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A right circular cone is inscribed in a hemisphere so that the base of  [#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) $$2:1$$

Diagnostic Test
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.
Updated.
Math Expert V
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Posts: 55618
A right circular cone is inscribed in a hemisphere so that the base of  [#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. Attachment: 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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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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Math Expert V
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A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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

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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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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 V
Joined: 02 Sep 2009
Posts: 55618
Re: A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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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 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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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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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 10080 times ]

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Avinash Verma
Math Expert V
Joined: 02 Sep 2009
Posts: 55618
Re: A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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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 the base of  [#permalink]

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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

Math Expert V
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Posts: 55618
Re: A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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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

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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Posts: 3
Re: A right circular cone is inscribed in a hemisphere so that the base of  [#permalink]

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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

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.

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

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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.

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

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