$23 \cdot 25$ As shown, $P Q$ is a diagonal of a cube. If the length of $P Q$ is $a$, then the surface area of the cube is
(A) $2 a^{2}$.
(B) $2 \sqrt{2} a^{2}$.
(C) $2 \sqrt{3} a^{2}$.
(D) $3 \sqrt{3} a^{2}$.
(E) $6 a^{2}$.
(32nd American High School Mathematics Examination, 1981)
[Solution] As shown in the figure, let $s$ be the side length of the cube, and $T$ and $R$ be two vertices of the cube.
In $\triangle P Q R$ and $\triangle P R T$, by the Pythagorean theorem, we have
$$
a^{2}-s^{2}=(P R)^{2}=s^{2}+s^{2} \text {, }
$$

which means $\square$
$$
a^{2}=3 s^{2} .
$$

Therefore, the surface area of the cube is $6 s^{2}=2 a^{2}$. Hence, the answer is $(A)$.