Question

A wedding cake has two layers, as shown. Each layer is in the shape of a cube. The bottom of the cake and the area where the two cakes meet is not frosted. What is the area of the cake that is frosted? Show and explain your work.

Answer 1

The area of the cake that is frosted is 464 in²

What is the area of the cake that is frosted?

Bottom cake = 10 inches

Top cake = 6 inches

The lateral area of the bottom cube is 4 faces, each of which is a 10-inch square.

Lateral area = 4 × s²

= 4 × (10 in)²

= 400 in²

Top cube

The top area is the difference in area between a 10-inch square and a 6-inch square;

= (10 in)² - (6 in)²

= (100 -36) in²

= 64 in²

Therefore,

Area of the cake frosted = the sum of the lateral area and the top frosted area.

Area of the cake frosted = 400 in² +64 in²

= 464 in²

Answer 2

Explanation:

Hello there, I think your question missed key information, allow me to add in and hope it will fit the orginal one.

Let assume the side of the cube is a units

=>The surface area of the cude is:

• A =
  `6a^(2)`

Given that The bottom of the cake and the area where the two cakes meet is not frosted.

=> the are of a cube that is frosted in the first layer

A' = A -
`a^(2)`

=
`6a^(2)` -
`a^(2)`

=5
`a^(2)`

=> the are of a cube that is frosted in the second layer

A'' =
`6a^(2)` -
`a^(2)` -
`a^(2)` (the bottom and the top of it are not frosted)

= 4
`a^(2)`

=> the total the area of the cake that is frosted is:

TA = A' + A''

= 5
`a^(2)` + 4
`a^(2)`

= 9
`a^(2)` square units

To solve this type of question, you just need to substitute the actual side of the cube into the expression 9
`a^(2)` . Hope it will find you well.