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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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09 Jul 2012, 04:41
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Difficulty:

15% (low)

Question Stats:

68% (01:48) correct 32% (01:03) wrong based on 761 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 11201 times ]

Regards,
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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
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Hello from the GMAT Club BumpBot!

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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 12003 times ]

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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!
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Kudos [?]: 106204 [0], given: 11609

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 8481 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
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Re: A right circular cone is inscribed in a hemisphere so that [#permalink]

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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

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 [ 36.85 KiB | Viewed 406 times ]
Math Expert
Joined: 02 Sep 2009
Posts: 38897
Followers: 7738

Kudos [?]: 106204 [0], given: 11609

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

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20 Apr 2017, 21:28
avinashvpec wrote:
Why cone can't be like the image below?
Attachment:
cone in hemisphere.png

We are told that the cone is "inscribed" in a hemisphere, not just inside, so the vertex must be on the circumference.
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Re: A right circular cone is inscribed in a hemisphere so that   [#permalink] 20 Apr 2017, 21:28
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