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Recently I started thinking about the properties of the following shape, which I like to call the "Crossed Trapezium". It has two parallel edges, which are joined by two crossing lines.

These comments were left on the original blog post:

Five Triangles 6 April 2016:

Related factoid: in the figure you dub a “crossed trapezium”, the two triangles that were removed from the original trapezium have equal area.

Also, this useful figure pops up in a lot of ratio problems we tweet about, such as “method 4” solution in this octagon problem:

David Butler 6 April 2016:

Interestingly, not only are they equal area, but their areas together are hab/(a+b). Most interesting.

Tom A 13 September 2017:

The relationship between the formulas for the ordinary and crossed trapezia is rather elegant. Unfortunately, the formula for the ordinary trapezium isn’t numerically well-behaved: if the two bases are the same length, then the a-b term is zero, and the result is undefined; if the two bases are nearly the same length, a-b is very small, and calculations on a computer may be inaccurate.

An alternative formula without that problem is:

a = 1/2 * ((b^2 + a^2 + 2ab) / (b + a)) * h

Which is still similar to the formula for the crossed trapezium, but doesn’t have an obvious (to me) geometric interpretation! Something about the triangles cut out of the sides?

Peter 19 September 2017:

Thanks for sharing this interesting work. I first started thinking of areas for crossed trapezia whilst preparing lessons on motion in a straight line for my students. I have been looking at the vel-time graph for a period of constant acceleration where the sign of the velocity changes. The area trapped between the graph and the t-axis is a crossed trapezium. Interestingly, if you consider the area in the first triangle to be postive and the second to be negative (consistent with calculating displacements), the formula for the area of the trapezium is unchanged from the canonical one.