Question

a sphere has a radius of 16 inches. a horizontal plane passes through the center of the sphere.
part 1
describe the cross section formed by the plane and the sphere.
it is a circle with a radius of 16 in.
part 2 out of 2
describe the cross sections formed as the plane intersects the interior of the sphere but moves away from the center.
the cross sections will be circles, but their radii will (select) as the plane moves away from the sphere’s center.

Explanation:

Step1: Analyze Part 1

A horizontal plane passing through the center of a sphere: By the definition of a sphere, a plane through its center will intersect it in a circle with radius equal to the sphere's radius. The sphere has radius 16 in, so the cross - section is a circle with radius 16 in.

Step2: Analyze Part 2

When a plane intersects a sphere and moves away from the center, we can use the Pythagorean theorem in the right triangle formed by the sphere's radius (\(R\)), the distance from the center of the sphere to the plane (\(d\)), and the radius of the cross - section circle (\(r\)). The formula is \(r=\sqrt{R^{2}-d^{2}}\). As \(d\) (the distance from the center to the plane) increases (as the plane moves away from the center), \(r\) (the radius of the cross - section circle) will decrease because \(R^{2}-d^{2}\) will be smaller. So the radii of the cross - sections will decrease as the plane moves away from the sphere's center.

Answer:

Part 1: It is a circle with a radius of 16 in.
Part 2: decrease