MHB Max Area of Simple Quadrilateral w/ Sides $a$ & $b$: Justification

[lfdahl]

What is the largest possible area of a simple quadrilateral, two sides of which
have length $a$ and two sides of which have length $b$? Please justify your statement.

 

[castor28]

What is the largest possible area of a simple quadrilateral, two sides of which
have length $a$ and two sides of which have length $b$? Please justify your statement.

[sp]
The area of polygon with given sides is obviously largest when the polygon is convex.

Given a convex quadrilateral $ABCD$, we can interchange the length of two consecutive sides without changing the area. Indeed, we can flip the triangle $ABC$ with respect to the perpendicular bisector of $AC$ to get a quadrilateral $AB'CD$ with the same area and perimeter. As transpositions generate the whole symmetric group $S_4$, we may permute the lengths of the sides in any way without changing the solution. Note that this is true for any convex polygon.

In this case, we may therefore assume that equal sides are not adjacent; the quadrilateral is therefore a parallelogram with sides $a$ and $b$. The area will be maximum and equal to $ab$ when that parallelogram is a rectangle.

The original quadrilateral will be either a rectangle or a kite-shaped figure with two opposite right angles (a right triangle and its mirror image with respect to the hypotenuse).

Note: it can be shown that, for a quadrilateral with any given sides, the area is largest when the quadrilateral is cyclic; and that too can be generalized to any polygon. See here for a proof.
[/sp]

 

[lfdahl]

[sp]
The area of polygon with given sides is obviously largest when the polygon is convex.

Given a convex quadrilateral $ABCD$, we can interchange the length of two consecutive sides without changing the area. Indeed, we can flip the triangle $ABC$ with respect to the perpendicular bisector of $AC$ to get a quadrilateral $AB'CD$ with the same area and perimeter. As transpositions generate the whole symmetric group $S_4$, we may permute the lengths of the sides in any way without changing the solution. Note that this is true for any convex polygon.

In this case, we may therefore assume that equal sides are not adjacent; the quadrilateral is therefore a parallelogram with sides $a$ and $b$. The area will be maximum and equal to $ab$ when that parallelogram is a rectangle.

The original quadrilateral will be either a rectangle or a kite-shaped figure with two opposite right angles (a right triangle and its mirror image with respect to the hypotenuse).

Note: it can be shown that, for a quadrilateral with any given sides, the area is largest when the quadrilateral is cyclic; and that too can be generalized to any polygon. See here for a proof.
[/sp]

You are of course right, castor28! Thankyou very much for your participation and for a very "beyond the challenge" instructive answer.(Nod)

 

[kaliprasad]

Spoiler

we can chose the vertices ABCD of the quadrilateal so that (AB= a, BC= b) and then based on the sides(AD=a, CD= b) or (AD=b, CD =a).

so the 2 triangles ABC and ADC/ ACD(depending on the case above) are congruent. hence area same

area of the triangle ABC (AB and BC fixed) is maximum when angle B is $90^\circ$ and it is right angled triangle.

so is angle D.

So ABCD is cyclic quadrilateral and area is ab.

 

[lfdahl]

Spoiler

we can chose the vertices ABCD of the quadrilateal so that (AB= a, BC= b) and then based on the sides(AD=a, CD= b) or (AD=b, CD =a).

so the 2 triangles ABC and ADC/ ACD(depending on the case above) are congruent. hence area same

area of the triangle ABC (AB and BC fixed) is maximum when angle B is $90^\circ$ and it is right angled triangle.

so is angle D.

So ABCD is cyclic quadrilateral and area is ab.

Yes, you are of course right, kaliprasad!
Thankyou for your participation.