1. Identify the points of tangency and reflections:
Let $D, E, F$ be the points of tangency of the incircle with sides $BC, AC,$ and $AB$ respectively. The points $A', B', C'$ are the reflections of the incenter $I$ across the sides $BC, AC,$ and $AB$ respectively.

2. Similarity of triangles:
Note that triangle $A'B'C'$ is similar to triangle $DEF$ with a ratio of $2:1$. This is because the reflections of the incenter $I$ across the sides of the triangle form a triangle that is homothetic to the contact triangle $DEF$ with a homothety ratio of $2$.

3. **Area of triangle $DEF$:**
To find the area of triangle $DEF$, we first need to compute the area of the smaller triangles $EFI, DFI,$ and $DEI$.

4. **Area of triangle $EFI$:**
By the Pythagorean theorem, we have:
$$AF = AE = \sqrt{AI^2 - FI^2} = \sqrt{2 - 1} = 1$$
Let $G$ be the intersection of $AI$ and $EF$. Since triangle $FGI$ is similar to triangle $AFI$, we have:
$$\frac{FG}{AF} = \frac{FI}{AI} \implies \frac{FG}{1} = \frac{1}{\sqrt{2}} \implies FG = \frac{1}{\sqrt{2}}$$
Similarly,
$$\frac{GI}{FI} = \frac{FI}{AI} \implies \frac{GI}{1} = \frac{1}{\sqrt{2}} \implies GI = \frac{1}{\sqrt{2}}$$
Therefore, the area of triangle $EFI$ is:
$$\text{Area of } EFI = \frac{1}{2} \cdot EF \cdot GI = \frac{1}{2} \cdot \frac{2}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2}$$

5. **Area of triangles $DFI$ and $DEI$:**
By similar calculations, we find the areas of triangles $DFI$ and $DEI$:
$$\text{Area of } DFI = \frac{2}{5}, \quad \text{Area of } DEI = \frac{3}{10}$$

6. **Total area of triangle $DEF$:**
Summing the areas of the three smaller triangles, we get:
$$\text{Area of } DEF = \frac{1}{2} + \frac{2}{5} + \frac{3}{10} = \frac{6}{5}$$

7. **Area of triangle $A'B'C'$:**
Since triangle $A'B'C'$ is similar to triangle $DEF$ with a ratio of $2:1$, the area of $A'B'C'$ is:
$$\text{Area of } A'B'C' = 4 \times \text{Area of } DEF = 4 \times \frac{6}{5} = \frac{24}{5}$$

The final answer is $\boxed{\frac{24}{5}}$