The Heron triangle has integer sides and area. The Robbins pentagon is just the generalization: it also has integer sides and area. The example below has sides $78, 126, 66, 50, 32$ and area $A_R = \color{brown}{4392}$.
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Define the formula for the area of Heron triangles as,
$$A_H(a,b,c) = \frac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}$$
We can partition the above Robbins pentagon into three Heron triangles and find that indeed,
$$A_R = A_H(78,126,120) + A_H(32,120,104) + A_H(104,50,66) = \color{brown}{4392}$$
P.S. Notice that the five diagonals form a pentagram (a $5$-pointed star) and also form a small pentagon .
Question:
- For this example, what is the area $A_r$ of the small pentagon?
- In general, if $A_R$ is rational, then is $A_r$ also rational?
We can place the large pentagon on a coordinate plane such that the vertices are $$\{(0,0), (96/5, -128/5), (336/5, -198/5), (120,0), (96/5, 378/5)\}.$$
Then the coordinates of the small pentagon $p'$ would be $$\left\{\left(\tfrac{3072}{85}, -\tfrac{12672}{595}\right), \left(\tfrac{10893}{190}, -\tfrac{1512}{95}\right), \left(\tfrac{507}{10}, 0\right), \left(\tfrac{96}{5},0\right), \left(\tfrac{96}{5}, -\tfrac{396}{35}\right)\right\}.$$ The resulting area is $$A_r = \frac{6983658}{11305}.$$
but $p'$ is not a Robbins pentagon.
The answer to your second question is not likely to be known, because it is not yet known whether there exist any Robbins pentagons that have interior diagonals that are irrational, but this is not a rigorous argument by any means.