1. What is the probability that a rectangle with perimeter $36 \mathrm{~cm}$ has area greater than $36 \mathrm{~cm}^{2}$ ?
Answer: $\frac{\sqrt{5}}{3}$
Solution: Let $x$ and $y$ be the lengths of the sides of the rectangle. We are looking for the probability that $x y>36$ given that $2 x+2 y=36$. Equivalently, we compute the probability that $x(18-x)>36$ given $036 \Leftrightarrow x^{2}-18 x+360$. Thus, the probability we are computing is $\frac{(9+3 \sqrt{5})-(9-3 \sqrt{5})}{18-0}=\frac{\sqrt{5}}{3}$.