A rectangle is inscribed in a circle of radius 2 . Let P=(x,y) be the point in quadrant I that is a vertex of the rectangle and is on the circle. (a) Express the area A of the rectangle as a function of x . (b) Express the perimeter p of the rectangle as a function of x
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(a) The rectangle is inscribed in a circle of radius 2, so the diameter of the circle is 4. The diameter of the circle is also the diagonal of the rectangle.
Let's denote the sides of the rectangle as a and b, where a is the side along the x-axis and b is the side along the y-axis.
Since P=(x,y) is a vertex of the rectangle and is on the circle, we can say that a = 2x and b = 2y.
By the Pythagorean theorem, we have a^2 + b^2 = diameter^2, so (2x)^2 + (2y)^2 = 4^2.
Solving for y, we get y = sqrt(4 - x^2).
Therefore, the area A of the rectangle is A = ab = 2x * 2y = 4x * sqrt(4 - x^2).
(b) The perimeter p of the rectangle is p = 2a + 2b = 2(2x) + 2(2y) = 4x + 4sqrt(4 - x^2).
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