# A plane parallel to the base of the cone cuts off a cone with a base area of 4π from it.

**A plane parallel to the base of the cone cuts off a cone with a base area of 4π from it. Find the radius of the base of the original cone if the plane divides the volume of the cone in a ratio of 1: 7, counting from the vertex.**

Knowing the area of the base of the truncated cone, we determine its radius.

Ssection = π * OD1 ^ 2.

OD1 ^ 2 = Ssection / π = 4 * π / π = 4.

OD = 2 cm.

Since the plane divides the volume of the cone in a ratio of 1/7, the volume of the original cone is equal to: V1 = (7 + 1) * X cm3, and the volume of the cut off cone is V2 = 1 * X cm3.

Both cones are similar, and the volume ratio is: V2 / V1 = 1/8.

The ratio of the volumes of such cones is equal to the ratio of the cubes of their radii.

1/8 = O1D ^ 3 / AO ^ 3.

AO ^ 3 = 8 * O1D ^ 3 = 8 * 8 = 64.

AO = 8 cm.

Answer: The base radius of the original cone is 8 cm.