A cube with 2 inch sides is placed on a cube with 3 inch sides. Then a cube with 1 inch sides is placed on the 2 inch cube. What is the surface area of the three cube tower?

let's try the "Divide and Conquer" method, that is we break down the problem into pieces and solve each piece until we arrive at the answer.

the formula for the surface area of the cube is:

SA = 6s^2; s is the length of one side of the cube

for the 3-inch cube

SA = 6 (3^2)

SA = 6 (9)

SA = 54 in^2

note however, that the actual surface area is less the area occupied by the 2-inch cube place on the 3-inch cube, which is

A = s^2; since this is a square

A = 2^2

A = 4 in^2

the actual SA (aSA) therefore of the 3-inch cube is

aSA = 54 - 4 = 50 in^2

for the 2-inch cube

SA = 6s^2

SA = 6 (2^2)

SA = 6 (4)

SA = 24 in^2

to solve for the actual SA of the 2-inch cube, we subtract the area which is in contact with the 3-inch cube and the 1-inch cube

aSA = 24 - 2^2 - 1^2

aSA = 24 - 4 - 1

aSA = 19 in^2

for the 1-inch cube

SA = 6s^2

SA = 6 (1^2)

SA = 6 in^2

the actual surface area is less the area in contact with the 2-inch cube

aSA = 6 - 1^2

aSA = 5 in^2

the surface area of the three cube tower will just be the sum of the aSA of the three cubes

SA = 50 + 19 + 5

SA = 74 in^2 ans