2023 AMC 10A Problem 24

Let $d=\tfrac37.$ Extend the slanted edges of the blocks that meet a fixed side of the frame. Because all the relevant angles are $60^\circ,$ the extensions form an equilateral triangle of side $1$ at one end and an equilateral triangle of side $1-d$ at the other. Thus that frame side is partitioned into lengths $d,1,1,$ and $1-d,$ so its length is $d+1+1+(1-d)=3.$ A regular hexagon of side $t$ has area $\tfrac{3\sqrt3}{2}t^2.$ Therefore the uncovered area is the area of the side-$3$ frame minus the areas of the six unit blocks: $\tfrac{3\sqrt3}{2}\cdot 3^2$ $- 6\cdot\tfrac{3\sqrt3}{2}$ $= \tfrac{27\sqrt3}{2} - 9\sqrt3$ $= \tfrac{9\sqrt3}{2}.$ Therefore, the answer is C.