probably in the eye of the beholder.

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Tapazoid- an object of four sides. 360°

With only one set of sides being parallel. Your argument is overbuilt to the simplicity of rhe concept. There needn’t be any further eloquent description. Occams Razor- the simplest solution is most often the best solution.

I’m not sure how one definition is “simpler” than the other. Someone could claim that insisting that one pair of opposite sides be nonparallel is additional baggage that makes the

exclusive definitionmore complicated. It seems like you could make the argument either way.Can you clarify?