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The area of a "Quadrilateral"

Albert

Well-known member
Jan 25, 2013
1,225
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

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
I can tell already this is not possible, as the LHS is a number while the RHS is a vector...
 

Albert

Well-known member
Jan 25, 2013
1,225
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

Administrator
Staff member
Feb 24, 2012
13,775
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:

\(\displaystyle (a+b)(x+y)\) (preferred)

or

\(\displaystyle (a+b)\cdot(x+y)\)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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)
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.