, 2023

Solution:

Let $d$ be the length of the smaller diagonal of the smaller rhombus. Since the ratio of the areas is $9:1$, the ratio of the lengths is $3:1$. This means that the smaller diagonal of the larger rhombus (which is also the longer diagonal of the smaller rhombus) has length $3d$.

Therefore, the smaller rhombus has diagonal lengths $d$ and $3d$, so since it has area $1$, we have

$$\frac{1}{2} d (3d) = 1 \Longrightarrow d = \sqrt{\frac{2}{3}}$$

By the Pythagorean Theorem, the side length of the smaller rhombus is

$$\sqrt{\left(\frac{d}{2}\right)^2 + \left(\frac{3d}{2}\right)^2} = \sqrt{\frac{5d^2}{2}} = \sqrt{\frac{5}{3}}$$

The side length of the larger rhombus is three times this, i.e. $\sqrt{15}$.