Is there a geometric construction for the square root of the volume of a parallelotope?

A parallelotope is the higher dimensional analog of a parallelogram.

Now, what I want to know is if there's a way to construct an object with size equal to the square root of the volume of the parallelotope.

Simple example: I have a parallelogram with vertices at (0,0), (a, b), (c, d) and (a+c,d+b). How do I construct a line with length equal to the square root of the area of the parallelogram? IE: $$\sqrt{ad-bc}$$

• You seem to have two different questions here. The first is about higher dimensions than two. How do you define "construct" in that context? The second, about a general parallelogram, seems unrelated to the first. By "construct" do you mean with compass and straightedge? – Rory Daulton Aug 11 '15 at 18:27
• The square root of the volume of a parallelotope depends on the chosen units. – Christian Blatter Aug 11 '15 at 18:35
• @ChristianBlatter is correct. The square root of volume would have the unit $\mathrm{m}^{3/2}$. To get the unit $\mathrm{m}$ or a similar linear unit you would need something like the cube root of the volume. – Rory Daulton Aug 11 '15 at 18:55

The parallelogram is $ABDC$. Construct the perpendicular to line $\overleftrightarrow{CD}$ through point $B$ and let point $E$ be the intersection with line $\overleftrightarrow{CD}$. Use a circle to find point $F$, the point on ray $\overrightarrow{AB}$ for which $BE=BF$. Construct a semicircle that has segment $\overline{AF}$ as a diameter. Let the intersection of the perpendicular line through point $B$ and the semicircle be point $G$.
Then line segment $\overline{BG}$ is the desired length, the square root of the area of the parallelogram.
Why is that the desired length? Line segment $\overline{BE}$ is the altitude of the parallelogram where line segment $\overline{AB}$ is the base, so we want the square root of the product of $AB$ and $BE=BF$. Adding some auxiliary line segments and looking closely, we can see that triangles $\triangle ABG$ and $\triangle GBF$ are similar to each other, meaning that $BG$ truly is the square root of the product of $AB$ and and $BF$.