Question 203382
Better wording would be: What is the area of a square inscribed in a circle whose radius is 8 inches? If this is your problem then , the picture looks like:
{{{drawing(400, 400, -2, 2, -2, 2, circle(0, 0, sqrt(2)), line(-1, -1, 1, -1), line(1, -1, 1, 1), line(1, 1, -1, 1), line(-1, 1, -1, -1), line(0, 0, 1, 1), line(0, 0, 1, 0), locate(.4, .7, 8), locate(.2, .2, 45), locate(1.1, .5, x), locate(-.9, 0, 2x))}}}
Since the area of a square is the square (hence the term) of the side, we need to find the side of the square. As I hope you can see, finding the leg of the right triangle will help us find the side of the square.<br>
In all 45-45-90 triangles the ratio of the hypotenuse to a leg is {{{Sqrt(2)}}}. In the 45-45-90 right triangle shown, the hypotenuse is 8.  Substituting this into the ratio we can find the leg (using x for leg because "l" can be confused with "1"): {{{8/x = sqrt(2)}}}
Multiplying both sides by x
{{{8 = x*sqrt(2)}}}
Divide both sides by sqrt(2):
{{{8/sqrt(2) = x}}}
Now the leg, x, is 1/2 of the side of the square, s. So the side of the square is twice as much:
{{{s = 2(8/sqrt(2)) = 16/sqrt(2)}}}
{{{A = s^2 = (16/sqrt(2))^2 = 256/2 = 128}}} square inches.