$24 \cdot 6$ The area of the upper base of a frustum of a cone is 1, and the area of the lower base is 16. A plane parallel to the bases cuts this frustum, and the distance from this plane to the upper base is twice the distance from it to the lower base. The area of this section is 
(A) 4.
(B) 9.
(C) $\frac{5}{2}$.
(D) $\frac{25}{4}$.
(2nd "Hope Cup" National Mathematics Invitational, 1991)
［Solution］Let the areas of the upper base, lower base, and the section of the frustum be $S_{\text {upper }}, S_{\text {lower }}$, and $S_{\text {middle, then }}$ thus
$$
\begin{array}{c}
\sqrt{S_{\text {middle }}}=\frac{\sqrt{S_{\text {upper }}}+2 \sqrt{S_{\text {lower }}}}{3}=\frac{1+2 \cdot 4}{3}=3, \\
S_{\text {middle }}=9 .
\end{array}
$$

Therefore, the correct choice is $(B)$.