18.91 As shown in the figure, figures I and III are equilateral triangles, with areas of $32 \sqrt{3}$ and $8 \sqrt{3}$ square inches, respectively. Figure II is a square with an area of 32 square inches. If the length of segment $A D$ is reduced by $12.5 \%$, but the lengths of $A B$ and $C D$ remain unchanged, then the percentage decrease in the area of the square is
(A) $12 \frac{1}{2}$.
(B) 25 .
(C) 50 .
(D) 75 .
(E) $87 \frac{1}{2}$.
(19th American High School Mathematics Examination, 1968)
[Solution] If the side length of an equilateral triangle is $a$, then its area is
$$
S=\frac{\sqrt{3}}{4} a^{2} .
$$

Using the above formula for equilateral triangles I and III, we can find
$$
A B=8 \sqrt{2}, \quad C D=4 \sqrt{2} .
$$

Additionally, we can find the side length of the square $B C=4 \sqrt{2}$.
Thus,
$$
A D=A B+B C+C D=16 \sqrt{2} \text {. }
$$

The $12.5\%$ of $A D$ is $2 \sqrt{2}$. When $B C$ is reduced by $2 \sqrt{2}$, it remains $2 \sqrt{2}$, at which point its area is 8, meaning the area has decreased by $75\%$.
Therefore, the answer is $(D)$.