Chapter 4.3, Problem 31MS

Golden Pythagoras (H). If you have a Golden Rectangle with base length ${l}=\frac{1+\sqrt{5}}{2}$ and height length 1, the Pythagorean Theorem from Section 4.1 tells you that the length of the diagonal of your rectangle is $\sqrt{1^2+{\left(\frac{1+\sqrt{5}}{2}\right)}^2}$. Simplify this expression as much as possible.