1. As shown in Figure 1, the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$ divide it into four parts, with areas $S_{1}, S_{2}, S_{3}, S_{4}$, respectively. Given that $S_{1}>1, S_{2}>1$. Then $S_{3}+S_{4}$ ).
$(\mathrm{A})=2$
(B) $>2$
(C) $<2$
(D) Not necessarily
-,1.B.
It is known that $S_{1} S_{2}=S_{3} S_{4}$.
Therefore, $S_{3}+S_{4} \geqslant 2 \sqrt{S_{3} S_{4}}=2 \sqrt{S_{1} S_{2}}>2$.