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)
○ 0
○ $\frac{\sqrt{2}}{2}$
○ 1
○ $\frac{\sqrt{3}}{2}$

Explanation:

Step1: Recall the unit circle definition

The unit circle is defined as the set of all points \((x, y)\) in the plane such that \(x^2 + y^2=1\), and its radius \(r = 1\) (centered at the origin \((0,0)\)).

Step2: Analyze the right triangle vertices

One vertex is at the origin \((0,0)\), one is on the \(x\)-axis (so its coordinates are \((a, 0)\) where \(a>0\) and \(a < 1\) since it's inside the unit circle), and one is on the unit circle (so its coordinates are \((x,y)\) with \(x^2 + y^2 = 1\)). The right angle of the right triangle is at the origin (because one leg is along the \(x\)-axis from \((0,0)\) to \((a,0)\) and the other leg is from \((0,0)\) to \((x,y)\) on the unit circle, and the hypotenuse is the line segment connecting \((a,0)\) to \((x,y)\)? Wait, no—wait, actually, let's re - examine. Wait, the right triangle: vertices at origin \((0,0)\), on \(x\)-axis \((x_1,0)\), and on unit circle \((x_2,y_2)\). The right angle is at the origin? Wait, no, maybe the right angle is at the point on the \(x\)-axis? Wait, no, let's think again. The unit circle has radius 1. The hypotenuse of the right triangle: in a right triangle inscribed in a circle with the hypotenuse as the diameter, but here the circle is the unit circle. Wait, no, the three points: origin \((0,0)\), a point on \(x\)-axis \((a,0)\) (inside unit circle, so \(a<1\)), and a point on unit circle \((x,y)\). The right angle: since it's a right triangle, and one vertex is at origin, one on \(x\)-axis, one on unit circle. The hypotenuse should be the distance from the origin to the point on the unit circle? Wait, no. Wait, the legs: one leg is from origin to \((a,0)\) (length \(a\)), another leg is from origin to \((x,y)\) (length \(\sqrt{x^2 + y^2}\)). But since \((x,y)\) is on the unit circle, \(\sqrt{x^2 + y^2}=1\). Wait, but the right angle: if the right angle is at the origin, then the hypotenuse is the distance between \((a,0)\) and \((x,y)\). But that can't be. Wait, maybe I made a mistake. Wait, the problem says "a right triangle has vertices at the origin, on the unit circle, and on the \(x\)-axis inside the unit circle". Let's consider the coordinates: let \(O=(0,0)\) (origin), \(A=(a,0)\) (on \(x\)-axis, \(0 < a<1\)), and \(B=(x,y)\) (on unit circle, so \(x^2 + y^2 = 1\)). The triangle is right - angled. Let's assume the right angle is at \(A\). Then \(OA\) is along \(x\)-axis, \(AB\) is vertical? No, \(B\) is on unit circle. Wait, maybe the right angle is at \(O\). Then \(OA\) is along \(x\)-axis (length \(a\)), \(OB\) is a line from origin to \((x,y)\) on unit circle (length 1, since \(x^2 + y^2 = 1\)), and \(AB\) is the hypotenuse. But that would mean \(OA^2+OB^2 = AB^2\), but \(OA=a\), \(OB = 1\), \(AB=\sqrt{(x - a)^2+y^2}=\sqrt{x^2-2ax + a^2+y^2}=\sqrt{(x^2 + y^2)-2ax + a^2}=\sqrt{1-2ax + a^2}\). But this seems complicated. Wait, maybe the right angle is at the point on the \(x\)-axis. So \(OA\) is horizontal (from \(O\) to \(A=(a,0)\)), \(AB\) is vertical (from \(A\) to \(B=(a,y)\)), and \(OB\) is the hypotenuse. But \(B=(a,y)\) is on the unit circle, so \(a^2 + y^2=1\). Then the hypotenuse \(OB\) has length \(\sqrt{a^2 + y^2}=1\) (since \(B\) is on unit circle). Ah! That makes sense. So regardless of where \(A\) is on the \(x\)-axis (as long as \(B=(a,y)\) is on the unit circle), the hypotenuse is the distance from the origin to \(B\), which is the radius of the unit circle, so length 1.

Answer:

1 (corresponding to the option "1")