$23 \cdot 12$ A regular pyramid has a base that is an equilateral triangle with side length 6, and each of the other edges is $\sqrt{15}$. The volume of this regular pyramid is
(A) 9.
(B) $\frac{9}{2}$.
(C) $\frac{27}{2}$.
(D) $\frac{9 \sqrt{3}}{2}$.
(E) None of these.
(24th American High School Mathematics Examination, 1973)
［Solution］The volume of a pyramid is equal to one-third of the base area times the height:
The base area of an equilateral triangle with a side length of 6 is $3 \cdot 3 \sqrt{3}=9 \sqrt{3}$.
With the pyramid's edge length $\sqrt{15}$ as the hypotenuse, and the distance from the center of the base to one of the vertices $2 \sqrt{3}$ as one leg of a right triangle, the height $h$ of the pyramid is the other leg of this right triangle.
Thus,
$$
\begin{array}{c}
h^{2}=(\sqrt{15})^{2}-(2 \sqrt{3})^{2}=3, \\
h=\sqrt{3},
\end{array}
$$

Then $\quad V=\frac{1}{3} \cdot$ (base area) (height) $=\frac{1}{3}(9 \sqrt{3}) \sqrt{3}=9$.
Therefore, the answer is $(A)$.