A certain convex hexagon is equilateral but not equiangular. Its opposite sides are also parallel, and separated by distances of 27, 31 and 38.

Find the area of this hexagon.

(In reply to re: Some thoughts by Jer)

Its also possible you are assuming the midlines of the parallel sides coincide at a single point O.  I'm not sure they must.

Assuming the midlines are the lines connecting the midpoints of the parallel  sides, I believe they must, and do.

Consider each pair of sets of parallel lines, which form 3 parallelograms say, P1,P2, P3. The midlines of P1 bisect each other at the centre of P1, say X. But since there is always a common midline between each pair of sets, this bisection must happen at the same point, so all these midlines must coincide at X; by construction O, in our case.

Regrettably this is not the same as saying that the perpendiculars to the sides through O pass through A and B, as I wrongly assumed. However, the concentric circles on O - while harmless in themselves -  run sufficiently close to A and B to invite the erroneous conclusion that the trick of the problem is that they do.

Reverting to an earlier save, and using a slightly modified construction, I find that lengths AT and BT are 7.625 and 13.625 respectively, giving an overall side length of 21.25, and an area of 1020 for the desired figure.

Edited on September 13, 2020, 6:45 am

Posted by broll on 2020-09-13 00:51:06