2. There are 81 squares of the same size. Arrange them into two rectangles so that their perimeters are the same. No squares should be left over.

(L. S. Koreschkova)
Answer: $3 \times 11$ and $6 \times 8$.

Solution. The sum of their areas is $33+48=81$, and their perimeters are indeed equal: $2 \cdot(3+11)=2 \cdot(6+8)=28$.

Note. There are no other suitable pairs of rectangles.

Criteria. A correct example is provided without calculations showing that it fits - 5 points. At least one of the sides is not an integer - 0 points.