Problem 2. The lengths of the adjacent sides of a rectangle are $\sqrt{404} \mathrm{~cm}$ and $\sqrt{909} \mathrm{~cm}$. Find the area of the circle and the length of the circumference of the circle that is circumscribed around a square whose area is equal to the area of the given rectangle.
Solution. The area of the given rectangle is

$$
P_{1}=\sqrt{404} \cdot \sqrt{909}=\sqrt{4 \cdot 101} \cdot \sqrt{9 \cdot 101}=2 \sqrt{101} \cdot 3 \sqrt{101}=606 \mathrm{~cm}^{2}
$$

If we denote the length of the side of the square that has the same area as the rectangle by $a$, then $a^{2}=606$, which means that $a=\sqrt{606} \mathrm{~cm}$. Therefore, the length of the diagonal of the square is

$$
d=a \sqrt{2}=\sqrt{606} \cdot \sqrt{2}=\sqrt{606 \cdot 2} \mathrm{~cm}=\sqrt{303 \cdot 2 \cdot 2}=2 \sqrt{303}
$$

Accordingly, the radius of the circle circumscribed around the square is

$$
r=\frac{d}{2}=\frac{2 \sqrt{303}}{2}=\sqrt{303} \mathrm{~cm}
$$

Finally, the length of the circle is $O=2 r \pi=2 \sqrt{303} \pi \mathrm{~cm}$, and the area is $P=\pi r^{2}=(\sqrt{303})^{2} \pi=303 \pi \mathrm{~cm}$.