2. For a rectangle, the sum of two sides is 11, and the sum of three sides is 19.5. Find the product of all possible different values of the perimeter of such a rectangle.
Solution:

Let the sides of the rectangle be $a$ and $b$.

If the sum of adjacent sides is 11, then the system describing the condition of the problem is $\left\{\begin{array}{l}a+b=11 \\ 2 a+b=19.5\end{array}\right.$, its solution is $a=8.5, b=2.5$, the perimeter of the rectangle is $P_{1}=22$.

If, however, the number 11 is equal to the sum of opposite sides, then there are two possible cases $\left\{\begin{array}{l}2 a=11 \\ 2 a+b=19.5\end{array}\right.$ or $\left\{\begin{array}{l}2 a=11 \\ a+2 b=19.5\end{array}\right.$.

The solution to the first system is $a=5.5, b=8.5$, the perimeter of the rectangle is $P_{2}=28$.

The solution to the second system is $a=5.5, b=7$, the perimeter of the rectangle is

$P_{3}=25$. The product of the three possible perimeter values is $P_{1} \cdot P_{2} \cdot P_{3}=22 \cdot 28 \cdot 25=15400$.

Online (correspondence) stage of the "Step into the Future" School Students' Olympiad in the subject of Mathematics

Answer: 15400.