Area of a hexagon from the distances of the opposite sides

The general problem is not easy at all: in some cases, it is best to take the problem poser point of view. How would you construct a simple equilateral hexagon with parallel opposite sides? My inspired guess was to exploit the $(3,4,5)$ Pythagorean triple:

enter image description here

and I was so lucky that I met the exact constraints on the distances of opposite sides.
From the above diagram it is trivial to get that the side length is $5$ and the wanted area is

$$ 25+25-2+6+6 = \frac{5}{2}(7+8+9) = \color{red}{60}.$$


Similar to Jack D'Aurizo's suggestion, consider embedding our hexagon into $\mathbb{C}$ by placing one vertex at $z=0$, another at $z=r$, and a third at $z=re^{i\alpha}$, where $r$ is the common length of the six sides.

a hexagon embedded in the complex plane

The leftmost vertex in the above figure is at $z=re^{i\alpha}$, so the vertex that forms the left endpoint of the top horizontal side is at $z=re^{i\alpha}+re^{i\pi-\beta}$, since the top-left side must be parallel to the bottom-right side. The imaginary part of this vertex will give us the distance from the bottom horizontal side to the top horizontal side, which we call $d_1$. So we have \begin{equation} d_1 = \Im(r(e^{i\alpha}+e^{i(\pi-\beta)})) = r(\sin\alpha + \sin(\pi-\beta)) = r(\sin\alpha+\sin\beta). \end{equation} Denoting the distance from the bottom-left side to the top-right side by $d_2$ and the distance from the top-left side to the bottom-right side by $d_3$, we similarly see that \begin{equation} d_2 = r(\sin\alpha+\sin\gamma) \quad\text{and}\quad d_3 = r(\sin\beta+\sin\gamma). \end{equation} (Edit: We didn't have to embed into $\mathbb{C}$ to get these equations. We could also have drawn a horizontal line through the leftmost vertex and vertical lines through the other two left-hand vertices, then computed the heights of the two resulting triangles.) Now we can start getting tricky. Since we expect $d_1,d_2$, and $d_3$ to determine many (perhaps all) of the properties of our hexagon, let's compute the angles from these distances. We have \begin{align*} d_1+d_2-d_3 &= 2r\sin\alpha\\ d_1-d_2+d_3 &= 2r\sin\beta\\ -d_1+d_2+d_3 &= 2r\sin\gamma, \end{align*} so \begin{align*} \alpha &= \arcsin\left(\frac{d_1+d_2-d_3}{2r}\right)\\ \beta &= \arcsin\left(\frac{d_1-d_2+d_3}{2r}\right)\\ \gamma &= \arcsin\left(\frac{-d_1+d_2+d_3}{2r}\right). \end{align*} Because the opposite sides of our hexagon are parallel, opposite angles must agree. This means that $4\pi=2\alpha+2\beta+2\gamma$, so \begin{equation} 2\pi = \arcsin\left(\frac{d_1+d_2-d_3}{2r}\right) + \arcsin\left(\frac{d_1-d_2+d_3}{2r}\right) + \arcsin\left(\frac{-d_1+d_2+d_3}{2r}\right). \end{equation} We can get some traction here by using the following addition formula: \begin{align*} \arcsin x &+ \arcsin y + \arcsin z\\ &= \arcsin\left(x\sqrt{1-y^2}\sqrt{1-z^2} + y\sqrt{1-x^2}\sqrt{1-z^2} + z\sqrt{1-x^2}\sqrt{1-y^2}-xyz\right). \end{align*} Applying this to our above equation and multiplying through by $8r^3$ yields \begin{align*} 0 = (d_1 &+d_2 - d_3)\sqrt{4r^2-(d_1-d_2+d_3)^2}\sqrt{4r^2-(-d_1+d_2+d_3)^2}\\ &+(d_1 - d_2 + d_3)\sqrt{4r^2-(d_1+d_2-d_3)^2}\sqrt{4r^2-(-d_1+d_2+d_3)^2}\\ &+(-d_1 + d_2 + d_3)\sqrt{4r^2-(d_1+d_2-d_3)^2}\sqrt{4r^2-(d_1d_2+d_3)^2}\\ &-(d_1+d_2-d_3)(d_1-d_2+d_3)(-d_1+d_2+d_3). \end{align*} This equation can be rearranged to produce \begin{align*} 1&= \frac{\sqrt{4r^2-(d_1-d_2+d_3)^2}}{d_1-d_2+d_3}\frac{\sqrt{4r^2-(-d_1+d_2+d_3)^2}}{-d_1+d_2+d_3} + \frac{\sqrt{4r^2-(d_1+d_2-d_3)^2}}{d_1+d_2-d_3}\frac{\sqrt{4r^2-(-d_1+d_2+d_3)^2}}{-d_1+d_2+d_3} + \frac{\sqrt{4r^2-(d_1+d_2-d_3)^2}}{d_1+d_2-d_3}\frac{\sqrt{4r^2-(d_1-d_2+d_3)^2}}{d_1-d_2+d_3}. \end{align*} This seems like a good place to hand things over to a computer algebra system. I asked Mathematica to solve \begin{equation} 1 = \frac{\sqrt{4r^2-x^2}}{x}\frac{\sqrt{4r^2-y^2}}{y}+\frac{\sqrt{4r^2-x^2}}{x}\frac{\sqrt{4r^2-z^2}}{z} + \frac{\sqrt{4r^2-y^2}}{y}\frac{\sqrt{4r^2-z^2}}{z} \end{equation} for $r$ in terms of $x$, $y$, and $z$. The result was \begin{equation} r = \frac{xyz}{\sqrt{-x^4+2x^2y^2-y^4+2x^2z^2+2y^2z^2-z^4}} = \frac{xyz}{\sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}}. \end{equation} We can then make the substitutions $x=d_1+d_2-d_3$, $y=d_1-d_2+d_3$, and $z=-d_1+d_2+d_3$ to obtain the following expression for $r$ in terms of the distances: \begin{equation} \boxed{r = \frac{(-d_1 + d_2 + d_3) (d_1 + d_2 - d_3) (d_1 - d_2 + d_3)}{\sqrt{(d_1+d_2+d_3)(3d_1-d_2-d_3)(3d_2-d_3-d_1)(3d_3-d_1-d_2)}}.} \end{equation} (Thanks to Blue for simplifying these expressions!) The area of the hexagon will thus be \begin{equation} \boxed{A = \frac{1}{2}r(d_1+d_2+d_3) = \frac{(-d_1 + d_2 + d_3) (d_1 + d_2 - d_3) (d_1 - d_2 + d_3)\sqrt{d_1+d_2+d_3}}{2\sqrt{(3d_1-d_2-d_3)(3d_2-d_3-d_1)(3d_3-d_1-d_2)}}.} \end{equation} It's worth pointing out that in our case we have $d_1=9$, $d_2=8$, and $d_3=7$ (up to some permutation of labels), so \begin{equation} r = \frac{(6)(8)(10)}{\sqrt{(24)(12)(8)(4)}} = \frac{480}{\sqrt{9216}} = \frac{480}{96} = 5, \end{equation} as concluded by Jack. This solution is not nearly as inspired or pleasant to look at as Jack's, but I think it provides the generality you hoped for.


Edit. Here's some of the work we asked Mathematica to do. Let's rewrite the equation as \begin{equation} 1 = \sqrt{\left[4\left(\frac{r}{x}\right)^2-1\right]\left[4\left(\frac{r}{y}\right)^2-1\right]} + \sqrt{\left[4\left(\frac{r}{x}\right)^2-1\right]\left[4\left(\frac{r}{z}\right)^2-1\right]} + \sqrt{\left[4\left(\frac{r}{y}\right)^2-1\right]\left[4\left(\frac{r}{z}\right)^2-1\right]}. \end{equation} We can move the rightmost root to the LHS and square to obtain \begin{equation} \left(1 - \sqrt{\left[4\left(\frac{r}{y}\right)^2-1\right]\left[4\left(\frac{r}{z}\right)^2-1\right]}\right)^2 = \left[4\left(\frac{r}{x}\right)^2-1\right]\left(\sqrt{\left[4\left(\frac{r}{x}\right)^2-1\right]\left[4\left(\frac{r}{y}\right)^2-1\right]} + \sqrt{\left[4\left(\frac{r}{x}\right)^2-1\right]\left[4\left(\frac{r}{z}\right)^2-1\right]}\right)^2. \end{equation} Solve this for $4(r/x)^2$ and you'll find that \begin{equation} \frac{4x^2r^2}{y^2z^2} = 4\left(\frac{r}{y}\right)^2 + 4\left(\frac{r}{z}\right)^2 + 2\sqrt{4\left(\frac{r}{y}\right)^2-1}\sqrt{4\left(\frac{r}{z}\right)^2-1} - 2. \end{equation} Now isolate the remaining radicals and square again. After factoring out $\frac{4r^2}{y^4z^4}$ you'll have an expression from which we can easily find $r^2$.


Extend every other edge of the hexagon (of side-length $s$) to determine three points $A$, $B$, $C$; define $a := |BC|$, $b := |CA|$, $c := |AB|$.

Note that the parallel edges of the hexagon give rise to similar triangles. The figure labels segments according to the appropriate proportions.

enter image description here

Now, consider the hexagon's opposite-edge distance $k$. By similarity, we can write (with $T := |\triangle ABC|$)

$$\frac{|BB_c|}{|BC|} = \frac{k}{\text{altitude from $C$}} \quad\to\quad \frac{s(a+b)}{ab} = \frac{k}{2T/c} \quad\to\quad a b ck=2Ts(a+b) \tag{1} $$ Likewise, $$a b c h = 2 T s (b+c) \qquad a b c j = 2 T s (c+a) \tag{2}$$

Thus,

$$h : j : k \;=\; b+c:c+a:a+b \tag{3}$$

and we deduce that, for some $\lambda$,

$$a = (-h + j + k)\lambda \qquad b = (h-j+k)\lambda \qquad c = (h+j-k)\lambda \tag{4}$$ so that, by Heron's Formula, $$\begin{align} T &=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} \\[4pt] &= \frac{\lambda^2}{4}\sqrt{(h+j+k)(3h-j-k)(3j-k-h)(3k-h-j)} \end{align} \tag{5}$$ Finally, substituting $(4)$ and $(5)$ back into $(1)$:

$$s =\frac{(-h+j+k)(h-j+k)(h+j-k)}{\sqrt{(h+j+k)(3h-j-k)(3j-k-h)(3k-h-j)}} \tag{6}$$

The hexagon's area follows immediately. $\square$