A right circular cone is inscribed in a hemisphere so that the base of

Question

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

Bunuel · Accepted Answer

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.

Answer: B.

Show Spoiler

Attachment:

Cone.png [ 12.65 KiB | Viewed 68525 times ]

Bunuel · Answer

Check other  3-D Geometry Questions  from our  Special Questions Directory .

essarr · Answer

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.

lol.

nsspaz151 · Answer

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.

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.

Dienekes · Answer

Hi Bunuel, how did you conclude the underlined in above statement? Thanks!

Bunuel · Answer

Consider the cross-section. We'd have an isosceles triangle inscribed in a semi-circle: Untitled.png

rohit8865 · Answer

height and radius are equal ...follow fig.

ans B

avinashvpec · Answer

Why cone can't be like the image below? cone in hemisphere.png

Bunuel · Answer

We are told that the cone is "inscribed" in a hemisphere, not just inside, so the vertex must be on the circumference.

kotharishakti · Answer

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

Am I missing something here. Please help

Bunuel · Answer

Why are you equating volumes? The volume of a cone is obviously less than the volume of the hemisphere it is inscribed in.

kotharishakti · Answer

Oh yes!!
That's Correct Bunuel.

I am doing it wrong! Such a silly one.

Thanks a lot for your reply!

ScottTargetTestPrep · Answer

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.

Answer: B

dcummins · Answer

This is what stumped me. I thought the base of the cone needed to only touch the base of the sphere, not extend to the edges of the sphere. So in my situation my height was the radius.

If we are told that the bases "coincide" do we presume they align equally?

jabhatta2 · Answer

Bunuel

Out of curiosity -->

Is the vertex angle (the one at the very top) -- can i assume that angle is a 90 degree angle ?

Per my understanding I can because the triangle is inscribed and the vertex angle thus HAS to be 90 degrees given the base of the triangle is the diameter of the sphere

Please let me know if my understanding is accurate ?

Thank you !

jabhatta2 · Answer

Experts -  chetan2u ,  Bunuel ,  generis ,  MahmoudFawzy

Just wondering if you could reply to the above post right above this one ?

jabhatta2 · Answer

Hi  Bunuel

How can one be sure the right angled cone isn't like this per the picture i have drawn ?

Angle BAC is 90 degree per my picture

If any issues with the picture, you should be able to zoom in

Thank you!

Bunuel · Answer

That's not a right circular cone. A right circular cone is the one where the vertex is above the base center point..

gmatchile · Answer

Having present hemisphere is half of a sphere.

If we inscribe a circular cone in a hemisphere, the base of the cone must coincide with the base of the hemisphere, both are identical circumferences.

Thus the radius of the base of the circular cone is the same radius as the base of the hemisphere.

Since the hemisphere is half that of a sphere, the height of the sphere coincides with the height of the cone.

The height of the sphere is the radius of the base of the hemisphere.

Therefore, the radius of the sphere is equal to the height of the cone with a circular base.

Answer B

A right circular cone is inscribed in a hemisphere so that the base of

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A right circular cone is inscribed in a hemisphere so that the base of

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards