Let $ ABC$ be a right angled triangle of area 1. Let $ A'B'C'$ be the points obtained by reflecting $ A,B,C$ respectively, in their opposite sides. Find the area of $ \triangle A'B'C'.$
1. Given that \( \triangle ABC \) is a right-angled triangle with area 1. Let \( A', B', C' \) be the reflections of \( A, B, C \) in their opposite sides respectively.
2. Since \( A' \) is the reflection of \( A \) across \( BC \), the line segment \( AA' \) is perpendicular to \( BC \) and \( AA' = 2AH \), where \( H \) is the foot of the altitude from \( A \) to \( BC \).
3. Similarly, \( B' \) is the reflection of \( B \) across \( AC \), and \( C' \) is the reflection of \( C \) across \( AB \). Therefore, \( AB' = AB \) and \( AC' = AC \).
4. Notice that \( \angle B'AC' = 90^\circ \), so \( AH' = AH \), where \( H' \) is the foot of the altitude from \( A \) to \( B'C' \).
5. By angle chasing, we can show that \( A, H, H' \) are collinear. Thus, the height of \( \triangle A'B'C' \) is \( 3AH \).
6. The area of \( \triangle A'B'C' \) can be calculated as follows:
   \[
   [A'B'C'] = \frac{1}{2} B'C' \cdot 3AH
   \]
7. Since \( B'C' = BC \) and \( \frac{1}{2} BC \cdot AH = [ABC] = 1 \), we have:
   \[
   [A'B'C'] = \frac{3}{2} BC \cdot AH = 3 \times \frac{1}{2} BC \cdot AH = 3 \times 1 = 3
   \]

The final answer is \(\boxed{3}\).