2. (20 points) As shown in the figure, the small square $E F G H$ is inside the large square $A B C D$, the total area of the shaded parts is 124 square centimeters, $E, H$ are on side $A D$, and $O$ is the midpoint of line segment $C F$. Then the area of quadrilateral $B O G F$ is $\qquad$ square centimeters.
【Analysis】As shown in the figure below: Connect $G C$, let: the side lengths of the squares $A B C D$ and $E H G F$ be $a$ and $b$ respectively. It has been verified that: the area of trapezoid $F G C B$ is equal to half of the shaded area; it has also been verified that the area of quadrilateral $B O G F$ is half of the trapezoid area, which allows us to solve the problem.
【Solution】As shown in the figure above: Connect $G C$, let: the side lengths of the squares $A B C D$ and $E H G F$ be $a$ and $b$ respectively, then the area of trapezoid $F G C B$ is $=(b+a)(b-a) \div 2=\left(b^{2}-a^{2}\right) \div 2$;

The shaded area $=\left(b^{2}-a^{2}\right)$, which means the area of trapezoid $F G C B$ is half of the shaded area; $O$ is the midpoint of line segment $C F$. Since if $\triangle B O F$ and $\triangle C O F$ are considered with $C O$ and $F O$ as their bases, they are triangles with equal bases and heights, so their areas are equal; similarly, the areas of $\triangle G O F$ and $\triangle G O$ are equal.

That is: the area of quadrilateral $B O G F$ is half of the trapezoid area,
the area of quadrilateral $B O G F$ = half of the trapezoid area = one quarter of the shaded area = $124 \div 4=31$ (square centimeters).

Therefore: The answer should be 31.