In a cylinder, a section with an area of 12√3, parallel to the axis, cuts off an arc of 120 degrees from the base circumference.

| January 12, 2021 | education

In a cylinder, a section with an area of 12√3, parallel to the axis, cuts off an arc of 120 degrees from the base circumference. Find the volume of the cylinder if its height is 4 cm.

The section parallel to the axis of the cylinder is a rectangle AВСD.
The cross-sectional area is equal to: Savsd = AB * AD = 12 * √3 cm2.
AB = 12 * √3 / 4 = 3 * √3 cm.
Since the degree measure of the arc AB is equal to 1200, the central angle AOB = 1200.
In an isosceles triangle AOB, the height OH is also the median and bisector of the triangle, then AH = BH = AB / 2 = (3 * √3) / 2 cm, and the angle AOH = AOB / 2 = 120/2 = 60.
SinAOB = AH / AO.
AO = R = AH / Sin AOB = ((3 * √3) / 2) / (√3 / 2) = 3 cm.
Determine the base area.
Sb = n * R2 = n * 9 cm2.
Let’s define the volume of the cylinder.
V = Sosn * AD = n * 9 * 4 = 36 * n cm3.
Answer: The volume of the cylinder is 36 * n cm3.



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