A section parallel to the axis of the cylinder cuts off an arc of 120 degrees from the base circumference.

A section parallel to the axis of the cylinder cuts off an arc of 120 degrees from the base circumference. Find the area of the lateral surface of the cylinder if the sectional area is Q.

Let’s construct the radii of the circle OA and OB.
The cross-sectional area is equal to: Ssection = AB * AD = Q cm2.
The angle AOB, by condition, is equal to 1200. By the cosine theorem, we determine the length of the radius of the circle.
AB ^ 2 = ОА ^ 2 + ОВ ^ 2 – 2 * ОА * ОВ * Cos120 = 2 * R2 – 2 * R2 * Cos120 = 3 * R2.
R = AB / √3 cm.
The area of the lateral surface of the cylinder is: Sbok = L * AD = 2 * π * R * AD = 2 * π * (AB / √3) * AD = 2 * π * Q / √3 cm2.
Answer: The area of the lateral surface of the cylinder is 2 * π * Q / √3 cm2.