Question

Nate baked 3 cakes using a circular cake pan of diameter 8 inches. He plans to make a layered cake with one cake on top of the other with frosting only in between the cakes and on top of the third cake. Katrina made 5 cakes using a circular cake pan of diameter 6 inches. She plans to make a layered cake with each cake on top of the other and frosting only in between the cakes and on top of the fifth cake. The height of frosting in both their cakes is the same.

Who requires more frosting?

Answer 1

Nate. He has to frost an area of 602.88 square inches, and Katrina has to frost an area of 141.3 square inches.

Katrina. She has more layers in her cake than Nate, and the radius of her set of cakes is only 1 inch less than the radius of Nate’s cakes.

Katrina. She has to frost an area of 94.2 square inches, and Nate has to frost an area of 75.36 square inches.

Nate. He has to frost an area that is 9.42 square inches greater than the area Katrina has to frost.

Answer 2

lol before i read this let me do 1 thing lol

Answer 3

@ryan123345 ok sure :)

Answer 4

This is an area problem. The area of a circle is \[A=\pi r^2\]Do you still need more help? What answer do you obtain.

Answer 5

would it effect Katrina's cake cuz she has more layers in her cake? I'm thinking the second one..?

Answer 6

You can not compare Nate and Katrina until you find how much frosting Nate needs, and how much frosting Katrina needs. What are these two values?

Answer 7

i'm kinda not really sure where to start..

Answer 8

Start piece by piece, or in this case, person by person.

Nate has how many layers of frosting?\[\]

Answer 9

50.24

Answer 10

then would u multiply that by 3 for each cake?

Answer 11

Nate's cake has 3 layers of frosting. how much area of frosting will each layer take?

Answer 12

150.72?

Answer 13

Your answers are not very descriptive. 150.72 what? Units are important. Observe:\[A_{L}=\pi r^2\]\[d=8\ in\]\[r=4\ in\]\[A_{L}=\pi (4\ in)^2=16\pi\ in^2\]
So the area of a single layer is 16pi. That means that the area of the 3 layers is 3 times this, or \[3\ Layers=3(16\pi\ in^2)\]\[Nate=48\pi\ in^2\]

Answer 14

Now compare this to Katrina's frosting. Katrina has 5 layers, d = 6 in | r = 3 in, with each layer having an area of \[A_{L}=\pi r^2\]\[A_{L}=\pi(3\ in)^2\]\[A_{L}=9\pi\ in^2\]

There are 5 layers, so Katrina has \[6(9\pi)=54\pi\ in^2\]

Katrina uses more frosting.

Answer 15

ooohkay!! i guess it just makes more sense if i see it. thank you so much :)