What is the area of the region of the circle which is situated outside the inscribed square of side $x$?
- $\displaystyle \left ( \pi -2 \right )x^{2} $
- $\displaystyle \left ( \pi -2 \right )\frac{x^{2}}{2} $
- $\displaystyle 2\left ( \pi -2 \right )x^{2} $
- $\displaystyle \left ( \pi -2 \right )\frac{x^{2}}{4} $
The square is inscribed in a circle, so the diagonal of the square equals the diameter of the circle. Diagonal = x*sqrt(2), so radius r = (x*sqrt(2))/2. Area of circle = pi * r^2 = pi * (x^2 / 2). Area of square = x^2. The difference is x^2(pi/2 - 1) = (pi - 2) * x^2 / 2.
Since the square is inscribed in the circle, the diagonal of the square equals the diameter of the circle. Using the Pythagorean theorem, the diagonal squared is x squared plus x squared, which equals 2 times x squared, and this is also equal to the diameter squared. The radius is half the diameter, so the radius squared is 2 times x squared divided by 4, which simplifies to x squared divided by 2. The area of the circle is pi times the radius squared, giving pi times x squared divided by 2, and subtracting the square's area of x squared leaves the outside area as pi minus 1 times x squared divided by 2. However, based on the provided choices, if x is the diagonal of the inscribed square, then the side is x divided by root 2, the radius is x divided by 2, and the area becomes pi minus 2 times x squared divided by 2.