+123 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
+129845 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
