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.)
Let the edge of the cube be 1.
Each truncation would remove a right tetrahedron with an edge of 1/2. The volume of the tetrahedron is (1/6)*(1/2)*(1/2)*(1/2)=1/48.
The total volume of all eight removed tetrahedra is 8*(1/48)=1/6. Then, the volume of the truncated cube is 1-1/6=5/6
Solution
