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Cone Geometry

Expert replies
by pepeprepa » Wed Aug 13, 2008 5:10 am
PS with a cone
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Source: — Problem Solving |

Re: Cone Geometry

by parallel_chase » Wed Aug 13, 2008 9:12 am
pepeprepa wrote:PS with a cone
Is the answer 1:4? Let me know if this is correct, so that I can post the solution.
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by sudhir3127 » Wed Aug 13, 2008 9:49 am
IMO 1:7 D

let me know so that i will post my solution...
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by pepeprepa » Wed Aug 13, 2008 9:54 am
That is 1:7
If you want another geometry one I posted one in Data Sufficiency called "Hexagon Geometry", it is nice.
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by sudhir3127 » Wed Aug 13, 2008 10:01 am
this is how i approached the problem...
let r be the radius of the base of the orginal cone and h be the height...

thus the smaller cone will have a radius of r/2 and height h/2.

volume of the cone = 1/3 pi*r^2h

1/3*pi (r/2)^2*h/2= pi*r^2*h/24...................................1


the remaining part will

1/3pi*r^2h- pi*r^2*h/24

= 7*pi*r^2*h/24.....................................................2

ratio of 1 by 2

(pi*r^2*h/24)/(7*pi*r^2*h/24)

which is 1:7.

please let me know if u have any doubts..
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by pepeprepa » Wed Aug 13, 2008 10:16 am
The thing that annoyed me in this question is something which is obvious for Sudhir I think; the link between x and R (names on the original graph).
Indeed, you can see that the two triangles in the cone are similar so the ratio is the same between AB:AC and x:R that is 1:2
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