2024 AMC 10A Problem 16

The areas of the eleven pieces sum to $200.$ Let the enclosing rectangle have height $h$ and width $w=AB.$ Similar figures have corresponding side lengths in the square-root ratio of their areas. From the diagram, the area-$32$ piece contributes its short side to the left edge, of length $h\sqrt{\frac{32}{200}}=\tfrac25h.$ Above it, the area-$36$ piece contributes its long side, of length $w\sqrt{\frac{36}{200}}=\tfrac{3\sqrt2}{10}w.$ These two segments make the full height, so $h=\tfrac25h+\tfrac{3\sqrt2}{10}w,$ which simplifies to $h=\frac{w}{\sqrt2}.$ Since $wh=200,$ we get $\frac{w^2}{\sqrt2}=200$ and $w=\sqrt{200\sqrt2}=10\sqrt[4]{8}.$ Therefore, the answer is D.