18.95 In an isosceles right $\triangle ABC$, there are two ways to inscribe a square, as shown in Figure (1). The area of the inscribed square in Figure (1) is $441 \mathrm{~cm}^{2}$. Then, in the same $\triangle ABC$, the area of the inscribed square in Figure (2) is 
(A) 378.
(B) 392.
(C) 400.
(D) 441.
(E) 484.
(38th American High School Mathematics Examination, 1987)
[Solution] Let $AB = a$, then $AC = \sqrt{2}a$. Suppose the side length of the square in Figure (3) is $x$, and the side length of the square in Figure (4) is $y$. From the given information,
$$
x^{2} = 441 \text{, so } x = 21.
$$

Using the properties of an isosceles right triangle, it is not difficult to see from the right figure:

and
then
$$
\begin{array}{c}
a = AB = 2x = 42. \\
y = \frac{1}{3} AC = \frac{1}{3} \cdot 42 \sqrt{2} = 14 \sqrt{2}.
\end{array}
$$

Therefore, the choice is $(B)$.
$$
S_{\Pi} = y^{2} = 392.
$$