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
Hello, xiukhung!
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 $\displaystyle 2r.$
It is $\displaystyle r$ units above the table.
The center of the fifth sphere forms a square-base pyramid.
All of its edges have length $\displaystyle 2r.$
We find that the height of this pyramid is $\displaystyle \sqrt{2}\,r.$
The top of the fifth sphere is $\displaystyle r$ units above the pyramid.
Hence, the top of that sphere is: .$\displaystyle r + \sqrt{2}\,r + r \:=\:\left(2 + \sqrt{2}\right)r$ units high.
Answer: .$\displaystyle 10\left(2+\sqrt{2}\right)\text{ units.}$
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:
$\displaystyle \sin(30^\circ) = \frac12 = \frac8x$
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.