Suppose you truncate a cube such that this truncation of a vertex takes away 1/8 of the original area from each of 3 square faces and creates a new equilateral triangle. If you did this to all 8 vertices, what would the volume be? (Only use geometric formulas/reasoning for this problem.)
(In reply to
Solution by Brian Smith)
Just a notational note.
The resulting figure is a cuboctahedron.
A truncated cube is usually considered the figure where the tetrahedra have edge length of less than half.
The exact length removed is 1-√(2)/2
this creates octagonal sides.
-Jer
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Posted by Jer
on 2004-03-04 12:25:16 |
re: Solution
