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

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

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09 Jul 2012, 04:41
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Difficulty:

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Question Stats:

69% (01:47) correct 31% (01:02) wrong based on 697 sessions

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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) $$2:1$$

Diagnostic Test
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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09 Jul 2012, 04:42
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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.

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

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09 Jul 2012, 06:28
1
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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 [ 6.18 KiB | Viewed 10583 times ]

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

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13 Jul 2012, 03:11
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 [ 23.74 KiB | Viewed 11309 times ]

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

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25 Sep 2013, 16:09
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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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

Hi Bunuel, how did you conclude the underlined in above statement? Thanks!
Math Expert
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Posts: 37560
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Kudos [?]: 99291 [0], given: 11010

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

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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

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 [ 2.44 KiB | Viewed 7865 times ]

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

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08 May 2014, 13:29
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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22 Jul 2015, 20:12
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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11 May 2016, 13:32
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, 13:32
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# A right circular cone is inscribed in a hemisphere so that

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