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

Step-by-step 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.