2. (1992 National High School Competition Question) Let the areas of the four faces of a tetrahedron be $S_{1}, S_{2}, S_{3}, S_{4}$, and their maximum value be $S$. Let $\lambda=\frac{\sum_{i=1}^{4} S_{i}}{S}$, then $\lambda$ must satisfy ( ).
A. $2<\lambda \leqslant 4$
B. $3<\lambda<4$
C. $2.5<\lambda \leqslant 4.5$
D. $3.5<\lambda<5.5$
2. A. Reason: Since $S_{i} \leqslant S (i=1,2,3,4)$, we have $\sum_{i=1}^{4} S_{i} \leqslant 4 S$, and thus $\lambda=\frac{\sum_{i=1}^{4} S_{i}}{S} \leqslant 4$. In particular, when the tetrahedron is a regular tetrahedron, the equality holds in the above expression, thereby negating B.

Consider any regular triangular pyramid where the plane angles of the dihedral angles between the lateral faces and the base are all $45^{\circ}$. Let $S_{4}$ represent the area of the base of this regular triangular pyramid. Then, $S=S_{4}=\left(S_{1}+S_{2}+S_{3}\right) \cdot \cos 45^{\circ}=\frac{\sqrt{2}}{2}\left(S_{1}+S_{2}+S_{3}\right)$, so $S_{1}+S_{2}+S_{3}=\sqrt{2} S, S_{1}+S_{2}+S_{3}+S_{4}=(1+\sqrt{2}) S$. At this point, $\lambda=\frac{\sum_{i=1}^{n} S_{i}}{S}=1+\sqrt{2}<2.5$, thereby negating C and D.