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A right circular cone is inscribed in a hemisphere so that the base of

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New post Updated on: 24 Mar 2019, 02:48
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A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

(A) \(\sqrt{3}:1\)

(B) \(1:1\)

(C) \(\frac{1}{2}:1\)

(D) \(\sqrt{2}:1\)

(E) \(2:1\)

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Question: 20
Page: 23
Difficulty: 600

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Originally posted by Bunuel on 03 Apr 2012, 16:49.
Last edited by Bunuel on 24 Mar 2019, 02:48, edited 2 times in total.
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A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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New post 09 Jul 2012, 04:41
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SOLUTION

A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

(A) \(\sqrt{3}:1\)

(B) \(1:1\)

(C) \(\frac{1}{2}:1\)

(D) \(\sqrt{2}:1\)

(E) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.
Image

Answer: B.

Attachment:
Cone.png
Cone.png [ 12.65 KiB | Viewed 3221 times ]

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Re: A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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New post 23 Nov 2012, 02:10
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Not sure if anyone else was scratching their heads wondering why the height couldn't be less than the radius, but just in case... it has to do with the word "inscribed": to draw within a figure so as to touch in as many places as possible <a regular polygon inscribed in a circle>

lol.
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New post 09 Jul 2012, 04:42
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A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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New post 08 Jun 2013, 22:36
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It's the combination of "right circular cone" and "inscribed". The cone shares a base with the hemisphere and it's tip touches the top of the hemisphere. If you had a right circular cone of a height to radius ratio of anything other than 1:1 you would not be able to inscribe it in a hemisphere. Essentially, what you've described is impossible for this particular problem.

Also, you cannot have a hemisphere of the type you've described. It wouldn't be a hemisphere. A sphere's radius is the same throughout. A hemisphere is half of a sphere (so it's height must equal it's radius.
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Re: A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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New post 06 May 2014, 12:57
Bunuel wrote:
SOLUTION

A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

(A) \(\sqrt{3}:1\)

(B) \(1:1\)

(C) \(\frac{1}{2}:1\)

(D) \(\sqrt{2}:1\)

(E) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.
Attachment:
Cone.png


Answer: B.

Hi Bunuel, how did you conclude the underlined in above statement? Thanks!
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Re: A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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New post 07 May 2014, 04:25
Dienekes wrote:
Bunuel wrote:
SOLUTION

A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

(A) \(\sqrt{3}:1\)

(B) \(1:1\)

(C) \(\frac{1}{2}:1\)

(D) \(\sqrt{2}:1\)

(E) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.
Attachment:
The attachment Cone.png is no longer available


Answer: B.

Hi Bunuel, how did you conclude the underlined in above statement? Thanks!


Consider the cross-section. We'd have an isosceles triangle inscribed in a semi-circle:
Attachment:
Untitled.png
Untitled.png [ 2.44 KiB | Viewed 18155 times ]

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A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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New post 19 Jul 2016, 11:30
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height and radius are equal ...follow fig.

ans B
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Re: A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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New post 20 Apr 2017, 19:04
Bunuel wrote:
Dienekes wrote:
Bunuel wrote:
SOLUTION

A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

(A) \(\sqrt{3}:1\)

(B) \(1:1\)

(C) \(\frac{1}{2}:1\)

(D) \(\sqrt{2}:1\)

(E) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.
Attachment:
The attachment Cone.png is no longer available


Answer: B.

Hi Bunuel, how did you conclude the underlined in above statement? Thanks!


Consider the cross-section. We'd have an isosceles triangle inscribed in a semi-circle:
Attachment:
The attachment Untitled.png is no longer available


Why cone can't be like the image below?
Attachment:
File comment: cone in hemisphere
cone in hemisphere.png
cone in hemisphere.png [ 36.85 KiB | Viewed 10080 times ]

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Re: A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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New post 20 Apr 2017, 21:28
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Re: A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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New post 07 Jul 2017, 06:06
Bunuel

It makes sense that the height of the cone will be equal to the radius of the hemisphere.

However, when I tried equating the volumes for both the cone and hemisphere, the solution I am getting is 2 : 1

1\3 x pie x r^2 x h = 1/2 x 4/3 x pie x r^3

h = 2r


Am I missing something here. Please help
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Re: A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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New post 07 Jul 2017, 06:23
kotharishakti wrote:
Bunuel

It makes sense that the height of the cone will be equal to the radius of the hemisphere.

However, when I tried equating the volumes for both the cone and hemisphere, the solution I am getting is 2 : 1

1\3 x pie x r^2 x h = 1/2 x 4/3 x pie x r^3

h = 2r


Am I missing something here. Please help


Why are you equating volumes? The volume of a cone is obviously less than the volume of the hemisphere it is inscribed in.
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Re: A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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New post 07 Jul 2017, 07:20
Bunuel wrote:
kotharishakti wrote:
Bunuel

It makes sense that the height of the cone will be equal to the radius of the hemisphere.

However, when I tried equating the volumes for both the cone and hemisphere, the solution I am getting is 2 : 1

1\3 x pie x r^2 x h = 1/2 x 4/3 x pie x r^3

h = 2r


Am I missing something here. Please help


Why are you equating volumes? The volume of a cone is obviously less than the volume of the hemisphere it is inscribed in.




Oh yes!!
That's Correct Bunuel.

I am doing it wrong! Such a silly one.

Thanks a lot for your reply!
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Re: A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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New post 27 Apr 2018, 10:03
NvrEvrGvUp wrote:
A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

A. \(\sqrt{3} : 1\)

B. \(1 : 1\)

C. \(\frac{1}{2} : 1\)

D. \(\sqrt{2} : 1\)

E. \(2 : 1\)


If a right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere, then the height of the cone is exactly the radius of the hemisphere. So the ratio is 1:1.

Answer: B
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