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The height of the solid cone above is 18 inches and the radius of the

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The height of the solid cone above is 18 inches and the radius of the  [#permalink]

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New post 06 Dec 2017, 22:01
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The height of the solid cone above is 18 inches and the radius of the base is 8 inches. A cut parallel to the circular base is made completely through the cone so that one of the two resulting solids is a smaller cone. If the radius of the base of the small cone is 2 inches, what is the height of the small cone, in inches?

(A) 2.5
(B) 4
(C) 4.5
(D) 9.0
(E) 12

Attachment:
2017-12-07_0955_003.png
2017-12-07_0955_003.png [ 10.85 KiB | Viewed 525 times ]

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Re: The height of the solid cone above is 18 inches and the radius of the  [#permalink]

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New post 07 Dec 2017, 06:10
Both the cones are similar triangles. So the ratio of the sides should be equal.
18/16 = x/4 => x=4.5
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Re: The height of the solid cone above is 18 inches and the radius of the  [#permalink]

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New post 07 Dec 2017, 10:45
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Bunuel wrote:
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The height of the solid cone above is 18 inches and the radius of the base is 8 inches. A cut parallel to the circular base is made completely through the cone so that one of the two resulting solids is a smaller cone. If the radius of the base of the small cone is 2 inches, what is the height of the small cone, in inches?

(A) 2.5
(B) 4
(C) 4.5
(D) 9.0
(E) 12

Attachment:
The attachment 2017-12-07_0955_003.png is no longer available

Attachment:
ppp.png
ppp.png [ 16.99 KiB | Viewed 343 times ]

When a cone is cut horizontally, the original and the smaller cone contain similar triangles.

In the diagram, ∆ ABC and ∆ EFG, both one-dimensional "slices" of their respective cones, are right triangles (the height is perpendicular to the radius).

Both right triangles share the same angle at the top of the cone (changing the size of the circular base does not affect the tip of the cone).

Both share the same angle where radius meets the cone's side (that slope does not change).

By similar triangle properties, the original cone's ratio of height to radius, in inches, equals the ratio of the new cone's height to radius, in inches:

\(\frac{H}{R} = \frac{h}{r}\)

\(\frac{AB}{AC} =\frac{EG}{EF}\)

\(\frac{18}{8} = \frac{h}{2}\)

\(36 = 8h\)

\(9 = 2h\)

\(h = \frac{9}{2} =\)


\(4.5\) inches

Answer C
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Re: The height of the solid cone above is 18 inches and the radius of the &nbs [#permalink] 07 Dec 2017, 10:45
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