An equilateral trapezoid with bases of 12 cm and 18 cm and a height of 4 cm rotates around

| September 1, 2021 | education

An equilateral trapezoid with bases of 12 cm and 18 cm and a height of 4 cm rotates around a large base. Find the volume of the body of revolution.

Let ABCD be a given trapezoid. BC = 12 cm, AD = 18 cm. CH = BE = 4 cm.

When rotating, you get such a figure, consisting of two equal cones and a cylinder.

Let’s find the volume of the cylinder.

Vts = Sosn * h.

The base of the cylinder will be a base with a radius equal to CH, R = 4 cm, so the area of ​​the base is equal to: Sbn = nR² = n * 4² = 16p.

The height of the cylinder is equal to the segment NOT, which means that it is equal to BC, h = 12 cm.

Hence, the volume of the cylinder is equal to: Vc = 16p * 12 = 192p cm².

Find the volume of the cone:

Vk = 1/3 * Sosn * h.

The height of the cone is equal to the segment DH = AE = (18 – 12): 2 = 3 cm.

The base of the cone is equal to the base of the cylinder: Sbn = 16p.

Hence the volume of the cone is:

Vk = 1/3 * 16p * 3 = 16p cm².

Let’s calculate the volume of the resulting figure:

V = Vts + 2 * Vk = 192p + 16p * 2 = 224p cm².



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