An isosceles triangle, in which the base is equal to the 4 root of 3, and the angle during rotation is 120

| January 27, 2021 | education

An isosceles triangle, in which the base is equal to the 4 root of 3, and the angle during rotation is 120 degrees, rotates around the straight line containing the base. Find the surface area of the body of revolution.

The figure of rotation of an equilateral triangle near its base has two equal-sized cones.

Let’s build the height of the ВO. Since ABC is isosceles, then ВO is also the median and bisector, then AO = AC / 2 = 2 * √3 cm, angle ABO = 120/2 = 600.

In a right-angled triangle AOB, the angle OAB = (90 – 60) = 300.

CosAOB = AO / AB.

AB = AO / CoAOB = 2 * √3 / (√3 / 2) = 4 cm.

By the Pythagorean theorem, BO2 = AB2 – AO2 = 16 – 12 = 4. BO = R = 2 cm.

The area of the base of the cone is not taken into account.

Scon = π * ВO * AB = π * 2 * 4 = 8 * π cm2.

Sfig = 2 * Scon = 16 * π cm2.

Answer: The area of the rotation figure is 16 * π cm2.



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