(2) For the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$ with foci $F_{1}, F_{2}$, if a point $P$ on the ellipse makes $P F_{1} \perp P F_{2}$, then the area of $\triangle P F_{1} F_{2}$ is $\qquad$.
(2) 9 Hint: It is known that $F_{1} F_{2}=8, P F_{1}+P F_{2}=10$, so $\left(P F_{1}+\right.$ $\left.P F_{2}\right)^{2}=10^{2}$, in the right-angled $\triangle P F_{1} F_{2}$, $P F_{1}^{2}+P F_{2}^{2}=8^{2}$, from the above two equations, we get $S_{\triangle P F_{1} F_{2}}=\frac{1}{2} P F_{1} \cdot P F_{2}=9$.