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The volume of regular octahedron PABCDQ is 243. Rotating octahedron PABCDQ about its diagonal $\overline{PQ},$ the path of the octahedron forms a new 3D shape. What is the volume of this 3D shape?

https://latex.artofproblemsolving.com/e/c/1/ec166c676365a480c3be6bb4646ec8c18068c507.png

 Apr 8, 2019
 #1
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Never done one like this before.....but....I believe that the rotation of this will produce two cones

 

The  volume of an octahedron is given by

 

V  = √2/3  * edge length^3

243* 3 / √2  =  edge length^3

729/√2  = edge length ^3

9 / (2)^(1/6)  = edge length 

 

The distance from any vertex to its non-adjacent vertex  = √2 edge  = 9√2 / {2)^(1/6)  = 9∛2

 

The radius of each cone will be (1/2) of this  =  4.5∛2

And this will also be the height of each cone

 

So....the volume of the 3D figure will be  the volume of the two cones =

 

2 [ (1/3) pi * radius^2 * height ]  =

 

(2pi)  ( 4.5∛2)^2 (4.5∛2) / 3  =

 

2pi  [ 4.5]^3 * 2 / 3  =

 

(4pi] * [91.125] / 3  =

 

4pi (243/8)  =

 

(243/ 2) pi units^3

 

cool cool cool

 Apr 9, 2019

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