Plug In for the radius, n, and solve for x. Let’s make n = 3: The area of the base of the cylinder is now 9π, and the circumference of the base is 6π. The ribbon itself is a rectangle, and we now know both its area, which is the same as the area of the base, and its length, which is the same as the circumference of the base. Now we can solve for x, which is the other side of the rectangle: 6πx = 9π, so x =
9π
6π
, or
3
2
. Our value for n is greater than our value for x, so Quantity B is greater