1.37. The perimeter of a right triangle \(ABC (\angle C = 90^\circ)\) is 72 cm, and the difference between the lengths of the median \(CM\) and the altitude \(CK\) is 7 cm. Find the length of the hypotenuse.
1.37. According to the condition, $\angle C=90^{\circ}, A B+B C+A C=72 \text{ cm}, C M$ is the median, $C K$ is the altitude, $C M-C K=7$ cm (Fig. 1.31); we need to find $A B$. Since $M$ is the center of the circumscribed circle,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-430.jpg?height=352&width=729&top_left_y=324&top_left_x=228)

Fig. 1.31

$A M=M B=M C=\frac{1}{2} A B$. We use the equality $A B^{2}=B C^{2}+$ $+A C^{2}$. Note that $B C^{2}+A C^{2}=(B C+A C)^{2}-2 B C \cdot A C$, hence $A B^{2}=(72-A B)^{2}-2 A B \cdot C K$, since $B C \cdot A C=A B \cdot C K=2 S$, where $S$ is the area of the triangle. Thus, we arrive at the equation $A B^{2}=(72-A B)^{2}-2 A B\left(\frac{1}{2} A B-7\right)$, or $A B^{2}+130 A B-5184=0$, from which $A B=-65+\sqrt{9409}=32$ cm (the second root of the equation does not satisfy the condition).

Answer: $32 \text{ cm}$