Question

A special right triangle drawn in Quadrant I intersects the unit circle at point (x, y)
where y=2√2
. What is the value of x in point (x, y)
? Use the equation of the unit circle to determine
the missing coordinate.(1 point)
Responses

−2√2
negative the fraction with numerator square root of 2 and denominator 2

12
1 half

2√2
the fraction with numerator square root of 2 and denominator 2

3√2

Answer 1

First note: y = 2√2 > 1 is impossible on the unit circle, so you must mean y = √2/2. Using x^2 + y^2 = 1:

x^2 + (√2/2)^2 = 1
x^2 + 1/2 = 1
x^2 = 1/2
x = ±√(1/2) = ±√2/2.

Since the point is in Quadrant I, x > 0, so x = √2/2.