Mathematics

Question

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? Show your work.

2 Answer

  • Answer:

    Step-by-step explanation:

    5 sides of the top cube is exposed.

    so we get 1*1*5 = 5 in ^2

    the second cube has 4 sides exposed also, so 2^2 * 4 = 16 in ^2, but also it has one side with the cube on top, so we have 4-1 = 3 on that side, so the overall is 19 in ^2

    Then we have for the 3rd cube 5 sides exposed, so we have 3^2 * 5 = 45. We also have the area of the 2 in cube on it, so we get 3^2 - 2^2 = 9-4 =5.

    So the overall is 45+5+19+5 = 55+19 =74

    I hope im right sorry if im not!

  • The surface area of the three cube tower is 79sq inch.

    What is surface area?

    The quantity of space that covers the outside of a three-dimensional form is called surface area.

    Total Surface area of a cube = 6 * (a^2) (a is the side of the cube)

    Side length of the bottom-most cube = 3 inch

    Total sides = 6 sides

    T.S.A = 6a^2 = 6.(3)^2 = 54 sq inch

    Side length of cube in the middle = 2 inch

    Total sides = 5 (bottom side is already considered in the 3 inch cube's area)

    T.S.A = 5a^2 = 5(2)^2 = 20 sq inch

    Side length of the topmost cube = 1 inch

    Total sides = 5 (bottom side is already considered in the 2 inch cube's area)

    T.S.A = 5(1)^2 = 5 sq inch

    The total surface area of the tower is 79 sq inch.

    Learn more about surface area on:

    https://brainly.com/question/16519513

    #SPJ2

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