25. Given that $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$, the common chord intersects the line segment $O_{1} O_{2}$ at point $G$. If $AB=48$, the radii of $\odot O_{1}$ and $\odot O_{2}$ are $30$ and $40$ respectively, then the area of $\triangle A O_{1} O_{2}$ is ( ).
(A) 600
(B) 300 or 168
(C) 168
(D) 600 or 168
25.D.

Since $O_{1} G=\sqrt{A O_{1}^{2}-A G^{2}}=\sqrt{30^{2}-24^{2}}=18$,
$$
O_{2} G=\sqrt{A O_{2}^{2}-A G^{2}}=\sqrt{40^{2}-24^{2}}=32 \text {, }
$$

then $S_{\triangle 1 O_{1} O_{2}}=\frac{1}{2} A G\left(O_{2} G \pm O_{1} G\right)=600$ or 168.