1. A smaller square was cut out from a larger square, one of its sides lying on the side of the original square. The perimeter of the resulting octagon is $40 \%$ greater than the perimeter of the original square. By what percentage is its area less than the area of the original square?
Answer: By $64 \%$. Solution. Let $ABCD$ be the original square with side 1 (and area 1), $KLMN$ be the cut-out square with side $x$, where $KL$ lies on $AB$ (point $K$ is closer to $A$ than $L$). Then the perimeter of the octagon $AKNMLBCD$ exceeds the perimeter of the square $ABCD$ by $KN + LM = 2x = 0.4 \cdot 4AB = 1.6AB$. That is, $x = 0.8AB$, and the area of the cut-out square is $0.64$ of the area of the original.