2005 AMC 10A Problem 19

When lowered, the rotated square’s two lower edges rest on the inner top corners of the adjoining squares, which are at height $1.$ The bottom vertex is centered between those corners, so each corner is horizontally $\tfrac12$ unit from it. A lower edge has slope $1$ in magnitude, so it rises $\tfrac12$ unit on the way to a corner. Therefore the bottom vertex is at height $1-\tfrac12=\tfrac12.$ Point $B$ is the opposite vertex, a full vertical diagonal of length $\sqrt2$ higher. Its height is therefore $\tfrac12+\sqrt2.$ Thus, the correct answer is D.