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# The volume of a cone is S × H/3, where S is the area of the base and H

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Joined: 02 Sep 2009
Posts: 52285
The volume of a cone is S × H/3, where S is the area of the base and H  [#permalink]

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29 Jul 2015, 02:27
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The volume of a cone is S × H/3, where S is the area of the base and H is the height. According to the figure above, if AB = BC, what is the ratio of the volume of the right section of the cone (with AB as height) to the volume of the left section of the cone (with BC as height)? (Note: Figure not drawn to scale.)

A. 1 : 3
B. 1 : 4
C. 1 : 6
D. 1 : 7
E. 1 : 8

Kudos for a correct solution.

Attachment:

cone-1.gif [ 7 KiB | Viewed 1924 times ]

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The volume of a cone is S × H/3, where S is the area of the base and H  [#permalink]

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Updated on: 29 Jul 2015, 05:09
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Bunuel wrote:

The volume of a cone is S × H/3, where S is the area of the base and H is the height. According to the figure above, if AB = BC, what is the ratio of the volume of the right section of the cone (with AB as height) to the volume of the left section of the cone (with BC as height)? (Note: Figure not drawn to scale.)

A. 1 : 3
B. 1 : 4
C. 1 : 6
D. 1 : 7
E. 1 : 8

Kudos for a correct solution.

Attachment:
cone-1.gif

Ratio of Height of smaller and bigger cone = 1/2

CONCEPT: Ratio of Volume of similar solids = (Ratio of Height)^3

So Ratio of Volume of similar solids = (Ratio of Height)^3 = (1/2)^3 = 1/8

i.e. Volume of Right Section (Smaller Cone) = 1
then, Volume of Bigger Cone (including Smaller Cone) = 8
i.e. Volume of Left Section (Frustum) = Volume of Bigger Cone - Volume of Smaller Cone = 8-1 = 7
i.e. volume of Left Section = 8-1 =7

i.e. Volume of Right Section / Volume of Right Section = 1 / 7

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Originally posted by GMATinsight on 29 Jul 2015, 03:47.
Last edited by GMATinsight on 29 Jul 2015, 05:09, edited 1 time in total.
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Re: The volume of a cone is S × H/3, where S is the area of the base and H  [#permalink]

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29 Jul 2015, 03:51
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Bunuel wrote:

The volume of a cone is S × H/3, where S is the area of the base and H is the height. According to the figure above, if AB = BC, what is the ratio of the volume of the right section of the cone (with AB as height) to the volume of the left section of the cone (with BC as height)? (Note: Figure not drawn to scale.)

A. 1 : 3
B. 1 : 4
C. 1 : 6
D. 1 : 7
E. 1 : 8

Kudos for a correct solution.

Let AB = BC = h. Then AC = 2h
Let the line extended from B meet at D and from C meet at E as shown in fig below

Attachment:

cone-1.gif [ 6.52 KiB | Viewed 1748 times ]

Triangle ABD and Triangle ACE are similar triangles ( ∠BAD = ∠ CAE , ∠B = ∠C =90)
Thus $$\frac{AB}{AC}$$ = $$\frac{1}{2}$$
And $$\frac{BD}{CE}$$ = $$\frac{1}{2}$$ ==> $$\frac{x}{R}$$ = $$\frac{1}{2}$$
R = 2x

volume of the right section of the cone (with AB as height) = π$$r^2$$ * $$\frac{AB}{3}$$
= π$$x^2$$ * $$\frac{h}{3}$$

volume of the left section of the cone (with BC as height) = volume of the left section of the cone (with AC as height) - volume of the right section of the cone (with AB as height)

volume of the left section of the cone (with AC as height) = π$$r^2$$ * $$\frac{AC}{3}$$
=π$$R^2$$ * $$\frac{2h}{3}$$
=π$$2x^2$$ * $$\frac{2h}{3}$$
=π$$x^2$$ * $$\frac{8h}{3}$$

Thus volume of the left section of the cone (with BC as height) = π$$x^2$$ * $$\frac{8h}{3}$$ - π$$x^2$$ * $$\frac{h}{3}$$
= π$$x^2$$ * $$\frac{7h}{3}$$

Thus Ratio = 1:7
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29 Jul 2015, 04:09
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Here's my solution

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math solution.PNG [ 17.8 KiB | Viewed 1732 times ]

Math Expert
Joined: 02 Sep 2009
Posts: 52285
Re: The volume of a cone is S × H/3, where S is the area of the base and H  [#permalink]

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17 Aug 2015, 08:44
Bunuel wrote:

The volume of a cone is S × H/3, where S is the area of the base and H is the height. According to the figure above, if AB = BC, what is the ratio of the volume of the right section of the cone (with AB as height) to the volume of the left section of the cone (with BC as height)? (Note: Figure not drawn to scale.)

A. 1 : 3
B. 1 : 4
C. 1 : 6
D. 1 : 7
E. 1 : 8

Kudos for a correct solution.

Attachment:
cone-1.gif

800score Official Solution:

1) We can create two similar triangles using AC and AB as the heights, R and x as the bases, and the edge of the cone as the hypotenuse. These triangles are similar because the angles for both are the same. The sides of similar triangles are proportional to each other.

2) Because AB = BC, and AC = AB + BC, then 2(AB) = AC.

3) Since the triangles are similar, and because 2(AB) = AC, 2x = R.

4) The volume of the left half of the cone equals the volume of the entire cone minus the volume of the right half of the cone, so the ratio we are looking for is:
(volume of right half) : (volume of cone – volume of right half).

5) Volume of the entire cone = S × H/3
= (πR²) × (AC)/3
= (π(2x)²) × (2AB)/3
= 8πx²AB/3

6) Volume of the right half of the cone = S × H/3
= πx²AB/3

7) Plug these values into the ratio equation above and solve:
πx²AB/3 : (8πx²AB/3 – πx²AB/3)
πx²AB/3 : 7πx²AB/3
1 : 7.

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Re: The volume of a cone is S × H/3, where S is the area of the base and H  [#permalink]

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07 Nov 2015, 11:51
if AB is equal to 1/2H then x must be R/2
let's find the volume of the small cone with the base on the B
we know that radius of the small circle is R/2 of the big, and height is 1/2 of the H.
volume must be (R/2)^2 * pi * H/3 * 1/2 = R^2/4 * pi * H * 1/6 or (R^2 * pi * H) / 24
this is the volume of the small cone.

the volume of the cone from B to C is volume of the big A-C cone minus the small cone A->B.
volume of the big cone is R^2*pi * H/3 or (8 * R^2 * pi * H) / 24
subtract from the whole, the small one
(8 * R^2 * pi * H) / 24 minus
(R^2 * pi * H) / 24

we get (7 * R^2 * pi * H) / 24 - volume of the cone B-C

volume of the A->B to the B-C is then:
(R^2 * pi * H) / 24
divided everything by
(7 * R^2 * pi * H) / 24

or (R^2 * pi * H) * 24 / 24 * (7 * R^2 * pi * H)
simplify by 24, R^2, pi, and H, and remain with 1/7.

tough and tricky one!!!
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Re: The volume of a cone is S × H/3, where S is the area of the base and H  [#permalink]

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16 Apr 2018, 10:38
One General idea i have,
whenever, we have two similiar triangles with sides in ratio 1:x ,
then ratio of the areas of the triangles will be of ratio 1 : x^2

now extending the concept to 3-D dimension, if there are two similiar cones of side ratio 1:x
then ratio of the volumes of cones will be of ratio 1:x^3

now we have small right cone and parent cone, small cone height/parent cone height = 1:2
so volume of small right cone/volume of parent cone = 1:8

so volume of remaining part of cone/volume of parent cone = 7:8
so volume of small right cone/volume of remaining part of cone = 1:7
Re: The volume of a cone is S × H/3, where S is the area of the base and H &nbs [#permalink] 16 Apr 2018, 10:38
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