smc

Given an equilateral triangle with side length $s$, the area is given by $\frac{s^2\sqrt{3}}{4}$. Setting this equation equal to the area of triangle $I$, $32\sqrt{3}$, we find that $s=8\sqrt{2}$. Because triangle $III$ is also equilateral, it is similar to trinagle $I$, and, because it has a quarter of the area of $I$, it has $\sqrt{\frac{1}{4}}=\frac{1}{2}$ of the side length. Thus, its sides have a length of $4\sqrt{2}$. Square $II$ initially has an area of $32$, so it's side length starts at $\sqrt{32}=4\sqrt{2}$. The initial length $AD$, therefore, is $16\sqrt{2}$. Because $AD$ decreases by $12.5$ %, it becomes $\frac{7}{8}$ of its inital value, which is $14\sqrt{2}$. Because the sides of the triangles remain unchanged, this decrease of $2\sqrt{2}$ must come from the side length of the square. Thus, the square's final side is $2\sqrt{2}$, which gives an area of $8$ square inches. $32$ to $8$ is a decrease of $75$ %. Therefore, our answer is $\boxed{75}$.