4
$\begingroup$

I've written a tree-like layout to help myself remember which polygons are sub-types of others, because I always get confused. I was just wondering if this is right:

|quadrilateral
    |parallelogram
        |rectangle
            |square
            |oblong
        |rhomboid
        |kite (corrected after rschwieb's answer, a rhombus is a kite)
            |rhombus
                 |square
    |trapezoid(AmE) / trapezium (BrE)
    |trapezium(AmE) / irregular quadrilateral

So a square is a rhombus and a parallelogram.

Also, I know that there are two definitions of "trapezoid." Under the inclusive definition "trapezoid" is immediately under "quadrilateral" in the tree and above parallelogram and kite. Under this definition all squares are trapezoids.

Is my tree correct, at least ignoring the difference in the trapezoid definition difference?

Edit: Thanks to rschwieb for helping me realise that a rhombus is a kite. There is also a nice Euler diagram Wikipedia

$\endgroup$
5
$\begingroup$

You need to see this post I wrote some time ago.

In short, the education system (at least in the US) has confused this issue and made it harder than it has to be.

There is a very natural hierarchy the depends on logical connections between quadrilaterals, and there is really no benefit to using the “exclusive version” of definitions.

I would argue for this picture for the main characters:

Image originally from www.andrews.edu/~calkins/math/webtexts/geom06.htm

Actually there is a little puzzle where you can figure out a new node to insert between "quadrilateral" and "kite" which also connects to "parallelogram," and I have never seen this shape mentioned in a textbook. It's just not common enough to encounter in normal life.

$\endgroup$
  • $\begingroup$ I see in your chart you've used the inclusive definition of trapezoid, that's all good. I see the difference between this chart and the "frankenstein" chart you referred to is that the kite is in a separate area on its own, but in yours it traces down to rhombus. In yours it shows a rhombus as being a kite. This seems unintuitive with my everyday notion of a kite, but Wikipedia says: "If all four sides of a kite have the same length (that is, if the kite is equilateral), it must be a rhombus." Also, the other chart doesn't link the kite to the rhombus at all. I think this chart is easier. $\endgroup$ – Zebrafish Apr 21 '18 at 2:21
  • $\begingroup$ There's just one thing I see wrong with it, a rectangle is an isosceles trapezoid? Are those two supposed to be connected with a line over on the right? $\endgroup$ – Zebrafish Apr 21 '18 at 2:49
  • $\begingroup$ "Rectangles and squares are usually considered to be special cases of isosceles trapezoids though some sources would exclude them." Wow, there are a few a discrepancies in definitions, that's not helping. $\endgroup$ – Zebrafish Apr 21 '18 at 2:52
  • $\begingroup$ @Zebrafish as I mentioned “exclusive definitions” aren’t as useful, they’re harder to state, and make proving things less convenient. $\endgroup$ – rschwieb Apr 21 '18 at 3:41
  • 1
    $\begingroup$ @Zebrafish yes, I would cal a rectangle an isosoles trapezoid. $\endgroup$ – rschwieb Apr 21 '18 at 3:43
0
$\begingroup$

Can't a kite also be a parallelogram, in the case where all sides are equal?

That of course depends on your definition of kite... I've rarely seen the term used at all. You can exclude that case specifically and your tree is then okay.

Wikipedia's Kite (geometry) article seems to include that in their special cases.

EDIT: In that special case, it can also be a rhombus or a square

$\endgroup$
  • 1
    $\begingroup$ A kite that’s also a parallelogram is a rhombus. $\endgroup$ – rschwieb Apr 21 '18 at 3:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.