10.215. In a right-angled triangle, the medians of the legs are $\sqrt{52}$ and $\sqrt{73}$. Find the hypotenuse of the triangle.
## Solution.

In $\triangle A B C$ (Fig. 10.25) $\angle A C B=90^{\circ}, B P$ and $A E$ are medians, $B P=\sqrt{52}$, $A E=\sqrt{73}$. Let $B C=x, A C=y$. Then $A B=\sqrt{x^{2}+y^{2}}$.

From $\triangle A C E\left(\angle A C E=90^{\circ}\right):$

$$
A C^{2}+C E^{2}=A E^{2} ; y^{2}+\frac{x^{2}}{4}=73
$$

From $\triangle B C P\left(\angle B C P=90^{\circ}\right)$ :

$$
B C^{2}+C P^{2}=B P^{2} ; x^{2}+\frac{y^{2}}{4}=52
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0726.jpg?height=561&width=492&top_left_y=77&top_left_x=753)

Fig. 10.25

We have $\left\{\begin{array}{l}4 y^{2}+x^{2}=292, \\ 4 x^{2}+y^{2}=208 ;\end{array} 5 x^{2}+5 y^{2}=500 ; x^{2}+y^{2}=100\right.$. Therefore, $A B=\sqrt{x^{2}+y^{2}}=10$.

Answer: 10.