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.
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