2022 AMC 10A Problem 21

View the rim from directly above. Place the bottom square at $(\pm\tfrac12,\pm\tfrac12,0).$ In a regular hexagon, an edge adjacent to the attached side has components $\tfrac12$ parallel and $\tfrac{\sqrt3}{2}$ perpendicular to that side. At a corner of the bottom square, the corresponding edges of two adjacent hexagons coincide. Comparing their horizontal components shows that the horizontal component of a unit vector perpendicular to an attached side within its hexagon is $\frac{1}{\sqrt3}.$ The opposite side of a regular hexagon is $\sqrt3$ units from the attached side, so its horizontal outward displacement is $1.$ Consequently, the four unit-length sides of the rim lie one unit beyond the four sides of the bottom square. Its top view is therefore a $3$-by-$3$ square with four isosceles right corner triangles of leg $1$ removed: Its area is $$3^2-4\left(\frac12\cdot1^2\right)=7.$$ Thus, B is the correct answer.