10.226. A circle is inscribed in a right-angled triangle. The point of tangency divides the hypotenuse in the ratio 2:3. Find the sides of the triangle if the center of the inscribed circle is $\sqrt{8}$ cm away from the vertex of the right angle.
Solution.

Let the radius of the circle be $r$, and the length of the hypotenuse be $5x$. Then (Fig. 10.35) $BC = 3x + r, AC = 2x + r$, since $BK = BL, AK = AM$. Since $\angle OCM = 45^{\circ}$, we have $r = OC / \sqrt{2} = \sqrt{8} / \sqrt{2} = 2 \text{ cm}$. For the area of triangle $ABC$, we have the expression $S = 0.5 \cdot AC \cdot BC$; on the other hand, $S = pr$, where $p = 0.5(AB + BC + AC)$. Therefore, $0.5(2x + 2)(3x + 2) = 2(5x + 2)$ or $3x^2 - 5x - 2 = 0$, from which $x = 2 \text{ cm}$ (the second root of the equation does not fit). Thus, $AB = 10 \text{ cm}, AC = 6 \text{ cm}, BC = 8 \text{ cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0732.jpg?height=662&width=438&top_left_y=884&top_left_x=823)

Fig. 10.35

Answer: 6, 8, and 10 cm.