A plane is drawn in the cylinder parallel to its axis, cutting off an arc of 120 degrees from the base circumference.

| May 13, 2021 | education

A plane is drawn in the cylinder parallel to its axis, cutting off an arc of 120 degrees from the base circumference. The height of the cylinder is 12 cm, the distance of the plane from the axis is 4 cm. Find the cross-sectional area.

A plane is drawn in the cylinder parallel to its CC₁ axis, giving a rectangle ABB₁A₁ in cross section, where the ABC points lie on the lower base, and the A₁, B₁, C₁ points – on the upper base. It is known that this plane cuts off an arc of 120 degrees from the base circumference, which means that the value of the central angle resting on this arc will be: ∠ACB = 120 °. The height of the cylinder AA₁ = 12 cm. The distance of the plane from the axis is equal to the height of the SC of the resulting isosceles triangle ABC, since AC = BC = r, where r is the radius of the base of the cylinder and SC = 4 cm. In the right-angled triangle AСK, the leg AK = СK ∙ tg 60 °, since ∠АСК = ∠АСВ: 2 = 120 °: 2 = 60 °, then AK = 4 ∙ √3 (cm). The cross-sectional area is S = AA₁ ∙ AB, which means that S = 12 ∙ 4 ∙ √3 = 48 ∙ √3 (cm ^ 2).
Answer: the cross-sectional area of ​​the cylinder is 48 ∙ √3 cm ^ 2.



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