+0  
 
0
106
3
avatar

An equilateral triangle of side length 6 is rotated about an altitude to form a cone. What is the volume of the cone?

 Nov 24, 2020
 #1
avatar+116126 
0

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

 

 

cool cool cool

 Nov 24, 2020
 #3
avatar
0

Hello, Phill!

Why don't you come back here and correct all your errors before they become mistakes?smiley

Guest Nov 25, 2020
 #2
avatar+1039 
+3

An equilateral triangle of side length 6 is rotated about an altitude to form a cone. What is the volume of the cone?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

 Nov 25, 2020

64 Online Users