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# If the volume of the cone is SH/3 where S is the area of

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If the volume of the cone is SH/3 where S is the area of [#permalink]

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21 Jan 2012, 04:07
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If the volume of the cone is SH/3 where S is the area of the base and H the height. AB=BC. What is the ratio of the volume of the cone (with BC as height) to the volume of the rest of the cone (with AB as height). (for graph see attached. Didn't know how to add picture to this post)

1. 1:3
2. 1:4
3. 1:6
4. 1:7
5. 1:8

Thanks for helping. I must have forgotten something about cone characteristics here...
[Reveal] Spoiler: OA

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graph.pptx [186.44 KiB]

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21 Jan 2012, 04:33
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Splendidgirl666 wrote:
If the volume of the cone is SH/3 where S is the area of the base and H the height. AB=BC. What is the ratio of the volume of the cone (with BC as height) to the volume of the rest of the cone (with AB as height). (for graph see attached. Didn't know how to add picture to this post)

1. 1:3
2. 1:4
3. 1:6
4. 1:7
5. 1:8

Thanks for helping. I must have forgotten something about cone characteristics here...

Note that the wording is not 100% accurate, plus it's highly unlikely you'll need this staff for the GMAT.

Attachment:

untitled.JPG [ 13.91 KiB | Viewed 4504 times ]
Anyway, notice that since AB=BC (AC=2*BC) then R=2r (from similar triangles property). Now, the area of the base of a little upper cone (cone BC) will be $$s=\pi*r^2$$ and as the area of the base of the bigger cone (cone AC) is $$S=\pi*R^2=\pi*4r^2=4s$$ then $$s=\frac{S}{4}$$.

Next, $$volume_{BC}=\frac{\frac{S}{4}*\frac{H}{2}}{3}=\frac{SH}{24}$$

$$volume_{AB}=volume_{AC}-volume_{BC}=\frac{SH}{3}-\frac{SH}{24}=\frac{7SH}{24}$$;

Ratio --> $$\frac{volume_{BC}}{volume_{AB}}=\frac{\frac{SH}{24}}{\frac{7SH}{24}}=\frac{1}{7}$$.

P.S. You can attach an image file as you attached ppt file.
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Re: If the volume of the cone is SH/3 where S is the area of [#permalink]

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29 May 2013, 01:25
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I think the easiest way to solve this problem is keeping in mind the idea that if any dimension (for example, height h) of similar figures are in ratio 1:x, then their areas (S) will be in ratio 1:x^2, and their volumes will be in ratio 1:x^3

First,
H(bc) - the height BC
H(whole cone) - the height of the whole cone
V(bc) - the volume of the smaller cone with the height BC
V(ab) - the volume of the larger cone with the height AB
V(whole cone) - the volume of the whole cone with the height AC

H(bc):H(whole cone) = 1:2, therefore V(bc):V(whole cone) = 1:8.

Let's V(bc) = x, then V(bc):V(whole cone) = x:8x.

Consequently, subtracting V(bc) from V(whole cone) we can calculate V(ab):
V(ab) = 8x-x=7x
and the ratio of the volume of the top half of the cone (with BC as height) to the volume of the bottom half of the cone (with AB as height) =
x:7x = 1/7

Last edited by thewind on 29 May 2013, 01:46, edited 2 times in total.
Manager
Joined: 07 May 2013
Posts: 104
Re: If the volume of the cone is SH/3 where S is the area of [#permalink]

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30 Nov 2013, 22:00
Buneul, actually this is a CAT question. However, here goes the solution. As, you said r=R/2 and AB=BC=H/2.

Now, volume of cone with height BC=$$Pi*r^2*BC$$----->1
Volume of frustum of a cone with height AB=$$\frac{Pi*AB*(R^2+r^2+rR)}{3}$$----->2
Substitute values, and do $$\frac{1}{2}$$ to get the answer.
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Re: If the volume of the cone is SH/3 where S is the area of [#permalink]

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21 Aug 2017, 01:30
Splendidgirl666 wrote:
If the volume of the cone is SH/3 where S is the area of the base and H the height. AB=BC. What is the ratio of the volume of the cone (with BC as height) to the volume of the rest of the cone (with AB as height). (for graph see attached. Didn't know how to add picture to this post)

1. 1:3
2. 1:4
3. 1:6
4. 1:7
5. 1:8

Thanks for helping. I must have forgotten something about cone characteristics here...

Let r be the radius of the cone. S = pi * r^2
So, radius of smaller cone = r/2
Volume of smaller cone = 1/3* pi * (r/2)^2 *H/2 = 1/3*pi*r^2*H/8
(AB= BC = H/2)
Volume of the rest of the cone = 1/3* pi * r^2 - 1/3*pi*r^2*H/8 = 1/3*7/8*pi*r^2*H

Volume of smaller cone/Volume of the rest of the cone = 1/7

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Re: If the volume of the cone is SH/3 where S is the area of [#permalink]

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25 Jan 2018, 11:57
One General concept,
if two triangles are similar, and sides of the two triangles are in ratio = k : 1
then the area of the two triangles are in the ratio = k^2 : 1
extending this concept to 3-D dimension, if two cones are similar(height1/height2 = radius1/radius), then the volume of the two cones are in the ratio = k^3 : 1

now given AB = BC => height of top of cone is half actual cone's height AB, sides are in ratio = 1:2
now applying above concept, volume of top cone to volume of actual cone = 1 : k^3 = 1 : 8

top cone is one part of 8 parts, remaining volume is 7 parts, => required ratio = volume of top cone : remaining volume = 1:7
Re: If the volume of the cone is SH/3 where S is the area of   [#permalink] 25 Jan 2018, 11:57
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