Arranging the sides in the order $3,15,11,15,3,15,11,15$, we obtain two overlapping, perpendicular inscribed rectangles of sides $3×h$ and $11×k$ respectively, where $h,k$ are to be determined. The Pythagorean theorem gives
Furthermore, the rectangles are aligned and have a common centroid. This gives
This immediately suggests finding Pythagorean triples $(a,b,30)$, of which there is only one: $18^2+24^2=30^2$. If we let $h-11=18$ and $k-3=24$, we see that the other equation $h^2-k^2=112$ is "miraculously" satisfied.
Thus $h=29$ and $k=27$. The octagon's area can then be calculated easily by adding the rectangles' areas, subtracting their intersection and adding the triangles' areas: