Find the radius of a circle in which an inscribed square has a side of 8 inches.

If the inscribed square has sides of 8in, the diameter of the circle is equal to the diagonal of the square.

d^2=x^2+x^2

d^2=2x^2

d=√(2x^2)

Since d=2r, r=d/2 so

r=(1/2)√(2x^2)

r=√((2x^2)/4)

r=√(x^2/2), since x=8

r=√(64/2)

r=√32  

r=√(16*2)

r=4√2 in (exact)

r≈5.66 in (to nearest hundredth of an inch)

What is the diagram u didn't post that