jmc

We first simplify the expression on the left hand side of the equation: $$\frac{13x}{41y}=\frac{{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{3}\right)}^2}{{\left(\frac{1}{4}\right)}^2+{\left(\frac{1}{5}\right)}^2}=\frac{\frac{1}{4}+\frac{1}{9}}{\frac{1}{16}+\frac{1}{25}}=\frac{\frac{9+4}{9\cdot 4}}{\frac{16+25}{16\cdot 25}}=\frac{13/36}{41/400}.$$ Since dividing by a fraction is the same thing as multiplying by its reciprocal, we have that $\frac{13x}{41y}=\frac{13}{36}\cdot \frac{400}{41}=\frac{13\cdot 400}{36\cdot 41}$. We can then multiply both sides of the equation by $\frac{41}{13}$ to get that $\frac{x}{y}=\frac{400}{36}$. Since $\sqrt{x}÷\sqrt{y}=\sqrt{\frac{x}{y}}$, to find $\sqrt{x}÷\sqrt{y}$, we can take the square root of both sides of the equation: $$\sqrt{\frac{x}{y}}=\sqrt{\frac{400}{36}}=\frac{20}{6}=\boxed{\frac{10}{3}}.$$