A sphere of radius r is cut by a plane at a distance of h from its cen

When a sphere with radius r is cut into two different pieces by a plane at a distance of h from its center, the total surface area of the two pieces exceeds the surface area of the sphere by the total area of the two circular cross sections of the two pieces. Furthermore, since the total surface area of the two pieces is 25% more than that of the sphere, the total area of the two circular cross sections must be 25% of the surface area of the sphere.

Since the two cross sections have the same area and they are circles, we can let the radius of each be R. Using the Pythagorean theorem, we can determine R, in terms of r and h, as √(r^2 - h^2). Thus the total area of the two circular cross sections is 2πR^2 = 2π[√(r^2 - h^2)]^2 = 2π(r^2 - h^2). Since this must be 25% of the surface area of the sphere, we can create the equation:

2π(r^2 - h^2) = ¼(4πr^2)

2π(r^2 - h^2) = πr^2

2r^2 - 2h^2 = r^2

r^2 = 2h^2

r^2/2 = h^2

√(r^2/2) = h

r/√2 = h

Answer: A