11.30. The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$. What is the smallest area that this quadrilateral can have if the area of triangle $AOB$ is 4 and the area of triangle $COD$ is 9?
11.30. Since $S_{A O B}: S_{B O C}=A O: O C=S_{A O D}: S_{D O C}$, then $S_{B O C} \cdot S_{A O D}=$ $=S_{A O B} \cdot S_{D O C}=36$. Therefore, $S_{B O C}+S_{A O D} \geqslant 2 \sqrt{S_{B O C} \cdot S_{A O D}}=12$, and equality is achieved if $S_{B O C}=S_{A O D}$, i.e., $S_{A B C}=S_{A B D}$, from which $A B \| C D$. In this case, the area of the quadrilateral is $4+9+12=25$.