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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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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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Re: A right circular cone is inscribed in a hemisphere so that [#permalink]
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09 Jul 2012, 06:28
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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 [ 6.18 KiB  Viewed 11418 times ]
Answer (B), Regards,



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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
SOLUTIONA 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 12250 times ]
Answer: B.
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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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Re: A right circular cone is inscribed in a hemisphere so that [#permalink]
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06 May 2014, 12:57
Bunuel wrote: SOLUTIONA 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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07 May 2014, 04:25
Dienekes wrote: Bunuel wrote: SOLUTIONA 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 crosssection. We'd have an isosceles triangle inscribed in a semicircle: Attachment:
Untitled.png [ 2.44 KiB  Viewed 8698 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
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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



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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: SOLUTIONA 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 crosssection. We'd have an isosceles triangle inscribed in a semicircle: 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 624 times ]
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Re: A right circular cone is inscribed in a hemisphere so that [#permalink]
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