[bonanova]

Imagine, or draw, or grab some play dough and make, a cube, 6" on a side.

Select one corner and trace its three edges back to their nearest vertex.

Slice off the corner of the cube, using the plane defined by these vertices.

Do the same with three other corners, none of which is adjacent to a previously removed corner.

What remains is a regular triangular pyramid, a tetrahedron.

Its vertices are the four un-cut corners of the cube.

What is the volume of this tetrahedron?

Slice the tetrahedron in such a way that the newly created face is a square.

What is the area of this square?

[Guest]

I guess one way to do it is to calculate the volume of the pyramids you cut off, then subtract it from the volume of the cube. It may not be the easiest way, but I'll try.

The base of each pyramid is an equilateral triangle, and its side equals the diagonal of the cube's face, which is 6√2.

the distance from any vertex of the base to its center is 2/3 of the triangle's height, which, in turn, is the side multiplied by the sine of 60. h = 6√2*√3/2 = 3√6. 2/3 of that would be 2√6. We'll call that a.

Now we imagine a right triangle where the legs are a and H (H is the height of the pyramid) and the hypotenuse is one of the edged of the pyramid that is also an edge of the cube(let's call it l, l = 6). Using the Pythagorean theorem, we get: H² = l² - a²

H² = 36 - 24

H = √12 = 2√3

Now we just calculate the volume of the pyramid:

V = Sb*H/3

Sb = (6√2)²√3/4 = 18√3

V = 18√3 * 2√3/3 = 36*3/3 = 36 cubic inches

Since we cut off 4 of those, we reduced the cube's volume by 4*36 = 144 cubic inches

The initial volume is 6³ = 216, so what we get in the end is 72 cubic inches.

As for the second part of the puzzle, I have no idea how you would make a square face show up on a tetrahedron. You'd have to slice it more than once, right? I'll come back later and find out

[k-man]

[spoiler=That's easy ]

The volume of a tetrahedron with the edge length a equals a³*sqrt(2)/12

a is a diagonal of a square with the side of 6", so it equals 6*sqrt(2). Plugging the value of a in the formula of the volume we get the volume of 6³*sqrt(2)³*sqrt(2)/12 = 216*4/12 = 72 cubic inches.

The get the square section of the tetrahedron we need to slice every side down its middle, so the side of the square will be a/2, and therefore the area of the square is a²/4. Pluggin the value of a we get 36*2/4 = 18 square inches.