11、Divide a rope of length 31 cm into three segments, with each segment's length being an integer. Any two segments can be taken to form the length and width of a rectangle, resulting in three possible rectangles. The maximum value of the sum of the areas of these three rectangles is $\qquad$ square centimeters.
【Analysis】Suppose the rope is divided into three segments of lengths $a$, $b$, and $c$, then the sum of the areas of these three rectangles is $a b + a c + b c = \frac{2(a + b + c)^{2} - \left[(a - b)^{2} + (a - c)^{2} + (b - c)^{2}\right]}{6}$. Given that $a + b + c = 31$, the sum of the areas is maximized when the differences between $a$, $b$, and $c$ are minimized, i.e., when $a$, $b$, and $c$ are approximately $10$, $10$, and $11$. The sum of the areas is then $10 \times 10 + 10 \times 11 + 10 \times 11 = 320$ (square centimeters).