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?

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