Eight cubes,each with an edge of length one inch,are positioned together to create a large cube. What is the difference in the surface area of the large cube and the sum of the surface areas of the small cubes? A) 24 in^2 B) 16 in^2 C) 12 in^2 D) 8 in^2 E) 0 in^2

A cube itself has 6 sides, each with surface area of 1 (L x W which are each 1) So each cube has an area of 6 in^2. 8 cubes, so 48 in^2 In the mega-cube, each cube only shows 3 of its sides. Therefore, each cube only has 3 in^2 represented. 8 cubes, 3x8 =- 24. 48 - 24 = 24 in^2

so, A

Surface area of large cube = 6*2^2 = 24 square inches Surface area of small cube = 6*1^2 = 6 square inches Sum of small cubes surface area = 6*8 = 48 square inches ----------------- So the difference is 48 - 24 = 24 square inches So the answer is A

ty

small cube = area ? large cube= area?

The surface area of each small cube is 6 in^2. There is 8 of them. So that's 48 in ^2.

Large cube area surface = 24 small is 6

Surface area of large is 24 in^2

the surface area of the large cube is 24 square inches the surface of one small cube is 6 square inches

ok, I get now,ty

A

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