MHB How Can the Area of a Quadrilateral Be Bounded by Its Side Lengths?

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The discussion focuses on proving the inequality \(4S \leq (a+b)(x+y)\) for quadrilateral ABCD, where \(S\) is the area and \(a, b, x, y\) are the lengths of its sides. The participants clarify that both sides of the inequality are numerical values, not vectors, and suggest using the product notation \((a+b)(x+y)\) for clarity. The area \(S\) is expressed as the sum of the areas of triangles formed within the quadrilateral, leading to the conclusion that \(4S\) can be bounded by the sum of the products of the sides. It is noted that the proof holds for convex quadrilaterals, with a caveat regarding re-entrant shapes. The discussion effectively establishes a mathematical relationship between the side lengths and the area of a quadrilateral.
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Quadrilateral ABCD

given :$\overline{AB}=a ,\overline{BC}=y ,\overline{CD}=b

\,\, and \,\, \overline{AD}=x$

prove :

$4S \leq (a+b)\times(x+y)$

(where S is the area of ABCD)
 
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I can tell already this is not possible, as the LHS is a number while the RHS is a vector...
 
Prove It said:
I can tell already this is not possible, as the LHS is a number while the RHS is a vector...
a,b,x,y are all numbers
(they are the length of respective segment)
the RHS is also a number
here $"\times"$
consider it as (a+b) multiplied by (x+y)
 
Albert said:
a,b,x,y are all numbers
(they are the length of respective segment)
the RHS is also a number
here $"\times"$
consider it as (a+b) multiplied by (x+y)

Albert, I would reserve usage of the $\LaTeX$ \times command for scientific notation (between the mantissa and radix) and for the vectorial cross-product to avoid potential confusion in the future. I would write either:

$$(a+b)(x+y)$$ (preferred)

or

$$(a+b)\cdot(x+y)$$
 
Albert said:
Quadrilateral ABCD

given :$\overline{AB}=a ,\overline{BC}=y ,\overline{CD}=b

\,\, and \,\, \overline{AD}=x$

prove :

$4S \leq (a+b)(x+y)$

(where S is the area of ABCD)
[sp]Write $\theta,\phi,\alpha,\beta$ for the angles $ABC,CDA,DAB,BCD$ respectively. Then $$S = \text{area of triangle } ABC + \text{area of triangle } ACD = \tfrac12ay\sin\theta + \tfrac12bx\sin\phi,$$ and also $$S = \text{area of triangle } ABD + \text{area of triangle } BCD = \tfrac12ax\sin\alpha + \tfrac12by\sin\beta.$$ Add those equations to get $$4S = ay\sin\theta + bx\sin\phi + ax\sin\alpha + by\sin\beta.$$ But each of those sines lies between 0 and 1, and so $$4s \leqslant ay + bx + ax + by = (a+b)(x+y).$$ Afterthought: Strictly speaking, that proof only works if the quadrilateral is convex. But a re-entrant quadrilateral obviously has smaller area than the quadrilateral obtained by pushing the re-entrant part out so as to form a convex quadrilateral with the same sides.[/sp]
 
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