Question - Area of a Circle with an Inscribed Square

To solve this problem, we need to relate the area of the square to the area of the circle. First, let's find the side length of the square. As we know, the area of a square is given by the formula \( A = s^2 \), where \( s \) is the side length of the square.

Given that the area of the square is 4, we have \( 4 = s^2 \). To find the side length \( s \), we take the square root of both sides, yielding \( s = \sqrt{4} = 2 \). Now, we know that the side length of the square is 2.

Since the square is inscribed in the circle, the diagonal of the square will be equal to the diameter of the circle. To find the length of the diagonal, we use the Pythagorean theorem in one of the right triangles that are formed by the diagonal in the square. We have \( Diagonal^2 = s^2 + s^2 \), which simplifies to \( Diagonal^2 = 2s^2 \).

Substituting \( s = 2 \), we get \( Diagonal^2 = 2 * 2^2 = 2 * 4 = 8 \). So, the length of the diagonal is \( Diagonal = \sqrt{8} = 2\sqrt{2} \), which is also the diameter of the circle.

The radius \( r \) of the circle is half of the diameter, therefore \( r = \frac{Diagonal}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2} \). The area \( A \) of the circle is given by \( A = \pi r^2 \).

Substituting our value for the radius, we find \( A = \pi (\sqrt{2})^2 = \pi * 2 = 2\pi \).

Therefore, the area of the circle is \( 2\pi \), which corresponds to option B.