7. In the Cartesian coordinate system, the area of the figure determined by the inequality $|x|+|y|+|x+y| \leqslant 2$ is

保留源文本的换行和格式，直接输出翻译结果如下：

7. In the Cartesian coordinate system, the area of the figure determined by the inequality $|x|+|y|+|x+y| \leqslant 2$ is
Answer:
As shown in the figure, where $A(1,0), B(0,1), C(-1,1), D(-1,0)$, $E(0,-1), F(1,-1)$.
Therefore, the area of hexagon $A B C D E F$ is 3.