7. The area of a right triangle is 1, and its hypotenuse is $\sqrt{5}$. Find the cosine of the acute angle between the medians of the triangle drawn to its legs.

![](https://cdn.mathpix.com/cropped/2024_05_06_331dee27c7cee369cafag-34.jpg?height=616&width=959&top_left_y=757&top_left_x=340)
# Solution:

Let the legs of the triangle be $a$ and $b$. Then $\left\{\begin{array}{c}\frac{a b}{2}=1, \\ a^{2}+b^{2}=5,\end{array} \Leftrightarrow\left\{\begin{array}{c}a=\frac{2}{b}, \\ \frac{4}{b^{2}}+b^{2}=5\end{array}\right.\right.$. Solving the equation $b^{4}-5 b^{2}+4=0$, we get: $b^{2}=4$ or $b^{2}=1$. Then $b=2$ and $a=1$ or $b=1$ and $a=2$. By the Pythagorean theorem, we find the medians $A N$ and $C K:$

$$
A N^{2}=1+1=2 \Rightarrow A N=\sqrt{2}, C K^{2}=4+\frac{1}{4}=\frac{17}{4} \Rightarrow C K=\frac{\sqrt{17}}{2}
$$

In triangle $A O K$, we have: $A K=\frac{1}{2}, \quad K O=\frac{\sqrt{17}}{6}, \quad A O=\frac{2 \sqrt{2}}{3}$. By the cosine rule, $\cos \alpha=\frac{A O^{2}+K O^{2}-A K^{2}}{2 A O \cdot K O}=\frac{\frac{8}{9}+\frac{17}{36}-\frac{1}{4}}{2 \cdot \frac{\sqrt{17}}{6} \cdot \frac{2 \sqrt{2}}{3}}=\frac{10}{2 \sqrt{34}}=\frac{5 \sqrt{34}}{34}$.

Answer: $\frac{5 \sqrt{34}}{34}$.