0
$\begingroup$

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: $$\sqrt{ad-bc}$$

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Rory Daulton Aug 11 '15 at 18:27
  • $\begingroup$ The square root of the volume of a parallelotope depends on the chosen units. $\endgroup$ – Christian Blatter Aug 11 '15 at 18:35
  • $\begingroup$ @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. $\endgroup$ – Rory Daulton Aug 11 '15 at 18:55
4
$\begingroup$

Here is how to do a construction with compass and straightedge of a line segment with length equal to the square root of the area of a given parallelogram.

enter image description here

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 $BF$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.