Problem One
Call the radius of the second can = r
Call the height of the second can = h
Then the radius of the first can = 1/3 r
The height of the first can = 3*h
A1 / A2 = (2*pi*(1/3r)*(3h)] / [2*pi * r * h]
Here's what will cancel. The twos on the right will cancel. The 3 and 1/3 will multiply to one. The 2 r's will cancel. The h's will cancel. Finally, the pis will cancel
Result A1 / A2 = 1/1
The labels will be shaped differently, but they will occupy the same area.
Problem Two
It seems like the writer of the problem put some lids on the new solid that were not implied by the question.
If I understand the problem correctly, looking at it from the top you are sweeping out a circle for the lid on top and bottom, plus the center core of the cylinder.
One lid would be pi r^2 = pi w^2 and so 2 of them would be 2 pi w^2
The region between the lids would be 2 pi r h for the surface area which is 2pi w h
Put the 2 regions together and you get
Area = 2 pi w^2 + 2 pi w h
Answer: Upper left corner <<<<< Answer
96+264 = 360
360 \ 24 = 15
Therefore he received 15 boxes
No because it fails the vertical line test hope it helps
