## 3D Trapezoids

A few days ago, I was discussing the area formula for a trapezoid with students who had just derived it for themselves. The discussion led to a neat generalization of the trapezoid to higher dimensions different from the one that envisions the 3D trapezoid as a pyramidal frustum.

The standard formula for the area of a trapezoid is $\frac{h}{2}(b_1+b_2),$ where $h$ is the height of the trapezoid and $b_1$ and $b_2$ are the lengths of the trapezoid’s two bases. We observed that the formula for area does not depend on where the bases are located provided that they slide along parallel lines separated by a distance of $h$.

All three trapezoids have the same base lengths and altitude and so they all have the same area.

This is an instance of Cavalieri’s principle (which I exploited in this visual proof of the Pythagorean theorem). We also saw that the area formula can be interpreted as the product of the average of the lengths of the bases and the distance $h$.

When you imagine sliding the bases left and right, it might make you think of a trapezoid in the following way: Take two line segments. Align them both horizontally so that they are parallel to each other (but not on the same line). Now draw a line segment that connects the leftmost endpoints and another line segment that connects the rightmost endpoints. The result is a trapezoid and its area is equal to the product of the average length of the two line segments and their vertical separation.

This perspective on the trapezoid suggests a neat generalization to higher dimensions.

For 3D, take three line segments instead of two. Align them horizontally so that they are mutually parallel, but make sure that they do not all lie in the same plane. Connect the three leftmost endpoints to form a triangle. Do the same for the rightmost endpoints. The result is a 3D solid which is not generally a triangular prism, skew or otherwise, because the three chosen segments need not have the same length. Here are two examples of this kind of 3D trapezoid:

What’s cool about this generalization of the trapezoid is that its volume formula is a natural generalization of the area formula for a trapezoid. The volume of this “3D trapezoid” is the product of the average length of the 3 chosen line segments and the perpendicular cross-sectional area. In other words, the average length of the bases in the 2D case is replaced with the average length of the “bases” in the 3D case and the altitude is replaced with a cross-sectional area.

Using the labeling in the figure above, the volume is $\overline{b}A,$ where $\overline{b} = \frac{b_1+b_2+b_3}{3}$ is the average length of the 3 chosen line segments and $A$ is the (perpendicular) cross-sectional area. (To get the right cross-sectional area, you may have to imagine that the bases are extended, just as in a skew trapezoid, there may not be an altitude contained within.)

This way of generalizing works in any dimension. In $N$-dimensions, pick any $N$ line segments you wish. Position them so that they are mutually parallel to each other and so that they are not all contained in a hyperplane. Join up the endpoints on the same ends of the segments to form two $N-1$-simplices that “cap off” our $N$-dimensional solid. The justification for this generalization is that no matter the dimension, the volume is given by the same formula: $\overline{b}A,$ where $\overline{b}$ is the average length of the $N$ chosen line segments and $A$ is the $N-1$-dimensional volume of the perpendicular cross-sectional area!

In this respect, this higher dimensional generalization of the trapezoid is nicer than the one that considers an $N$-dimensional trapezoid to be a pyramidal frustum. Also, with the pyramidal frustum, it isn’t clear what should be allowed for the base.