10.047. The perimeter of a right-angled triangle $ABC$ ( $\angle C=90^{\circ}$ ) is 72 cm, and the difference between the lengths of the median $CK$ and the altitude $CM$ is 7 cm (Fig. 10.47). Find the length of the hypotenuse.
Solution.

According to the condition $C K - C M = 7 \text{ cm}, P_{A B C} = 72 \text{ cm}, p = \frac{P_{A B C}}{2} = 36$. If the radius of the inscribed circle, then $r = p - A B, S = p \cdot r = p \cdot (p - A B)$. Let $A B = x, K$ be the center of the circumscribed circle, $C K = A K = K B = \frac{A B}{2} = \frac{x}{2}; S = 36 \cdot (36 - x); \frac{x}{2} - C M = 7$ (by condition) $C M = \frac{x}{2} - 7$; $S = \frac{x}{2} \left( \frac{x}{2} - 7 \right) \cdot \frac{x^2}{4} - \frac{7x}{2} = 36^2 - 36x, x^2 + 130x - 5184 = 0, x = 32$ cm.

Answer: $32 \text{ cm}$.