1 sphere on top of 4 spheres

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Apr 7th 2010, 07:47 PM
xiukhung

1 sphere on top of 4 spheres

hi, I have this question.
4 spheres of radius 10 are placed on a table so that each touches 2 others. another sphere is placed on top. Find the height of the top of this fifth sphere above the table.
thanks :)
Apr 7th 2010, 09:15 PM
Wilmer

Tangent Spheres -- from Wolfram MathWorld
Apr 8th 2010, 01:41 AM
xiukhung

Quote:

Originally Posted by Wilmer

Tangent Spheres -- from Wolfram MathWorld

hi, i've read it through but I dont know how this would help. thanks
Apr 8th 2010, 04:56 AM
Soroban

Hello, xiukhung!

Quote:

Four spheres of radius 10 are placed on a table so that each touches 2 others.
Another sphere is placed on top.
Find the height of the top of this fifth sphere above the table.

The centers of the four spheres form a square with side $2r.$
It is $r$ units above the table.

The center of the fifth sphere forms a square-base pyramid.
All of its edges have length $2r.$

We find that the height of this pyramid is $\sqrt{2}\,r.$

The top of the fifth sphere is $r$ units above the pyramid.

Hence, the top of that sphere is: . $r + \sqrt{2}\,r + r \:=\:\left(2 + \sqrt{2}\right)r$ units high.

Answer: . $10\left(2+\sqrt{2}\right)\text{ units.}$
Apr 8th 2010, 03:52 PM
xiukhung

Thanks a lot, what about this problem?
A sphere of radius length 8cm rests on top of a hollow inverted cone of height 15 cm whose vertical angle is 60 degrees. Find the height of the centre of the sphere above the vertex of the cone.
Apr 9th 2010, 01:17 AM
earboth

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Quote:

Originally Posted by xiukhung

Thanks a lot, what about this problem?
A sphere of radius length 8cm rests on top of a hollow inverted cone of height 15 cm whose vertical angle is 60 degrees. Find the height of the centre of the sphere above the vertex of the cone.

I'm not certain if I understand your question correctly.

If the attached image describes the situation completely then you'll get x by using the Sine function:

$\sin(30^\circ) = \frac12 = \frac8x$

Solve for x.

By the way: If you have a new question use a new thread. Otherwise you risk that no member of the forum will realize that you need further assistance.