An equilateral triangle is rotated about one of its altitudes, sweeping out a cone in space. If the height of the equilateral triangle is $3 \sqrt{3},$ then find the volume of the cone. A circular sector is rolled to form a cone. If the radius of the sector is $15$ and its central angle is $144^\circ,$ then find the height of the cone.

itsash Oct 27, 2023

#1**-3 **

Problem 1

The volume of a cone is given by the formula:

V = 1/3 * π * r^2 * h

where:

V is the volume of the cone

π is a mathematical constant with the approximate value of 3.14

r is the radius of the base of the cone

h is the height of the cone

In this case, the height of the cone is 3sqrt(3) and the radius of the base is equal to the height of the equilateral triangle, which is also 3sqrt(3).

Therefore, the volume of the cone is:

V = 1/3 * π * (3*sqrt(3))^2 * 3*sqrt(3) = 27*sqrt(3)*pi

Problem 2

The height of the cone is equal to the arc length of the sector. The arc length of a sector is given by the formula:

arc length = central angle / 360 degrees * 2 * π * radius

In this case, the central angle is 144 degrees and the radius is 15. Therefore, the arc length is:

arc length = 144 degrees / 360 degrees * 2 * π * 15

Simplifying, we get:

arc length = 30π

Therefore, the height of the cone is 30π.

bingboy Oct 27, 2023