Four, (25 points) Let the two intersection points of the functions $y=2x$ and $y=\frac{4}{x}$ be $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\left(x_{1}>x_{2}\right)$, and point $C(\sqrt{2},-2 \sqrt{2})$. Find the area of $\triangle ABC$.
From $\left\{\begin{array}{l}y=2 x, \\ y=\frac{4}{x},\end{array}\right.$ eliminating $y$ we get
$$
2 x=\frac{4}{x} \Rightarrow x^{2}=2 \Rightarrow x= \pm \sqrt{2} \text {. }
$$

Therefore, $A(\sqrt{2}, 2 \sqrt{2}), B(-\sqrt{2},-2 \sqrt{2})$.
Since point $C(\sqrt{2},-2 \sqrt{2})$, $\triangle A B C$ is a right triangle, and $\angle A C B=90^{\circ}$.
Obviously, $A C=4 \sqrt{2}, B C=2 \sqrt{2}$.
Thus, $S_{\triangle A C B}=\frac{1}{2} A C \cdot B C=8$.