8.1. Two rectangles with horizontal and vertical sides are given. The horizontal side of the second rectangle is $11 \%$ longer than the horizontal side of the first, and the vertical side of the second rectangle is $10 \%$ shorter than the vertical side of the first. a) Which rectangle has a larger area and by what percentage? b) Find the ratio of the sides of the first rectangle if it is known that its perimeter is $5 \%$ less than the perimeter of the second.
Answer: a) the area of the second rectangle is less than the area of the first by $0.1\%$, b) the ratio of the horizontal side to the vertical side is 290/109 ≈ 22.66. Solution. Let $a$ and $b$ be the horizontal and vertical sides of the first rectangle, respectively, then the sides of the second rectangle are $1.11a$ and $0.9b$. The area of the second rectangle is $1.11a \cdot 0.9b = 0.999ab$, which in percentage terms is $0.999ab / ab = 99.9\%$, i.e., $0.1\%$ less than the area of the first rectangle. b) The perimeter of the second rectangle is $2 \cdot (1.11a + 0.9b)$ and, by condition, the perimeter of the first rectangle, i.e., $2a + 2b$, is $0.95 \cdot 2 \cdot (1.11a + 0.9b)$. Equating these two values, we get $0.145b = 0.0545a$, i.e., $a / b = \frac{290}{109} \approx 2.66$. Comment. Note that the condition of the problem that the perimeter of the first rectangle is $5\%$ less than the perimeter of the second does not mean that the perimeter of the second rectangle is $5\%$ greater than the perimeter of the first, and with such a slightly different condition, the answer would be $a / b = 2.5$.