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Talk:Quadrilateral

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There could be tangential quadrilaterals which are also trapezoids. No? --203.186.238.243 9 July 2005 16:02 (UTC)

Indeed. Even tangential quadrilaterals which are also right-angled trapezia. (But not every tangential quadrilateral is a trapezium.) You're welcome to extend the diagram to add the new categories, though I think "cyclic right-angled trapezium" might be a bit obscure... Gdr 12:55:55, 2005-08-03 (UTC)

Shouldn't the isosceles trapezoid be between the trapezoid and parallelogram instead of between the trapezoid and the rectangle to be fully inclusive? An isosceles trapezoid has one+ pair of parallel sides and one+ pair of congruent sides, so wouldn't that all be inherited by a parallelogram? --Rockychat3 (talk) 02:58, 4 March 2014 (UTC)[reply]

To answer my own question, an isosceles trapezoid also has congruent diagonals, which a parallelogram does not, so not all properties are inherited. Symmetry is also not inherited. --Rockychat3 (talk) 04:02, 4 March 2014 (UTC)[reply]

The cell for whether an isosceles trapezoid has perpendicular diagonals says "Not in general", but the cell for whether a kite has congruent diagonals, which this same phrase fits equally, says "No". Any thoughts?? Georgia guy (talk) 20:58, 29 March 2013 (UTC)[reply]

I agree with GG, and changed the "not in general" entry back to "no". Special cases ought to be treated equally. Plus, nothing at Kite (geometry) nor isosceles trapezoid talk about these special case, probably because there's no new symmetry involved. If the articles note the cases, then perhaps a mark in the table could note that with wlink referencing why its noteworthy. Tom Ruen (talk) 21:11, 29 March 2013 (UTC)[reply]

I disagree because of the caveat I placed above the table. A kite can have equal diagonals only when it is a named subset of the set of kites. (See apology below.) An isosceles trapezoid can have perpendicular diagonals without being anything other than an isosceles trapezoid. Dbfirs 21:26, 29 March 2013 (UTC)[reply]

Perhaps we could compromise by replacing two "No"s with a note that, although the most general trapezoids and isosceles trapezoids do not have perpendicular diagonals, there is an infinite number of (non-similar) trapezoids and isosceles trapezoids that do have perpendicular diagonals and are not any other named quadrilateral. Dbfirs 22:09, 29 March 2013 (UTC)[reply]

What is a "kite with equal diagonals" called? If the equal diagonals have the same center, its a square, but otherwise, it's still just a kite to all the terminology I know. Tom Ruen (talk) 22:16, 29 March 2013 (UTC)[reply]

Yes, you are correct. (I'd missed seeing that! Apologies to Georgia Guy.) The whole table has issues. Have we made it too complicated with the notes? Dbfirs 22:34, 29 March 2013 (UTC)[reply]

I don't have a clear sense of what's most helpful. Perhaps every "no" has exceptions and could be explained, like "no, becomes a square", etc. With "no, in general" if there's no special name for the special cases! Tom Ruen (talk) 22:47, 29 March 2013 (UTC)[reply]

I think all the other cases are covered by the caveat above the table, but I'll not object if you want to make that adjustment. Dbfirs 22:53, 29 March 2013 (UTC)[reply]

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Is there a reason why we don't use the word quadragon instead of quadrilateral? After all it has the same meaning as hexagon (hex meaning 6) or pentagon (penta meaning 5)

So quadragon (quad meaning 4) should have some sense. — Preceding unsigned comment added by 185.69.145.97 (talk) 12:48, 2 September 2016 (UTC)[reply]

Because the word of that form that we (rarely) use is instead "tetragon". I don't know what history led to the different choice of root for the number four. —David Eppstein (talk) 15:57, 2 September 2016 (UTC)[reply]

-lateral is Latin and should use the Latin root for 4, hence quadrilateral. -gon is Greek and should use the Greek root, hence tetragon. In the case of trianglevstrigon (and using the latter name in polyhedra names usually implies you're a chemist), both Latin and Greek have tri- as the root for 3. Double sharp (talk) 03:07, 3 September 2016 (UTC)[reply]

Interesting comment Double sharp. If the word trigonometry (that studies the lengths and angles of triangles) stems from the old word 'trigon', then shouldn't the said word be renamed to ''triangleometry?'' Just a curious thought more than anything. I am not trying to instigate an edit war or anything. — Preceding unsigned comment added by 77.97.245.229 (talk) 18:54, 11 October 2016 (UTC)[reply]

If you want a language that is designed to be logical and consistent, then English is not for you. You could try Lojban, at http://jbo.wikipedia.org . —David Eppstein (talk) 20:05, 11 October 2016 (UTC)[reply]

David Eppstein, I was born and bred in the UK and English is my first language. Please don't ever insult me again. — Preceding unsigned comment added by 77.97.245.229 (talk) 22:49, 16 November 2016 (UTC)[reply]

He didn't. —Tamfang (talk) 20:40, 11 March 2023 (UTC)[reply]

If thats the case then why don't we call a triangle a 'trilateral' or a hexagon a 'hexalateral'? — Preceding unsigned comment added by 77.97.245.229 (talk) 23:36, 30 September 2016 (UTC)[reply]

Don't we? When I search Google for the definition of trilateral one of the results is a triangle. Same for trigon. Also, triangle is parallel to a third word for a quadrilateral or tetragon, namely "quadrangle". —David Eppstein (talk) 00:01, 1 October 2016 (UTC)[reply]

David Eppstein, I have known since I was a child that 'quadrangle' is an alternative name for 'quadrilateral'. Please do not take me for an idiot. — Preceding unsigned comment added by 77.97.245.229 (talk) 22:53, 16 November 2016 (UTC)[reply]

Again, "hex-" for 6 is a Greek prefix which would not usually be added to a Latin root... AnonMoos (talk) 17:36, 8 October 2016 (UTC)[reply]

I editted them.Could you check if it is right?The area of the Varignon parallelogram equals half the area of the original quadrilateral. This is true in convex, concave and crossed quadrilaterals provided the area of the latter is defined to be the difference of the areas of the two triangles it is composed of.
https://en.wikipedia.org/wiki/Bretschneider%27s_formula Sorry,I misunderstanded a part.It was right about the part.https://en.wikipedia.org/w/index.php?title=Quadrilateral&diff=1072455009&oldid=1072428222 ,and I editted a new part.this article's trigonometric formulas Couldn't it simplify like this?And I undid the not-organized part.--240D:1E:309:5F00:77B:76BC:5AAE:498E (talk) 03:24, 20 February 2022 (UTC)[reply]

And furthermore,we have this area formura.${\displaystyle S={\frac {a\sin A(d-c\cos D)+(d-a\cos A)c\sin D}{2}}}$I'll show the proof.As ${\displaystyle \theta }$ can make between extended lines,${\displaystyle S={\frac {1}{2}}pq\sin \theta }$ According to sine law and first cosine law,${\displaystyle \sin \theta _{1}={\frac {a|\sin A|}{q}},\ cos\theta _{1}={\frac {d-a\cos A}{q}}}$Similary,we can proof about ${\displaystyle \theta _{2}}$.According to angle sum identity,${\displaystyle S={\frac {1}{2}}pq\sin(\theta _{1}\pm \theta _{2})={\frac {a|\sin A|(d-cosD)\pm (d-a\cos A)c|\sin D|}{2}}}$(Both ${\displaystyle \pm }$ is to omit the case of convex and concave quadrilateral.)If we check the sign in each case,it is like above.And we have this formura.${\displaystyle S={\frac {1}{2}}{\sqrt {p^{2}q^{2}-((d-a\cos A)(d-c\cos D)-ac\sin A\sin D)^{2}}}}$— Preceding unsigned comment added by 219.104.224.225 (talk • contribs)

This is original research, which we cannot use on Wikipedia. You can't just do math on a talk page and use it as a citation - you've got to cite it to something from an independent publisher. - MrOllie (talk) 22:24, 30 May 2022 (UTC)[reply]