Two cubes have volumes of 64 cubic feet and 4,096 cubic feet. What is the ratio of the surface area of the smaller cube to the surface area of the larger cube?

Answer 1

Solution:

The volume of the 1st cube is 64 cubic ft , the side of the first cube S1 = ³√ 64 cubic ft = 4 ft

The volume of the 2nd cube is 4096 cubic ft , the side of the first cube S2 = ³√ 4096 = 16 ft

The cube has six equal faces, the surface area of the cube is 6 times the surface area of one of the faces

The surface area of 1st cube = A1 = 6 ( S1²) = 6( 4 ft )² = 96 ft²

The surface area of 2nd cube = A2= 6(S2)² = 6( 16 ft )² = 1536 ft²

The ratio of the surface area of the smaller cube to the surface area of the larger cube is

R = A1/A2 = 96 ft² / 1536 ft² = 0.0625 or the area of the smaller cube is 6.25% of the area of the larger cube

Answer 2

64 / 4096 = 4³ / 16³

^{Scale factor is 1 / 4}

Ratio or surface areas is 1² / 4²