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Given the length of the sides of a quadrilateral $a,b,c,d$ the area of the quadrilateral is less than or equal to $\frac{(a+b+c+d)^2}{16}$ i.e it is an upper bound of the area of any quadrilateral with side length $a,b,c,d$.

A question that came to my mind is if $\frac{(a+b+c+d)^2}{16}$ is an upper bound Then the least upper bound exits so what is the infimum and supremum of the area of quadrilateral (0 is an oblivious lower bound so the infimum exists too)?

I made a silly mistake thinking that $\frac{(a+b+c+d)^2}{16}$ is the supremum but it turns out I was wrong.

If $a=b, \ c=d$ "a kite" or if $a=c, \ d=b$ "a parallelogram" then it is easy to prove the infimum is $0$ and the supremum is just $a^2$ when it is a rhombus.

The infimum depends on $a,b,c,d$ in this examples the side lengths are $19,13,7,4$ and the minimum area is around $9.5$ so it is not $0$.

enter image description here

Conjecture: If $a<b<c<d$ and the area of the quadrilateral is $S$ then $\inf S $ is the area of the triangle $a,b,d-c$ which is

$$\sqrt{\frac{a+b+d-c}{2}\cdot\frac{a-b+d-c}{2}\cdot\frac{-a+b+d-c}{2}\cdot\frac{a+b-d+c}{2}}$$ This also works when $a=b=c=d, \ \ \ a=b, \ c=d \ \ \ ,a=c, \ d=b $.

I couldn't rigorously proof that claim but I am pretty sure it is correct.

In the previous example when side lengths are $19,13,7,4$ the naximum area is around $73.5$ but this time I have no Idea what kind of shape is that.

enter image description here

How to determine the infimum and the supremum of the area given $a,b,c,d$.

  • 2
    $\begingroup$ May we assume that the side lengths are given in order around the quadrilateral? $\endgroup$
    – paw88789
    Apr 10 at 21:39
  • $\begingroup$ It might depend on the values of $a,b,c,d$ but in general you can make some of the angels as small as possible and the others as close to $180$. But we could have some issues. If $b=d < c <a$ but $c+b >a$ we can't lay any two sides flat. $\endgroup$
    – fleablood
    Apr 10 at 21:43
  • 1
    $\begingroup$ Let three sides be $1$ and the last edge $0.0001$ Draw some pictures $\endgroup$
    – Will Jagy
    Apr 10 at 21:47
  • 1
    $\begingroup$ @WillJagy That should probably be an answer (with perhaps one more sentence). :) $\endgroup$ Apr 10 at 21:49
  • 2
    $\begingroup$ @WillJagy Note also, in your example the area (which is roughly that an equilateral triangle $\approx \frac{\sqrt{3}}4$ is a lot less than $\frac {(1+1+1+0.0001)^2}{16}$. So neither $0$ is the greatest lower bound (of all) nor is $\frac {(a+b+c+d)^2}{16}$ the least upper bound (of all). $\endgroup$
    – fleablood
    Apr 10 at 21:55

2 Answers 2


The supremum will be when the four nodes lie exactly on a circle: a cyclic quadrilateral. The area of a quadrilateral inscribed in a circle is given by Bretschneider’s formula (actually Brahmagupta's formula) as:Brahmagupta

Area = √[(s-a) (s-b) (s – c) (s – d)] where s is the semi-perimeter of the quadrilateral $\frac {a+b+c+d}{2}$

Your original for maximum area provides an example of this:enter image description here
EDIT: Proof
Bretschneider’s formula gives the area for any quadrilateral from lengths of sides:
Area $$ \sqrt{(s-a)(s-b)(s-c)(s-d)-abcd*cos^2((\alpha+\gamma)/2)} $$
where $\alpha$ and $\gamma$ are either pair of opposite angles in the quadrilateral.

a,b,c and d are all positive, $cos^2$ is always positive, therefore this will be a maximum when $cos^2$ = 0, i.e. when shape is cyclic quadrilateral. QED

The smallest area will occur when the $cos^2$ term is as large as possible (1). This happens when the quadrilateral is flattened and those angles will be 0+0 or $\pi$ + $\pi$.
If the quadrilateral cannot be flattened then a triangle will remain. If we label the sides a,b,c, and d then let c lie collinear with d. The angle between c and d is then 0. c and d are chosen such that the angle between the other two sides, a and b is the smallest of the possible choices. Increasing the angle between c and d also increases the angle between a and b increasing the area, so this is the minimum.

Apologies - the first site I linked to had errors.

  • $\begingroup$ This is a very good observation, It remains to prove these two calims. $\endgroup$
    – pie
    Apr 12 at 22:28
  • $\begingroup$ @pie First one down $\endgroup$
    – Rich
    Apr 12 at 23:56
  • $\begingroup$ @pie My other assumption was not correct. I have edited it out $\endgroup$
    – Rich
    Apr 13 at 0:49

Areas can get arbitrarily small under some circumstances, suach as edges (in order) $a,b,a,b$ which has maximal area as a rectangle, but can be squashed down as a parallelogram.

The example that came to mind was, essentially, an equilateral triangle. Let three edges have length $100$ but the final edge have length $1.$ From what I can see, the least area is when the $1$ lies flat against a neighboring edge, nonconvex, resulting triangle edges $100, 100, 99$

My first guess on the maximum area is isosceles triangle with sides $100, 100, 101$

triangle $100, 100, 99$ came out 4301.02. triangle $100, 100, 101$ came out 4358.75. The cyclic quadrilateral $100,100, 100, 1$ comes out 4387.9095.

  • $\begingroup$ The greatest area will be when the length 1 side is close to perpendicular to form a quadrilateral. The closer you can get to a circle, the greater the enclosed area. 4380 vs 4360 for the triangle. $\endgroup$
    – Rich
    Apr 12 at 21:36

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