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- #1

#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

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)

- Thread starter Albert
- Start date

- Thread starter
- #1

- Jan 25, 2013

- 1,225

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)

- Thread starter
- #3

- Jan 25, 2013

- 1,225

a,b,x,y are all numbersI can tell already this is not possible, as the LHS is a number while the RHS is a vector...

(they are the length of respective segment)

the RHS is also a number

here $"\times"$

consider it as (a+b) multiplied by (x+y)

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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 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)

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

or

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

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- #5

- Feb 7, 2012

- 2,799

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)