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

[Albert1]

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)

 

[Prove It]

I can tell already this is not possible, as the LHS is a number while the RHS is a vector...

 

[Albert1]

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)

 

[MarkFL]

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

$$(a+b)\cdot(x+y)$$

 

[Opalg]

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]