1. **Problem Statement:** We have a right triangle with vertices at the origin $(0,0)$, a point on the unit circle, and a point on the x-axis inside the unit circle. We need to find the hypotenuse of this triangle. 2. **Understanding the Unit Circle:** The unit circle is defined as the set of all points $(x,y)$ such that $$x^2 + y^2 = 1.$$ This means the radius of the circle is 1. 3. **Vertices of the Triangle:** - One vertex is at the origin $(0,0)$. - Another vertex is on the unit circle, say $(x,y)$, where $x^2 + y^2 = 1$. - The third vertex is on the x-axis between the origin and the circle, so it is at $(x,0)$ with $0 < x < 1$. 4. **Hypotenuse Identification:** The hypotenuse is the segment connecting the origin $(0,0)$ to the point on the unit circle $(x,y)$. 5. **Length of the Hypotenuse:** Since the point lies on the unit circle, the distance from the origin to $(x,y)$ is the radius of the circle. 6. **Distance Formula:** The distance $d$ between $(0,0)$ and $(x,y)$ is $$d = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}.$$ Since $(x,y)$ lies on the unit circle, $$x^2 + y^2 = 1,$$ so $$d = \sqrt{1} = 1.$$ **Final answer:** The hypotenuse of the triangle is 1.