2003 AMC 12A Problem 14

The square $ABCD$ has side $4.$ By the $90^\circ$ rotational symmetry, $KLMN$ is also a square. Each equilateral triangle on a side of length $4$ has height $2\sqrt3,$ so the diagonal $KM=4+2(2\sqrt3)=4+4\sqrt3.$ A square with diagonal $d$ has area $\dfrac12 d^2,$ so $[KLMN]=\dfrac12(4+4\sqrt3)^2$ $=32+16\sqrt3.$ Thus, the correct answer is D.