The area of the rectangle is 6x14=84 square units.

A rectangle is inscribed in the circle given by `x^2+y^2=58` . The side lengths are integers, and we are asked to find the area of the rectangle.

Let one of the vertices of the rectangle have x-coordinate x. Since the point lies on the circle, we find the y-coordinate to be `y=+-sqrt(58-x^2)` . Thus by symmetry the four points can be given by the following:

`(x,sqrt(58-x^2)),(-x,sqrt(58-x^2)),(x,-sqrt(58-x^2)),(-x,-sqrt(58-x^2))`

For the side lengths to be integral, we need `58-x^2` to be a perfect square. The domain for x is {1,2,3,4,5,6,7}. We find that if x=3, y=7, and if x=7, y=3. These solutions are symmetries of the same solution.

Thus we can use the rectangle with vertices (7,3),(-7,3),(-7,-3), and (7,-3). The side lengths are 6 and 14. The area is 6x14=84 square units.

See the attachment for a diagram. The rectangle ABCD is the solution described. The rectangle EFGH is the solution rectangle rotated by 90 degrees. There are an infinite number of such rectangles.