7. (3 points) Petya and Vasya are playing a game. Each has one move: Petya draws two $3 \times 3$ squares on the plane, and Vasya changes the position of one of the squares using a parallel translation. Vasya wins if, after his move, the area of the intersection of the squares is at least 7. Who wins with correct play?
Answer: Vasya.

Solution: With his move, Vasya only needs to align the centers of the two squares. Then their inscribed circles will also coincide. Thus, the area of the intersection of the squares will be at least the area of their inscribed circle, which is no more than $\frac{9 \pi}{4}$.

$$
\frac{9 \pi}{4}>\frac{9 \cdot 3.14}{4}>\frac{9 \cdot 3,(1)}{4}=7
$$

Of course, this is not the only strategy.