2. Given $a>0$, the graphs of the functions $f(x)=|x+2a|$ and $g(x)=|x-a|$ intersect at point $C$, and they intersect the $y$-axis at points $A$ and $B$ respectively. If the area of $\triangle ABC$ is 1, then $a=$ $\qquad$ .
2. 2 .

From the graphs of $f(x)$ and $g(x)$, we know that $\triangle ABC$ is an isosceles right triangle with base $a$, so its area is $\frac{a^{2}}{4}=1$. Therefore, $a=2$.