Free Questions | Re: S96-04

Is it reasonable to know that the area of the circle less the area of the square is 2 times the area outside of Quadrant 1? ((Circle - Square) / 2 + Square ) / Circle) is our answer. Seems faster than solving individual secions as it allows you to eliminate most of the variables at the end.

d^2 + d^2 = (Diameter of Circle)^2; sqrt(2d^2) = diameter, radius is sqrt(2d^2)/2

Area of circle is π(sqrt(2d^2)/2)^2 = π(2d^2/4) or πd^2/2;
Area of square is d^2
Area of the circular edges is πd^2/2
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