There is a cube with side eight inches. It is then cut into small cubes in which the side of each cube is two inches. Find the ratio of the sum of the surface area of all the smaller cubes with respect to the surface area of the larger cube.
1. 1 ∶ 4
2. 4 ∶ 1
3. 1 ∶ 6
4. 6 ∶ 1
1. 1 ∶ 4
2. 4 ∶ 1
3. 1 ∶ 6
4. 6 ∶ 1
Correct Answer – Option 2 : 4 ∶ 1
Given:
Big cube side is 8 inches and small cube side is 2 inches
Formula Used:
Volume of cube = \({\left( {side} \right)^3}\)
Total surface area of cube = \(6{\rm{ × }}{(side)^2}\)
Calculation:
Volume of big cube = 8× 8 × 8
⇒ 512
Volume of small cube = 2 × 2 × 2
⇒ 8
Number of small cube =(volume of big cube)/ (volume of small cube)
⇒ No. of cube = 512/8
⇒ No. of cube = 64
According to question,
Total surface area of big cube = \(6 × {(8)^2}\)
⇒ 6 × 64
Total surface area of small cube = 6 × \({(2)^2}\) × 64
⇒ 6 × 4 × 64
The ratio of total surface area of sum of small cubes and big cube
⇒ 6 × 4 × 64 ∶ 6 × 64
Required ratio = 4 ∶ 1
∴ Ratio of cube is 4 : 1