10.211. On the larger leg of a right triangle, a circle is constructed with this leg as its diameter. Determine the radius of this circle if the smaller leg of the triangle is 7.5 cm, and the length of the chord connecting the vertex of the right angle with the point of intersection of the hypotenuse and the circle is 6 cm.
Solution.

In $\triangle ABC$ (Fig. 10.21) $\angle ACB=90^{\circ}, BC>AC, AC=7.5$ cm, $N$ is the intersection point of the circle mentioned in the condition and the hypotenuse $AB, CN=6$ cm. $\angle CNB$ is an inscribed angle and subtends the diameter. Therefore, $\angle CNB=90^{\circ}$. From $\triangle ANC \quad (\angle ANC=90^{\circ}): AN=\sqrt{AC^{2}-CN^{2}}=4.5$ cm. Since $CN$ is the altitude of the right $\triangle ACB$, $\triangle ANC \sim \triangle CNB$. Therefore, $\frac{AN}{AC}=\frac{CN}{BC}, BC=\frac{AC \cdot CN}{AN}=10$ cm. The desired radius $R=\frac{1}{2} BC=5$ cm.

Answer: 5 cm.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0724.jpg?height=487&width=488&top_left_y=146&top_left_x=206)

Fig. 10.22

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0724.jpg?height=601&width=428&top_left_y=61&top_left_x=725)

Fig. 10.23