# Sphere in Cone Geometry

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• Mar 15th 2009, 10:15 PM
anonymous_maths
Sphere in Cone Geometry
http://i667.photobucket.com/albums/v...conesphere.jpg

That's a question on my textbook.... i can't solve it and my teacher is away. Anyone help?
Thanks.
• Mar 16th 2009, 01:05 AM
earboth
Quote:

Originally Posted by anonymous_maths
...
That's a question on my textbook.... i can't solve it and my teacher is away. Anyone help?
Thanks.

1. The cross-section of the cone is an equilateral triangle with AM = R.
In the top right triangle you get:

$R \cdot \tan(30^\circ) = r~\implies~R=\dfrac r{\tan(30^\circ)} = \dfrac r{\frac13 \sqrt{3}}~\implies~\boxed{R=r \cdot \sqrt{3}}$

2. Let denote x the distance in question. Use Pythagorean theorem:

$R^2+r^2=(r+x)^2~\implies~3r^2+r^2 = (r+x)^2~\implies~2r=r+x~\implies~x=r$

Unfortunately I have to leave now ...
• Mar 16th 2009, 08:08 AM
earboth
Quote:

Originally Posted by anonymous_maths
...
That's a question on my textbook.... i can't solve it and my teacher is away. Anyone help?
Thanks.

Let h denote the height of the cone. Then:

$h^2+R^2=(2R^2)~\implies~h=R\cdot \sqrt{3}$

The volume of a cone is calculated by:

$V=\dfrac13 \cdot \pi \cdot R^2 \cdot h$

That means:

$V=\dfrac13 \cdot \pi \cdot R^2 \cdot R \cdot \sqrt{3} = \dfrac{\pi \sqrt{3}}3 R^3$
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From a) you know that
$R=r \cdot \sqrt{3}~\implies~r=\dfrac{\sqrt{3}}3 R$

The volume of the sphere is calculated by:

$V=\dfrac43 \pi r^3$

That means:

$V=\dfrac43 \pi \left(\dfrac{\sqrt{3}}3 R\right)^3 = \dfrac{4 \pi \sqrt{3}}{27} R^3$
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The percentage is calculated by $percentage = \dfrac{partial\ amount}{total\ amount} \cdot 100$

$p=\dfrac{\dfrac{4\pi \sqrt{3}}{27} R^3}{\dfrac{\pi \sqrt{3}}3 R^3}\cdot 100 = \dfrac49 \cdot 100 \approx 44.4\%$