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There are two right circular cylinders, A and B. The

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There are two right circular cylinders, A and B. The [#permalink] New post 25 Jul 2008, 07:01
There are two right circular cylinders, A and B. The diameter of the base of the cylinder A is twice that of the base of the cylinder B. The height of the cylinder A is half that of the cylinder B. If the radius and the height of the cylinder B are r and h, respectively, what is the absolute value of the difference of the surface areas between the cylinder A and the cylinder B?

A: 4pi r^2
B: 6pi r^2
C: 8pi r^2
D: 10pi r^2
E: 14pi r^2

Please show your working..

also what is the ratio of the surface area of A to B?
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Re: Surface areas [#permalink] New post 25 Jul 2008, 07:25
My answer is B. We need to calculate only the surface difference of the circles (top and buttom), because the other parts of the cyliners are equal.

sorry I have no more time to explain further.
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Re: Surface areas [#permalink] New post 25 Jul 2008, 07:31
Dimensions of A are 2r and h/2
for B we have r and h
area= 2pi r(r+h)
A= 8pi r*r +2pi r*h
B= 2pi r*r + 2pi r*h
therefore A-B= 6pi r*r
Hence B

Ration wil be A/B i think i dont need to write that
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Re: Surface areas [#permalink] New post 25 Jul 2008, 07:58
agree with B...ratio of SA is 4:1
Re: Surface areas   [#permalink] 25 Jul 2008, 07:58
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There are two right circular cylinders, A and B. The

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