# skeleton cuboid

• Jan 6th 2011, 05:23 AM
peter2727
skeleton cuboid
is there a formula for working out how many cm cubes fit into say a 12*6*5 skeleton cuboid.
• Jan 6th 2011, 05:42 AM
Soroban
Hello, peter2727!

Quote:

Is there a formula for working out how many cm cubes
fit into, say, a 12 x 6 x 5 cuboid?

If your "cm cubes" are unit cubes (1 x 1 x 1),

. . the number of cubes is simply the product of those dimensions.

For a $12 \times 6 \times 5$ cuboid, there will be: . $12\cdot6\cdot5 \:=\:360$ unit cubes.

• Jan 6th 2011, 05:47 AM
Prove It
How many cubes fit in a row?
How many rows?
How many layers?
• Jan 6th 2011, 06:14 AM
peter2727
no sorry i meant to say a skeleton cuboid not a solid
• Jan 6th 2011, 08:21 AM
Soroban
He4llo, peter2727!

Quote:

i meant to say a skeleton cuboid, not a solid.

I saw the word "skeleton", but didn't know what it meant.
. . (And I'm still not sure.)

If you mean that the unit cubes form the outline of the cuboid,
. . we can invent a formula.

There are 8 cubes at the vertices (one in each "corner").

There are 4 lengths with $L-2$ cubes.
There are 4 widths with $W - 2$ cubes.
There are 4 heights with $H-2$ cubes.

Total number of cubes: . $N \;=\;8 + 4(L-2) + 4(W-2) + 4(H-2)$

. . which simplifies to: . $N \;=\; 4(L+W+H-4)$

• Jan 6th 2011, 08:27 AM
alexmahone
Quote:

Originally Posted by Soroban
He4llo, peter2727!

I saw the word "skeleton", but didn't know what it meant.
. . (And I'm still not sure.)

If you mean that the unit cubes form the outline of the cuboid,
. . we can invent a formula.

There are 8 cubes at the vertices (one in each "corner").

There are 4 lengths with $L-2$ cubes.
There are 4 widths with $W - 2$ cubes.
There are 4 heights with $H-2$ cubes.

Total number of cubes: . $N \;=\;8 + 4(L-2) + 4(W-2) + 4(H-2)$

. . which simplifies to: . $N \;=\; 4(L+W+H-4)$

That's not what he meant. He just means that the cuboid is hollow, not solid.

Edit: I stand corrected.
• Jan 6th 2011, 09:07 AM
peter2727
Soroban
thanks for that however someone else said it was 4(l+w+h) -16 can that be the case
• Jan 6th 2011, 09:10 AM
peter2727
8 corners
4 square rods of length (L – 2)
4 square rods of length (B – 2)
4 square rods of length (H – 2)
Let C = number of cubes in the model.
Then C = 8 + 4(L – 2) + 4(B – 2) + 4(H – 2)
C = 4(L + B + H) – 16 or C = 4(L + B + H – 4)
C = 4(12 + 6 + 10) – 16