A right triangle has vertices at the origin, on the unit circle, and on the x-axis inside the unit circle. What is the hypotenuse of the triangle? (1 point) The radius is □.

Answer 1

Let's denote the vertices of the right triangle as follows:

1. $A(0,0)$: The origin.
2. $B(x,0)$: A point on the x-axis inside the unit circle. Since it is inside the unit circle, $x$ must satisfy $-1<x<1$ and $y=0$.
3. $C(\cos (\theta ),\sin (\theta ))$: A point on the unit circle with angle $\theta$.

The right triangle formed consists of the following:

• The leg along the x-axis from point $A(0,0)$ to point $B(x,0)$.
• The leg from point $B(x,0)$ to point $C(\cos (\theta ),\sin (\theta ))$.
• The leg from point $A(0,0)$ to point $C(\cos (\theta ),\sin (\theta ))$.

Since we need to find the hypotenuse of the triangle, we can identify it as the longest side. The hypotenuse connects points $B$ and $C$.

To calculate the length of the hypotenuse $BC$, we use the distance formula:

$$BC=\sqrt{(x-\cos (\theta ){)}^2+(0-\sin (\theta ){)}^2}$$ This simplifies to:

$$BC=\sqrt{(x-\cos (\theta ){)}^2+{\sin }^2(\theta )}$$

Now we know that $C$ lies on the unit circle such that ${\cos }^2(\theta )+{\sin }^2(\theta )=1$, and point $B$ is bounded between $-1<x<1$.

To find the hypotenuse of the triangle, we need to express any necessary relationships. The maximum distance from $A$ to the unit circle, which has a radius of 1, is simply 1 unit. Given that all points $B$ and $C$ are constrained by positions related to $(0,0)$, the maximum length of hypotenuse $BC$ can be computed when both points $B$ and $C$ are farthest apart, which respects the triangle definitions.

Thus, the hypotenuse can be generally expressed in terms of the radius $r=1$, where:

$$\text{Hypotenuse}=1$$

Therefore, the final answer is:

The radius is $\boxed{{\displaystyle 1}}$.