Area of a hexagon from the distances of the opposite sides Problem. All the six sides of the hexagon in the attached figure have equal length, and the opposite sides are parallel. The distances of the three opposite pairs are $d_1=7\mathrm{cm}$, $d_2=8\mathrm{cm}$ and $d_3=9\mathrm{cm}$, respectively. Find the area of the hexagon.
Clearly, 
$$
\text{Area}=\frac{1}{2}x(d_1+d_2+d_3),
$$ 
where $x$ is the length of each of the sides. Hence, it suffices to find $x$ as a function of $d_1,d_2$ and $d_3$.
EDIT. The area in the particular case $d_1=7\mathrm{cm}$, $d_2=8\mathrm{cm}$ and $d_3=9\mathrm{cm}$ is provided in the answer of Jack D'Aurizio that follows. I was wondering whether it is possible to obtain an expression the area of the hexagon as a function of $d_1,d_2$ and $d_3$.

 A: 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:

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

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. $\square$
