In triangles ABC and DEF, DE = 4AB, EF = 4BC, and F D = 4CA. The area of
DEF is 360 units more than the area of ABC. Compute the area of ABC.
1. Given that in triangles \( \triangle ABC \) and \( \triangle DEF \), the sides of \( \triangle DEF \) are 4 times the corresponding sides of \( \triangle ABC \):
   \[
   DE = 4AB, \quad EF = 4BC, \quad FD = 4CA
   \]
   
2. Since the sides of \( \triangle DEF \) are 4 times the sides of \( \triangle ABC \), the ratio of the areas of similar triangles is the square of the ratio of their corresponding sides. Therefore:
   \[
   \frac{[DEF]}{[ABC]} = \left(\frac{4}{1}\right)^2 = 16
   \]
   This implies:
   \[
   [DEF] = 16[ABC]
   \]

3. We are given that the area of \( \triangle DEF \) is 360 units more than the area of \( \triangle ABC \):
   \[
   [DEF] = [ABC] + 360
   \]

4. Substituting \( [DEF] = 16[ABC] \) into the equation \( [DEF] = [ABC] + 360 \):
   \[
   16[ABC] = [ABC] + 360
   \]

5. Solving for \( [ABC] \):
   \begin{align*}
   16[ABC] - [ABC] &= 360 \\
   15[ABC] &= 360 \\
   [ABC] &= \frac{360}{15} \\
   [ABC] &= 24
   \end{align*}

The final answer is \( \boxed{24} \)