The height of the cone is 18 cm, and the radius of the base is 6 cm. The plane perpendicular to the axis

The height of the cone is 18 cm, and the radius of the base is 6 cm. The plane perpendicular to the axis of the cone intersects its lateral surface in a circle whose radius is 4 cm. Find the distance from the plane of the section to the plane of the base of the cone.

To calculate the distance from the plane to the base, consider the axial section of this cone.

Since the axial section of the cone is an isosceles triangle, for convenience we denote it by ∆ABS. The plane intersecting the lateral surface of the cone will be denoted by A1C1.

The point is the center of the base of the cylinder, point O1 is the center of the cutting plane.

Thus, the segment OO1 is the distance between the base and the plane.

This plane cuts off the triangle ∆A1BC1 from the axial section.

Triangles ∆ABS and ∆A1BC1 are similar since they have a common vertex ∠B.

Since the height of an isosceles triangle cuts it into two equal rectangular ones: ∆ABO and ∆СВО, consider one of them, ∆ABO.

Since ∆A1BO1 is similar to ∆ABO, we calculate the similarity coefficient, which is the ratio of the similar sides of the triangle:

k = OS / OS1 = ВС / ВС1 = VO / VO1;

k = 6/4 = 1.5.

Thus:

BO1 = BO / k;

BO1 = 18 / 1.5 = 12 cm.

Since the height ∆А1BC1 is 12 cm, the distance from the plane to the base is equal to:

OO1 = BO – BO1;

OO1 = 18 – 12 = 6 cm.

Answer: the distance from the plane to the base is 6 cm.