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 Unit Circle Chart
(- 1/2, (sqrt(3))/2)
(- (sqrt(3))/2, 1/2)
(- (sqrt(2))/2, (sqrt(2))/2)
(3pi)/4
(5pi)/6
150°
π 180°
(-1,0)
sqrt((2pi)/3)
135°
120°
(0,1)
pi/2
90 deg
60 deg
45 deg
Z*pi
30
pi/3
pi/4
pi/6 =
((sqrt(2))/2, (sqrt(2))/2)
(sqrt(3))/2, 1/2 )
(1,0)
360 deg
2pi
210 deg
(- (sqrt(3))/2, - 1/2)
(- (sqrt(2))/2, - (sqrt(2))/2)
225
(5n)/4
240 deg
270 deg
(4pi)/3 * x
(- 1/2, - (sqrt(3))/2)
(3pi)/2
(0,-1)
234014
bp
F4
FS
F6
F1
FB
F9
F10
F11
F12
조
144
m
(1/2, (sqrt(3))/2)
Q Search
330 ^ prime prime
315
(11pi)/6
(7pi)/4
(sqrt(3))/2, - 1/2 )
300 deg
(5pi)/3
(sqrt(2))/2, - (sqrt(2))/2 )
(1/2, - (sqrt(3))/2)
The Unit Circle Chart
(- 1/2, (sqrt(3))/2)
(- (sqrt(3))/2, 1/2)
(- (sqrt(2))/2, (sqrt(2))/2)
(3pi)/4
(5pi)/6
150°
π 180°
(-1,0)
sqrt((2pi)/3)
135°
120°
(0,1)
pi/2
90 deg
60 deg
45 deg
Z*pi
30
pi/3
pi/4
pi/6 =
((sqrt(2))/2, (sqrt(2))/2)
(sqrt(3))/2, 1/2 )
(1,0)
360 deg
2pi
210 deg
(- (sqrt(3))/2, - 1/2)
(- (sqrt(2))/2, - (sqrt(2))/2)
225
(5n)/4
240 deg
270 deg
(4pi)/3 * x
(- 1/2, - (sqrt(3))/2)
(3pi)/2
(0,-1)
234014
bp
F4
FS
F6
F1
FB
F9
F10
F11
F12
조
144
m
(1/2, (sqrt(3))/2)
Q Search
330 ^ prime prime
315
(11pi)/6
(7pi)/4
(sqrt(3))/2, - 1/2 )
300 deg
(5pi)/3
(sqrt(2))/2, - (sqrt(2))/2 )
(1/2, - (sqrt(3))/2)
Answers
GPT-4o mini
To find the length of the hypotenuse of a right triangle that has its vertices at the origin, on the unit circle, and on the x-axis inside the unit circle, we first need to identify the coordinates of these points.
1. **The Origin:** The first vertex is at the point (0, 0).
2. **On the Unit Circle:** The unit circle is given by the equation \( x^2 + y^2 = 1 \). For simplicity, let's denote a point on the unit circle as \( (x, y) \) where \( x = \cos(\theta) \) and \( y = \sin(\theta) \), for some angle \( \theta \).
3. **On the X-axis Inside the Unit Circle:** Let's denote the point on the x-axis as \( (a, 0) \), where \( -1 < a < 1 \).
Now we form a right triangle from these three points:
- Point 1: \( (0, 0) \) (the origin)
- Point 2: \( (\cos(\theta), \sin(\theta)) \) (on the unit circle)
- Point 3: \( (a, 0) \) (on the x-axis)
The hypotenuse of the triangle is the segment between points \( (0, 0) \) and \( (\cos(\theta), \sin(\theta)) \).
To find the length of the hypotenuse:
\[
\text{Length of hypotenuse} = \sqrt{(\cos(\theta) - 0)^2 + (\sin(\theta) - 0)^2} = \sqrt{\cos^2(\theta) + \sin^2(\theta)}
\]
By the Pythagorean identity, we know:
\[
\cos^2(\theta) + \sin^2(\theta) = 1
\]
Thus, we have:
\[
\text{Length of hypotenuse} = \sqrt{1} = 1
\]
Therefore, the length of the hypotenuse of the triangle is \( \boxed{1} \).
1. **The Origin:** The first vertex is at the point (0, 0).
2. **On the Unit Circle:** The unit circle is given by the equation \( x^2 + y^2 = 1 \). For simplicity, let's denote a point on the unit circle as \( (x, y) \) where \( x = \cos(\theta) \) and \( y = \sin(\theta) \), for some angle \( \theta \).
3. **On the X-axis Inside the Unit Circle:** Let's denote the point on the x-axis as \( (a, 0) \), where \( -1 < a < 1 \).
Now we form a right triangle from these three points:
- Point 1: \( (0, 0) \) (the origin)
- Point 2: \( (\cos(\theta), \sin(\theta)) \) (on the unit circle)
- Point 3: \( (a, 0) \) (on the x-axis)
The hypotenuse of the triangle is the segment between points \( (0, 0) \) and \( (\cos(\theta), \sin(\theta)) \).
To find the length of the hypotenuse:
\[
\text{Length of hypotenuse} = \sqrt{(\cos(\theta) - 0)^2 + (\sin(\theta) - 0)^2} = \sqrt{\cos^2(\theta) + \sin^2(\theta)}
\]
By the Pythagorean identity, we know:
\[
\cos^2(\theta) + \sin^2(\theta) = 1
\]
Thus, we have:
\[
\text{Length of hypotenuse} = \sqrt{1} = 1
\]
Therefore, the length of the hypotenuse of the triangle is \( \boxed{1} \).