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A right circular cone is inscribed in a hemisphere so that

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

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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
[Reveal] Spoiler: OA

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

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

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New post 09 Jul 2012, 05:28
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Bunuel 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) \(2:1\)

Hi,

Difficulty level: 600

As per below diagram,
Attachment:
ch.jpg
ch.jpg [ 6.18 KiB | Viewed 9840 times ]


Answer (B),

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

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New post 13 Jul 2012, 02:11
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
Cone.png [ 23.74 KiB | Viewed 10493 times ]


Answer: B.
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Re: A right circular cone is inscribed in a hemisphere so that [#permalink]

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New post 25 Sep 2013, 15:09
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Re: A right circular cone is inscribed in a hemisphere so that [#permalink]

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New post 06 May 2014, 11:57
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 [#permalink]

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New post 07 May 2014, 03:25
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 7122 times ]

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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 08 May 2014, 12:29
Height is same as the radius of the hemisphere. No calculations needed, I guess.
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Re: A right circular cone is inscribed in a hemisphere so that [#permalink]

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New post 22 Jul 2015, 19:12
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Re: A right circular cone is inscribed in a hemisphere so that [#permalink]

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New post 11 May 2016, 12:32
Hemisphere is just half of a sphere
if you can visualize the figure , you can easily deduce that height of cone will be equal to radius of hemisphere
So the ratio would be 1:1
correct answer - B
Re: A right circular cone is inscribed in a hemisphere so that   [#permalink] 11 May 2016, 12:32
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