How many faces?

[bonanova]

A regular tetrahedron (a solid with four faces) can be constructed from four unit-sided equilateral triangles.

Four similar triangles can also be assembled to have a common vertex and a unit-square base.

Adding that square creates a square pyramid (with five faces.)

These two solids can now be joined into another solid: by gluing together, vertex to vertex,

a triangular face from each of them.

How many faces does the joined solid have?

[plasmid]

The shortest and most intuitive (although least formal) proof of the answer that I can come up with.

Suppose you take two square pyramids and put them side-by-side such that their square faces lie in the same plane (say it's the x-y plane) and are touching at one edge, and have their vertices where the equillateral triangle faces meet lie on the same side of the square planes (say they're on the positive side of the z-axis).

Then draw a line linking the two vertices that lie above the x-y plane.

It should be obvious that the line linking the two vertices is the same length as the lines forming the edges of the square pyramids, and that the tetrahedron formed by the two vertices on the tops of the pyramides and by the two vertices where the bases of the square pyramids meet would form the tetrahedron described in the OP.

The answer at this point should be obvious.

[ThunderCloud]

Seven. Each shape "sacrifices" one of its faces to be joined to the other shape. (5 - 1) + (4 - 1) = 7.

[nakulendu]

Seven.

Tetrahedron - 4 faces (triangles)
The Square Pyramid - 5 faces (4 Triangles + 1 square)
Glue two together One triangle face to another. so two faces out fo total 9 hence seven - 6 triangles plus 1 square

[bonanova]

No correct answers yet.

[Yodell]

I'm guessing

some faces get on the same plane .. so my guess (picturing it in my mind is) is 5 faces .. the square base unit, two equilateral triangle faces, and two rhomboidal faces , each of them made of two equilateral triangles united on a vertex line...

Am I right ?

[k-man]

Prism with 5 faces

[BobbyGo]

My guess is 8.

I think the main thing is that the triangles in each of the two objects should be a different size. If they were the same size, the square pyramid would have a slightly shorter vertex than the tetrahedron. Joining the two triangles of dissimilar size would only cancel the smaller, not both.

[BobbyGo]

My guess is 8.

I think the main thing is that the triangles in each of the two objects should be a different size. If they were the same size, the square pyramid would have a slightly shorter vertex than the tetrahedron. Joining the two triangles of dissimilar size would only cancel the smaller, not both.

Too late to edit:

The last sentence should have read:

Joining two similar triangles with different sizes would only cancel the smaller, not both.

Edited April 8, 2013 by BobbyGo

[bonanova]

The shortest and most intuitive (although least formal) proof of the answer that I can come up with.

Suppose you take two square pyramids and put them side-by-side such that their square faces lie in the same plane (say it's the x-y plane) and are touching at one edge, and have their vertices where the equillateral triangle faces meet lie on the same side of the square planes (say they're on the positive side of the z-axis).

Then draw a line linking the two vertices that lie above the x-y plane.

It should be obvious that the line linking the two vertices is the same length as the lines forming the edges of the square pyramids, and that the tetrahedron formed by the two vertices on the tops of the pyramides and by the two vertices where the bases of the square pyramids meet would form the tetrahedron described in the OP.

The answer at this point should be obvious.

plasmid: Bingo.

It is a proof, even with plasmid's disclaimer.

The the front and rear faces of the pyramids are clearly coplanar, and points on these faces define two of the quadrilateral's faces.

To visualize, the center quadrilateral point is a hidden point. These are stick figures..

Yodell and k-man had the correct answer as well.