These are some “Always, Sometimes, Never” questions. Like, “Is it always, sometimes or never true that a rhombus is a parallelogram.”

What’s the fastest way to help these students?

(Thanks for the submission, Tina C!)

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These are some “Always, Sometimes, Never” questions. Like, “Is it always, sometimes or never true that a rhombus is a parallelogram.”

What’s the fastest way to help these students?

(Thanks for the submission, Tina C!)

## 10 replies on “A Parallelogram is a Rhombus”

A quadrilateral hierarchy of sorts can be helpful:

Quadrilateral

Trapezoid Parallelogram

Rectangle Rhombus

Square

There would be arrows connecting the Quadrilateral and Parallelogram and Quadrilateral and Trapezoid which would indicate sometimes a quadrilateral is a trapezoid and sometimes a quadrilateral is a parallelogram

There would be arrows connecting the Parallelogram and Rectangle and Parallelogram and Rhombus, again indicating that sometimes a parallelogram is a rectangle and sometimes a parallelogram is a rhombus

There would be arrows connecting the Rectangle to the Square and the Rhombus to the square, again indicating that sometimes a rectangle is square and sometimes a rhombus is a square.

If the shape is below another, then it is always the shape above it as well. So a rhombus is always a parallelogram, a square is always a rectangle, and always a parallelogram, and always a quadrilateral, etc.

If I could draw the arrows in this reply it would look better. Also, we have gone through the definitions of each special quadrilateral before we look at the hierarchy

I like Venn-type diagrams better than the arrow hierarchy, because then I can see overlap vs containment more visually.

Either way, I’d like to see what we can do beyond this specific example, to get students focused on the idea of definitions, what they’re for, and why inclusive definitions are so much more useful. We prove that a parallelogram’s diagonals bisect each other, and since a rhombus is a special kind of parallelogram, we know that a rhombus’s diagonals bisect each other, too, without having to do any further proof. Adding vectors makes a parallelogram picture, with no special exception of “or a rectangle if the vectors are perpendicular”. (Though maybe there is a special exception for the degenerate parallelograms you get if the vectors are parallel? Or if one vector is zero?)

The famous “Nature of Proof” course had a lot to say about this, starting with examples from real life like “sales tax will not be collected on food” that then leads to a long debate about what “food” is. After that, I think definitions of parallelogram and rhombus will seem easy by comparison!

I’m also curious what to do about the apparently common tendency of students to want to include too much in the definition. For me it’s easy to say “do you really want to have to prove all those things are true before you’re allowed to conclude it’s a parallelogram?” but first of all only I care about that, not the students, and second of all you can prove a bunch of theorems that will eliminate that problem anyway (as we ordinarily do, with the “if the diagonals bisect each other it’s a parallelogram” and all that sort of thing).

I did venn diagrams. I wonder if creating an arrow diagram before or after would help solidify what the subsets mean.

We play devil’s advocate (which kids are always disappointed to discover isn’t really a game) to get the most concise yet precise definitions possible.

WELL THEN If I HAVE NEVER SEEN SUCH A MORE FICTIONAL ANSWER ON THIS WEBPAGE, A PARALLELOGRAM IS OBVIOUSLY NOT A RHOMBUS AND THATS THE WAY ID LIKE TO KEEP IT THANK YOU AND HAVE A NCE DAY:)

umm

A parallelogram isn’t a rhombus, it’s the other way around.

Yes, a parallelogram is always a rhombus.

No a parallelogram is not always a rhombus. The best way to prove this would be to plot 4 points on a cartesian plain such as A (0,4) B (3,0) C (9,1) D (6,5) find the slope of ab and compare to cd. Then find slope of bc and compare to ad. You find that they are parallel. But to be a rhombus the diagonals have to be perpendicular aka slopes are negative reciprocals of one another. So find slope of bd and ac and compare. Do all this work and will realize it’s a parallelogram and not a rhombus as ac and bd are not negative reciprocals

Cartesian plane*

Parallelogram is not always a rhombus they are not congruent that’s a simple one see solved

________________________________who ever made this is an idiot lol I’m a freaking grade 4