1. ## skeleton cuboid

is there a formula for working out how many cm cubes fit into say a 12*6*5 skeleton cuboid.

2. Hello, peter2727!

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

3. How many cubes fit in a row?
How many rows?
How many layers?

4. no sorry i meant to say a skeleton cuboid not a solid

5. He4llo, peter2727!

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

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

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

6. 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 $\displaystyle L-2$ cubes.
There are 4 widths with $\displaystyle W - 2$ cubes.
There are 4 heights with $\displaystyle H-2$ cubes.

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

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

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

Edit: I stand corrected.

7. Soroban
thanks for that however someone else said it was 4(l+w+h) -16 can that be the case

8. 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

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# there is a frame of cuboid of length 6 units breath 5 units and height 7units the cuboid is only composed of skeleton of 210 cubes of side 1

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