If a parallelogram has sides of length $a$ and $b$, and diagonals of length $d$ and $e$, then we can find its area in the following way. By the polarization identity, we have $a b \cos\theta = \frac{1}{4}|d^2-e^2|$. Now the area is $a b \sin \theta$, which is equal to $\sqrt{a^2 b^2 – (a b \cos \theta)^2}$, or
$$\sqrt{a^2 b^2 – \frac{1}{16}(d^2-e^2)^2}.$$
This is not an exceptionally pretty formula, but maybe it's the best one can do.
How can such a formula be generalized to higher dimensions? In particular, what is the volume of a parallelpiped, given its edge and diagonal lengths?
Best Answer
Let the parallelepiped with a vertex $O$ at the origin have vertices $A$, $B$, $C$. Write $D = B+C$; $E= C+A$; $F=A+B$; and $G=A+B+C$. Further, assign these names to the lengths of edges and diagonals: $$a := |OA|\quad b := |OB| \quad c := |OC| \quad d := |AD| \quad e := |BE| \quad f := |CF| \quad g := |OG|$$
Note that a parallelepiped (and it properties) is completely determined by six quantities: the lengths of three edges meeting at a vertex, and the measures of the three angles between those edges. The total number of edge-lengths and diagonal-lengths is seven, so there's a dependency among these values, namely $$4 \;(\; a^2 + b^2 + c^2\;) = d^2 + e^2 + f^2 + g^2$$ Although one diagonal (say, $g$) is unnecessary, it helps to simplify the volume formula a bit:
$$\begin{align} 32 V^2 &= 32 a^2 b^2 c^2 + d^2 e^2 f^2 + g^2 ( d^2 e^2 + e^2 f^2 + f^2 d^2 )\\ &- 2 g^2 ( a^2 d^2 + b^2 e^2 + c^2 f^2 ) - 2 ( a^2 e^2 f^2 + d^2 b^2 f^2 + d^2 e^2 c^2 ) \end{align}$$ In the case of a rectangular parallelepiped, for which $d^2=e^2=f^2=g^2=a^2+b^2+c^2$, the formula reduces to $V = a b c$, as expected.
I derived the formulas using coordinates $$A = (a_x,0,0) \qquad B = (b_x, b_y, 0) \qquad C = (c_x, c_y, c_z)$$ Expressing the various distances in terms of these $$a^2 = A\cdot A = a_x^2 \qquad b^2 = B\cdot B = b_x^2 + b_y^2 \qquad c^2 = C\cdot C = c_x^2 + c_y^2 + c_z^2$$ $$d^2 = \overrightarrow{AD}\cdot \overrightarrow{AD}=\dots \quad e^2 = \overrightarrow{BE}\cdot \overrightarrow{BE} =\dots \quad f^2 = \overrightarrow{CF}\cdot \overrightarrow{CF}=\dots$$ along with $$V = a_x b_y c_z$$ I systematically eliminated $a_x$, $b_x$, $b_y$, $c_x$, $c_y$, $c_z$ from the system with the help of Mathematica's
Resultant[]function. (Appropriate applications of the Law of Cosines should work just as well, but the Resultant process lets me crank through the equations without having to think too hard. :)An $n$-dimensional parallelotope is determined by $\frac{1}{2}n(n+1)$ values. (These are the triangular numbers. You can see their relevance by noting the coordinatization done above considers vectors using $1$, $2$, $3$, ..., $n$ non-zero coordinates.) The figure has n characteristic edge-lengths and $2^{n-1}$ diagonals, so expressing volume in terms of these is certainly possible. With an increasing number of "extra" lengths, the resulting formula can take a variety of forms.