9. Given triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$, the areas of which are 1 and 2025, respectively. It is known that rays $A B$ and $A^{\prime} B^{\prime}$ are parallel and point in opposite directions (see figure). The same is true for pairs $B C$ and $B^{\prime} C^{\prime}, C A$ and $C^{\prime} A^{\prime} . A^{\prime \prime}, B^{\prime \prime}$ and $C^{\prime \prime}$ are the midpoints of segments $A A^{\prime}, B B^{\prime}$ and $C C^{\prime}$. Find the area of triangle $A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$.
Answer: 484.

Solution: Obviously, triangles $A B C A^{\prime} B^{\prime} C^{\prime}$ are similar with a coefficient of 45.

Three trapezoids are formed, in which $A^{\prime \prime} B^{\prime \prime}, B^{\prime \prime} C^{\prime \prime}$ and $C^{\prime \prime} A^{\prime \prime}$ are segments connecting the midpoints of the diagonals (i.e., equal to half the difference of the bases). Therefore, triangle $A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$ is also similar to $\triangle A B C$ and its coefficient is $(45-1) / 2=22$. Consequently, its area is $22^{2}=484$ times larger than the area of the small triangle.