Question 1201954
A circle with an area of pi has a radius of 1. Now consider the inscribed rectangle. The diagonal of this rectangle is the hypotenuse of a right triangle with the length and width as the other two sides. The hypotenuse of the rectangle is equal to 2, the diameter of the circle. Since the length is twice the width, by the Pythagorean theorem, we have 2^2 = W^2 + (2W)^2 = 5W^2.
Thus W = sqrt(4/5) = 2/sqrt(5)
Therefore, the area is W*L = 4/sqrt(5)*2/sqrt(5) = 8/5