1. **State the problem:** We have a right triangle ABC with a right angle at A. The hypotenuse BC is 36 cm, and the segment from A to the point where the inscribed circle touches the triangle is 10 cm. We need to find the radius of the inscribed circle. 2. **Recall the formula for the inradius (radius of the inscribed circle) of a right triangle:** $$r = \frac{a + b - c}{2}$$ where $a$ and $b$ are the legs of the right triangle, and $c$ is the hypotenuse. 3. **Identify known values:** - Hypotenuse $c = 36$ cm - The segment from $A$ to the tangent point of the circle is 10 cm. This segment corresponds to $s - a$, where $s$ is the semiperimeter and $a$ is one leg. 4. **Use the tangent length property:** The tangent length from a vertex to the point of tangency with the incircle equals $s -$ the side adjacent to that vertex. Given the tangent length from $A$ is 10 cm, then: $$10 = s - a$$ 5. **Express the semiperimeter:** $$s = \frac{a + b + c}{2}$$ So, $$10 = \frac{a + b + 36}{2} - a = \frac{a + b + 36 - 2a}{2} = \frac{b + 36 - a}{2}$$ Multiply both sides by 2: $$20 = b + 36 - a$$ Rearranged: $$b - a = -16$$ 6. **Use the Pythagorean theorem:** $$a^2 + b^2 = 36^2 = 1296$$ 7. **From step 5, express $b$ in terms of $a$:** $$b = a - 16$$ 8. **Substitute into Pythagorean theorem:** $$a^2 + (a - 16)^2 = 1296$$ Expand: $$a^2 + a^2 - 32a + 256 = 1296$$ Combine like terms: $$2a^2 - 32a + 256 = 1296$$ Subtract 1296: $$2a^2 - 32a + 256 - 1296 = 0$$ $$2a^2 - 32a - 1040 = 0$$ Divide entire equation by 2: $$a^2 - 16a - 520 = 0$$ 9. **Solve quadratic equation:** $$a = \frac{16 \pm \sqrt{(-16)^2 - 4 \times 1 \times (-520)}}{2} = \frac{16 \pm \sqrt{256 + 2080}}{2} = \frac{16 \pm \sqrt{2336}}{2}$$ Calculate $\sqrt{2336} \approx 48.34$: $$a = \frac{16 \pm 48.34}{2}$$ Two solutions: - $$a = \frac{16 + 48.34}{2} = 32.17$$ - $$a = \frac{16 - 48.34}{2} = -16.17$$ (discard negative length) 10. **Find $b$:** $$b = a - 16 = 32.17 - 16 = 16.17$$ 11. **Calculate semiperimeter $s$:** $$s = \frac{a + b + c}{2} = \frac{32.17 + 16.17 + 36}{2} = \frac{84.34}{2} = 42.17$$ 12. **Calculate inradius $r$:** $$r = s - c = 42.17 - 36 = 6.17$$ Or use the formula: $$r = \frac{a + b - c}{2} = \frac{32.17 + 16.17 - 36}{2} = \frac{12.34}{2} = 6.17$$ **Final answer:** The radius of the inscribed circle is approximately **6.17 cm**.