# A cube of 1 m edge length is crushed in order to get small cubes of edge length 1 mm. By which factor does the surface area increase?

### 6 Answers

- davidLv 75 months ago
There are created 1000 X 1000 X 1000 or 10^9 new small cubes

each of these has 6 sides 1mm X 1mm or total 6 mm^2 area for one or a total of 6x10^9 mm^2

the orig. cube had a surface of 6 X 1000^2 or 6x10^6 mm^2

6x10^9 mm^2 / 6x10^6 mm^2 = 10^3 or a factor of 1000 <<< answer

- Ian HLv 75 months ago
Side length of original cube was 1000 mm

Just one small cube has of edge length 1 mm.

Comparison of areas goes as square of side length ratios.

Surface area for one small cube only is 1/1,000,000 of original.

There will now be 1000,000,000 small cubes, each of 1/1,000,000 of original.

That explains why the factor for total surface area increase is 1000

- ted sLv 75 months ago
crushed ( ???) or cut ??....1 m = 10³ mm so there are 10^9 cubes of side 1 mm...each having 6 faces

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- PuzzlingLv 75 months ago
The original surface area is 6 m².

Along each edge:

1 m = 1000 mm

So you need to make 999 cuts along each of 3 dimension, for a total of 2997 cuts. Each cut exposes two faces for a total of 2 m² added to the surface area on each cut. So altogether you will be adding 5994 m² for a total of 6000 m² or an increase of 1000 times.

Answer:

1000 times

- billrussell42Lv 75 months ago
crushed? exactly what does this mean?

Guessing you mean the cube is of sheet metal, and it is cut up and formed into small cubes. But the total area has to be the same, which is 6 m²

or perhaps the cube is of solid metal (which cannot be crushed) and melted down and formed into smaller cubes?

the volume stays constant in that case, at 1 m³

how many smaller cubes are formed?

first convert volume into mm³

(1 m³) x (1000 mm/m)³ = 1e9 mm³

1e9 mm³ / 1 mm³ = 1e9 cubes.

each has a surface area of 6 mm²

total area = 6 mm²/cube x 1e9 cubes = 6e9 mm²

convert original area of 6 m² to mm²

6 m² x (1000 mm/m)² = 6e6 mm²

now divide

6e9 mm² / 6e6 mm² = 1000 ⬅

next time post a question that makes sense.