1326 ** Find the area of the region enclosed by the graph of the equation $|x-60|+|y|-\left|\frac{x}{4}\right|$.

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Parse: Substituting $-y$ into the known equation still yields $|x-60|+|y|=\left|\frac{x}{4}\right|$. Therefore, the graph of the known equation is symmetric about the $x$-axis. For this reason, we only need to find half of it, that is, the area enclosed by $\left\{\begin{array}{l}y=\left|\frac{x}{4}\right|-|x-60| \\ y \geqslant 0\end{array}\right.$. On the coordinate plane, draw the graphs of $y_{1}=\left|\frac{x}{4}\right|$ and $y_{2}=|x-60|$. They are two rays that are symmetric about the $y$-axis with the vertex at $(0,0)$ and symmetric about the line $x=60$ with the vertex at $(60,0)$, respectively. The difference between them and the $x$-axis in the upper half-plane encloses the region $\triangle D A B$. Solving $\left\{\begin{array}{l}y=\frac{x}{4}-(x-60), \\ y=0 .\end{array}\right.$ yields $B(80, 0)$. Solving $\left\{\begin{array}{l}y=\frac{x}{4}-(60-x), \\ y=0 .\end{array}\right.$ yields $A(48,0)$. Solving $\left\{\begin{array}{l}y=\frac{x}{4}, \\ x=60 .\end{array}\right.$ yields $D(60,15)$. Thus, $|A B|=|48-80|=32,|D C|=15 . S_{\triangle D A B}=\frac{1}{2} \times 32 \times 15=240$. Therefore, the area enclosed by the original equation is $2 S_{\triangle D A B}=480$.