The diameter of the circle is given as 7 inches. Therefore, its area will be \(\pi \times (\frac{7}{2})^2\)

\(\pi \times (\frac{7}{2})^2 = \frac{22}{7} \times (\frac{7}{2})^2 = \frac{22}{7} \times \frac{49}{4} = \frac{11 \times 7}{2} = \frac{77}{2}\)

The diagonal of the square is given as 7 inches. Since the diagonal is \(\sqrt{2}\) times the side of a square, the side of the square will be \(\frac{7}{\sqrt{2}\)

The area of a square is given by the length of the side squared, therefore \((\frac{7}{\sqrt{2}})^2 = \frac{49}{2}\)

The positive difference between the area of the square and that of the circle = \(\frac{49}{2} - \frac{77}{2} = -\frac{28}{2} = -14 = 14\) (positivized)

Quantity A = 14
Quantity B = 14

Answer is Choice C. Both quantities are equal.