Ytukyg

To add a derivation that puts the square root factor in a Heronian context ...

For $c := |overline{CD}| neq |overline{AB}| =: a$,
$$|square ABCD| = frac12 (a+c) cdot h = frac12 (a+c) cdot frac{2 |triangle AC^prime D|}{|overline{C^prime D}|} = frac{a+c}{|a-c|};|triangle AC^prime D|$$

Then, applying Heron's Formula to a triangle with side-lengths $b$, $d$, $c-a$, we have
$$|triangle AC^prime D| = frac14sqrt{((c-a)+b+d)(-(c-a)+b+d)((c-a)-b+d)((c-a)+b-d);}$$

A note for the case when only two sides are parallel,
just the set of four side lengths do not determine the area.
An additional information is needed to define,
which pair of sides are parallel.
An illustrative example for side lengths $19,23,29,31$:

This is how to calculate the area of a trapezoid when the four sides are known:

Hint (if we know the parallel sides):

From The picture:

take: $a=AB, b=BC, c=CD, d=DA,x=AE$

so we have:
$
h=ED=sqrt{d^2-x^2}=sqrt{b^2-(a-c-x)^2}=CF
$

solve for $x$ and find $h=sqrt{d^2-x^2}$

Find the area $A=frac {a+b}{2}h$

There can't be such a formula. The side lengths do not determine the area.

Think about all the rhombi with four sides of length $1$. They are all trapezoids (even parallelograms) with the same side lengths but different areas, which can be anything between $0$ and $1$.

-For the trapzoid abcd to have parallel sides it requires all these conditions to be set for uniqueness of area:

• $a+b+c>d$, $b+a<d$, $c+a<d$.

-Neverthless, a trapzoid with non parallel sides can not be defined just by his side lengths, but a fifth coordinate added should.

In this experience we show how come multiple trapzoid shapes can be formed with same lengths of edges.

Imagine we bring a fork and a knife to start dining on some digestable geometrical concepts :

A thread and two pins:

We instill the pins on some flat table:

Then take the fork and the knife, choose two fixed points in the string, then pull it from these points with those tools without changing the fulcrum points.

Distance of the chord from the pinpoints to the coordinates of fulcrums dosn't change, while the shape of the trapzoid changes infinitely!

now envisage that $h_1$ is figured by the fork, $h_2$ symbolised by the knife, h1 is directly relative to h2 always regarding the same side lengths. We will show in the following trigonometric relations:

• $costheta=b`/a$,
  
  $sintheta=(h_2-h_1)/a implies sqrt{1-(b`/a)^2}=(h_2-h_1)/a$


• $b``/d= sqrt{1-(h_2/d)^2}$


• $(b-b`-b``)/c= sqrt{1-(h_1/c)^2}$

Since there is 4 unknowns $b`,b``,h_1,h_2$ we can formulate $h_1$ in function of $h_2$ and 4 side constants.

Credits for the images goes to canstockphoto.com

I derived the formula for the area using the same procedure as is used to show Heron's Formula. Looking at the end formula, I realized that the formula follows quite simply from Heron's Formula.

Given a trapezoid with unequal parallel bases $a$ and $b$,

consider the triangle with base $|b-a|$ and sides $c$ and $d$:

Using $s=frac{|b-a|+c+d}2$, the area of the triangle is
$$
text{Area of Triangle}=sqrt{s(s-c)(s-d)(s-|b-a|)}
$$
The altitude of the trapezoid and the triangle are the same, so the area is proportional to the average of the lengths of the bases. That is,
$$
bbox[5px,border:2px solid #C0A000]{text{Area of Trapezoid}=frac{b+a}{|b-a|}sqrt{s(s-c)(s-d)(s-|b-a|)}}
$$

I think you mean Brahmagupta's formula, not Heron's formula. There is no formula for the area of trapezoid given the lengths of the sides, because the sides alone do not determine the area. This is true even for a parallelogram. Imagine a parallelogram made of four sticks, joined together by pins at the corners. Then you can slide it closed by moved the top side parallel to the bottom side. You'll get zero area when the top and bottom coincide, and maximum are when you have a rectangle.

In the case of Brahmagupta's formula, the quadrilateral is circumscribable, and you can't change the sides like that.

M Weiss put

$$A=frac{a+b}{2}csqrt{1-left( frac{(b-a)^2+c^2-d^2}{2c(b-a)} right)^2} .$$

I would rather have that answer symmetrical in $c$ & $d$, which would be

$$A=frac{a+b}{4(b-a)}sqrt{(b-a)^4+2(b-a)^2(c^2+d^2)+(c^2-d^2)^2} .$$

But it's just that to my taste it looks odd that it's not symmetrical in $c$ & $d$ ... my mind just protests that it ought to be!

Or

$$A=frac{a+b}{4(b-a)}sqrt{(b-a)^2((b-a)^2+2(c^2+d^2))+(c^2-d^2)^2} ,$$

even.

It depends on your definition of trapezoid. If it includes parallelograms, then No: here are two trapezoids, same side-lengths, different area: