Problem
Faces, edges, vertices and Euler's formula
A table has columns for the tetrahedron, the cube and the octahedron. The rows are sketch, face shape, faces, edges and vertices. Only a few cells are filled in. Fill in the rest of the table.
Reasoning · Grade 6-1 Regular Polyhedra
Your answer
Watch out Dividing by 3 works often, but the octahedron meets four at a vertex.
How to solve
Strategy Make a Systematic List — Counting edges by staring at a picture goes wrong because the hidden ones are easy to miss. So instead of counting edges directly I count something I definitely know — the sides of all the faces — and then correct for the sharing. Each edge belongs to 2 faces, so I divide by 2; each vertex belongs to as many faces as meet there, so I divide by that. This turns each column of the table into two small multiplications and a division, which is the same job three times over. Then Euler's formula, v + f - e = 2, is used at the end as an independent check on all three columns at once, since it must come out to 2 every time.
1STEP 1
Fill in the two easy cells from the names and the picture
The names give faces 4, 6 and 8.
tetra: f = 4, cube: f = 6, octa: f = 8
The Greek number hidden in each name tells you the face count for free: tetra 4, hexa 6, octa 8. Nothing needs to be counted for this row.
2.G.A.1Visualize Spatial Relationships2STEP 2
Draw the two missing sketches
Draw the two missing sketches.
A sketch is just line segments joining points you have decided on, and the dashed/solid choice is only a question of which lines the solid itself hides — the same convention already used in the cube that is drawn for you.
4.G.A.1Draw A Diagram3STEP 3
Count the edges by counting sides and halving
Sides halved give edges 6, 12, 12.
tetra 4 × 3 ÷ 2 = 6 cube 6 × 4 ÷ 2 = 12 octa 8 × 3 ÷ 2 = 12
Counting sides face by face is something you can do without ever seeing the back of the solid, and the divide-by-2 is just 'each edge was counted from both sides', which is exactly the sharing idea from Grade 3 multiplication and division.
3.OA.C.7Identify SubproblemsCounting every face's sides and halving gives the number of edges, without ever seeing the back of the solid.
Why?
Two faces meet along every edge, so each edge is counted once from each of them and the sweep gives exactly double.
🧱Counted twice, so halveIf every item got counted from both of its two ends, the total is exactly double the truth, so halving fixes it.
Why?
The three counts are tied together by the fact that vertices minus edges plus faces always makes 2, so any one can be checked against the others.
🧱Euler's formula for solidsFor any solid with flat faces, vertices minus edges plus faces always comes out to 2, so a count you cannot see can be recovered.
4STEP 4
Count the vertices the same way
The same trick gives vertices 4, 8, 6.
tetra 12 ÷ 3 = 4 cube 24 ÷ 3 = 8 octa 24 ÷ 4 = 6
Exactly the same sharing idea as for edges, with a different divisor: the reason edges divide by 2 is that 2 faces meet along an edge, and the reason vertices divide by 3 or 4 is that 3 or 4 faces meet at a corner.
3.OA.C.7Identify Subproblems5STEP 5
Check every column with Euler's formula
Every column gives 2 for vertices plus faces minus edges.
4 + 4 - 6 = 2 8 + 6 - 12 = 2 6 + 8 - 12 = 2
One little sum has to land on 2 for every solid, so it catches a mistake in any of the three rows at once — and because it works for every polyhedron, it can also be used to find a count you could not see.
6.EE.B.6Look For A Pattern6STEP 6
Write out the finished table
The finished table checks out.
& tetra & cube & octa ; faces & 4 & 6 & 8 ; edges & 6 & 12 & 12 ; vertices & 4 & 8 & 6
Laying the answers out as a table makes the pattern jump out: the tetrahedron has as many vertices as faces, while the cube and the octahedron have their face and vertex counts swapped over.
4.G.A.1Make A Systematic ListAnswer
4, 6, 4 / 6, 12, 8 / 8, 12, 6
8 × 3 ÷ 2 = 12
Every count comes out a whole number, which it has to: 12 divided by 2, 24 divided by 2, 12 divided by 3, 24 divided by 3 and 24 divided by 4 all divide exactly, and a fraction anywhere would have meant a wrong face count or a wrong number of faces per vertex. The sizes are sensible too — the octahedron has more faces than the cube but fewer vertices, which matches the picture, since its corners are sharp points where 4 triangles meet while the cube's 8 corners are blunt. Euler's formula gives 2 in all three columns, an independent check that ties faces, edges and vertices together. And the tetrahedron's answers can simply be counted straight off a drawing: 4 triangles, 6 edges, 4 corners, all visible at once.
Takeaway
Count the sides of every face and then divide by how many faces share each part — 2 for an edge, 3 or 4 for a corner — and Euler's v + f - e = 2 tells you whether you got it right.
- Fill in the two easy cells from the names and the picture
- Draw the two missing sketches
- Count the edges by counting sides and halving
- Count the vertices the same way
- Check every column with Euler's formula
- Write out the finished table
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