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

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In summary: So the inequality is true in general.In summary, by using the given lengths of the sides of quadrilateral ABCD, the inequality $4S \leq (a+b)(x+y)$ can be proved using basic trigonometry and the fact that sines are bounded by 0 and 1. This holds true for both convex and re-entrant quadrilaterals.
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Albert1
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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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  • #2
I can tell already this is not possible, as the LHS is a number while the RHS is a vector...
 
  • #3
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)
 
  • #4
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:

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

or

\(\displaystyle (a+b)\cdot(x+y)\)
 
  • #5
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]
 

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

What is a quadrilateral?

A quadrilateral is a 2-dimensional shape with four straight sides and four angles. It is a type of polygon and can have a variety of different shapes and sizes.

How do you calculate the area of a quadrilateral?

The formula for calculating the area of a quadrilateral depends on the type of quadrilateral. For a rectangle or square, you can multiply the length by the width. For a parallelogram, you can multiply the base by the height. For a trapezoid, you can add the lengths of the two parallel sides and multiply by the height, then divide by 2.

Can you use any formula to find the area of a quadrilateral?

No, the formula used to find the area of a quadrilateral depends on the type of quadrilateral. Using the wrong formula will result in an incorrect answer.

What is the difference between a regular and an irregular quadrilateral?

A regular quadrilateral has all equal sides and angles, while an irregular quadrilateral has varying sides and angles. Regular quadrilaterals include squares and rectangles, while irregular quadrilaterals can be any other shape.

How does the area of a quadrilateral relate to its perimeter?

The area and perimeter of a quadrilateral are related, but not directly. The perimeter is the sum of all the sides of the quadrilateral, while the area is a measure of the space inside the shape. In general, a larger perimeter does not necessarily mean a larger area, as the shape of the quadrilateral can vary greatly.

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