1953 AMC 12 Problem 15

Let the original square have side $s$

The final square’s diagonal equals the circle’s diameter, which is $s$

Let the original square have side $s.$ The inscribed circle has diameter $s.$ That diameter is the diagonal of the largest square cut from the circle, so the final square has side $\frac{s}{\sqrt2}$ and area $\frac{s^2}{2}.$ The material left after both cuts is this final square, so the total waste is $s^2-\frac{s^2}{2}=\frac{s^2}{2}.$

Thus, the correct answer is B.