Question

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 \( \boxed{1} \).