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Can somebody please check my list of conditions for certain quadrilaterals or suggest some more possibilities? Please remember they must be applicable to vectors. Thanks!
To prove quadrilateral is a
Square:
Opposite sides congruent and all sides meet at right angles.
Opposite sides congruent and diagonals are perpendicular.
Diagonals are perpendicular and of the same length
Rhombus:
Diagonals are perpendicular
Opposite sides congruent
Rectangle:
All sides right angles
Parallelogram:
Opposite sides are congruent
Diagonal bisect each other.
Trapezium:
One pair of congruent sides
Cyclic Quadrilateral
Opposite angles are complementary.
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One cool theorem, altough maybe not so useful, is ptolemys theorem.
A quadrilateral ABCD is cyclic iff:
|AC||BD|=|AB||CD|+|BC||AD|
http://en.wikipedia.org/wiki/Ptolemy%27s_theorem
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Thanks Kurre, that looks interesting
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I came across that formula just a few days ago!
In general, the formula is an inequality:
|AC||BD|≤|AB||CD|+|BC||AD|, for all quadrilaterals.
Equality occurs when the quadrilateral is cyclic, as you say.
There's another cool formula that lets you work out the area of a cyclic quadrilateral knowing just its side lengths.
Let s be the semiperimeter of a cyclic quadrilateral.
ie. s = (a+b+c+d)/2, where a to d are the side lengths.
Then (Area)² = (s-a)(s-b)(s-c)(s-d).
You can also get the area of any triangle from its side lengths by using this formula.
(A triangle is a cyclic quadrilateral which has a side of length 0 )
Why did the vector cross the road?
It wanted to be normal.
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Thanks,
So are all the conditions I posted correct and complete in their reasons?
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regarding
Rhombus:
Diagonals are perpendicular
Opposite sides congruent
Diagonals are perpendicular, in and of itself, does not necessarily prove that a quadrilateral is a rhombus; it could also be a kite(not even a parallelogram)
if Diagonals are perpendicular and Opposite sides congruent then you a rhombus you have!
Last edited by kuadratic (2009-02-28 05:49:37)
teacher resources--my online edu dir
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Can somebody please check my list of conditions for certain quadrilaterals or suggest some more possibilities? Please remember they must be applicable to vectors. Thanks!
Square:
Opposite sides congruent and all sides meet at right angles. Opposite sides congruent and diagonals are perpendicular.
Diagonals are perpendicular and of the same length
Rhombus:
Diagonals are perpendicular Opposite sides congruent
Rectangle:
All sides right anglesParallelogram:
Opposite sides are congruent
Diagonal bisect each other.Trapezium:
One pair of congruent sidesCyclic Quadrilateral
Opposite angles are complementary.
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Thanks a heap Jane!
I'll have to change my list
Last edited by Identity (2009-03-09 22:33:44)
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...I believe the opposite sides in a cyclic quadrilateral are actually supplementary (add to 180 degrees), not complementary (add to 90 degrees)
hi Identity,
{squares} are a subset of {rectangles}, yes? Because a quadrilateral with 90 degree corners and all sides equal is a square but is also a rectangle (a quad. with 90 degree corners).
So many years ago, I thought I'd try to make a large Venn diagram to show all quadrilaterals with the right subsets and intersections. Oh boy, that was quite a task and it got just a bit too complicated.
And what about {trapeziums} ? Must they have exactly one pair of parallels or at least one pair of parallels.
At that time I gave up my quest for that diagram. But you've reminded me of it, so maybe I'll try again.
Question. What exactly do you mean by congruent?
Do you mean equal in length and parallel?
Because that makes your parallelogram definition ok but not your trapezium. I'm confused.
How about the quadrilateral shown below?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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