An equilateral triangle of side length 6 is rotated about an altitude to form a cone. What is the volume of the cone?
The slant height of the cone will = the height of the equilateral triangle
This slant height = sqrt [ 6^2 - 3^2 ] = sqrt [ 36 - 9 ] = sqrt (27 )
The circumference of the cone = side of the triangle = 6
We can find the radius, r, of the cone as
circumference = 2pi *r
6 = 2pi r
r = 6/ (2pi) = 3/pi
Heigt of cone = sqrt [ [sqrt (27)]^2 - (3/pi)^2 ] = sqrt [ 27 - 9/ pi^2 ] = sqrt [ 27pi^2 - 9] / pi =
sqrt [ 9 (3 pi^2 - 1 ) ] /pi = (3/pi) sqrt ( 3pi^2 - 1 )
Volume of cone = (1/3) radius^2 * height =
(1/3) ( 3/pi)^2 * (3/pi) sqrt ( 3pi^2 - 1) =
(9/ pi^3)sqrt (3pi^2 -1) ≈ 1.55 units^3
An equilateral triangle of side length 6 is rotated about an altitude to form a cone. What is the volume of the cone?
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Cone radius r = 3
Slant height hs = 6 ( Height h = √27 )
Cone volume V = 1/3 * pi * r2 * sqrt(hs2 - r2) ==> V = 48.97258284 u3
or
V = r2 * pi * h / 3 ===> V = 48.97258284 u3