# The total surface area of the cone is 108. Parallel to the base of the cone

**The total surface area of the cone is 108. Parallel to the base of the cone, a section is drawn, dividing the height in half. Find the total surface area of the trimmed cone.**

The axial section of the cone is an isosceles triangle.

Triangles ABC and DBE are isosceles and are similar in two angles, angle B is common, angle BDE = BAC as the corresponding angles at the intersection of parallel lines DE and AC secant AB.

The coefficient of similarity of triangles is 1/2.

AB = 2 * BD, AO = 2 * DO1.

The area of the large cone is: S1 = π * AO * (AO + AB) = π * 2 * DO1 * (2 * DO1 + 2 * BD) = 4 * π * DO1 * (DO1 + BD) = 108 cm2.

The area of the smaller cone is: S2 = π * DO1 * (DO1 + BD).

S1 / S2 = π * 4 * DO1 * (DO1 + BD) / π * DO1 * (DO1 + BD) = 4.

S2 = S1 / 4 = 27 cm2.

Answer: The area of the cut off cone is 27 cm2.