Let $ABCDEF$ be a convex hexagon and denote by $A',B',C',D',E',F'$ the middle points of the sides $AB$, $BC$, $CD$, $DE$, $EF$ and $FA$ respectively. Given are the areas of the triangles $ABC'$, $BCD'$, $CDE'$, $DEF'$, $EFA'$ and $FAB'$. Find the area of the hexagon.

[i]Kvant Magazine[/i]
1. **Identify the given areas and the midpoints:**
   Let the areas of the triangles \( \triangle ABC', \triangle BCD', \triangle CDE', \triangle DEF', \triangle EFA', \triangle FAB' \) be given. Denote the midpoints of the sides \( AB, BC, CD, DE, EF, \) and \( FA \) by \( A', B', C', D', E', \) and \( F' \) respectively.

2. **Express the areas of the triangles in terms of the hexagon's area:**
   Since \( A' \) is the midpoint of \( AB \), the area of \( \triangle ABC' \) can be expressed as:
   \[
   [ABC'] = \frac{[ABC] + [ABD]}{2}
   \]
   Similarly, for the other triangles:
   \[
   [BCD'] = \frac{[BCD] + [BCE]}{2}
   \]
   \[
   [CDE'] = \frac{[CDE] + [CDF]}{2}
   \]
   \[
   [DEF'] = \frac{[DEF] + [DEA]}{2}
   \]
   \[
   [EFA'] = \frac{[EFA] + [EFB]}{2}
   \]
   \[
   [FAB'] = \frac{[FAB] + [FAC]}{2}
   \]

3. **Sum the areas of the triangles:**
   Adding up all the areas of the triangles:
   \[
   [ABC'] + [BCD'] + [CDE'] + [DEF'] + [EFA'] + [FAB']
   \]
   Substituting the expressions from step 2:
   \[
   \frac{[ABC] + [ABD]}{2} + \frac{[BCD] + [BCE]}{2} + \frac{[CDE] + [CDF]}{2} + \frac{[DEF] + [DEA]}{2} + \frac{[EFA] + [EFB]}{2} + \frac{[FAB] + [FAC]}{2}
   \]

4. **Factor out the common factor:**
   \[
   \frac{1}{2} \left( [ABC] + [ABD] + [BCD] + [BCE] + [CDE] + [CDF] + [DEF] + [DEA] + [EFA] + [EFB] + [FAB] + [FAC] \right)
   \]

5. **Group the areas to form the hexagon:**
   Notice that each area of the hexagon \( ABCDEF \) is counted twice in the sum:
   \[
   [ABC] + [BCD] + [CDE] + [DEF] + [EFA] + [FAB] + [ABD] + [BCE] + [CDF] + [DEA] + [EFB] + [FAC]
   \]
   This sum is actually twice the area of the hexagon \( ABCDEF \):
   \[
   2 \times [ABCDEF]
   \]

6. **Simplify the expression:**
   \[
   \frac{1}{2} \times 2 \times [ABCDEF] = [ABCDEF]
   \]

7. **Conclusion:**
   The sum of the areas of the triangles \( \triangle ABC', \triangle BCD', \triangle CDE', \triangle DEF', \triangle EFA', \triangle FAB' \) is equal to the area of the hexagon \( ABCDEF \).

The final answer is \( \boxed{ [ABCDEF] } \).