# please can anybody give the height of the pile

• Dec 19th 2010, 05:59 AM
rcs
please can anybody give the height of the pile

Four spheres, each of radius 1.5, are placed in a pile with three at
the base and the other on top. If each sphere touches the other three
spheres, give the height of the pile.

thanks
• Dec 19th 2010, 06:10 AM
Prove It
If you were to join each of the centres of the spheres, you get an equilateral triangular pyramid.

Each of the edges of the pyramid has length $\displaystyle = 2\times \textrm{radius} = 3\,\textrm{units}$.

Start by drawing the pyramid and see if you can use trigonometry to find the height of the pyramid.
• Dec 21st 2010, 03:20 AM
HallsofIvy
Nice! Of course, that gives the height from the center of a bottom sphere to the center of the top sphere. You will need to add twice the radius to find the actual height of the pile.
• Dec 22nd 2010, 10:27 PM
earboth
Quote:

Originally Posted by rcs

Four spheres, each of radius 1.5, are placed in a pile with three at
the base and the other on top. If each sphere touches the other three
spheres, give the height of the pile.

thanks

... only in case you don't want to use trigonometry:

1. Draw a sketch of the tetrahedron with the centers of the spheres as vertices as ProveIt suggested.

2. The footpoint of the height of this tetrahedron equals the centroid of the equilateral base triangle.
Calculate the length of a median using Pythagorean theorem:

$\left(\frac12 \cdot (2r)\right)^2+k^2=(2r)^2$

3. Use Pythagorean theorem.
One edge of the tetrahedron has the length 2r. Then the height h of the tetrahedron is:

$\left(\dfrac{2r}3 \sqrt{3}\right)^2+h^2=(2r)^2$

4. To get the final result have a look at HallsofIvy's post.
• Dec 23rd 2010, 07:49 PM
bjhopper
Hello
The height of the pile is 1.5rad 3 +3. Obtained using 30-60-90 triangle relationships

bjh
• Dec 24th 2010, 04:40 AM
bjhopper
height of pile
Hello.

My answer was wrong.I see why after looking more closely at Earboth's solution

bjh
• Jan 23rd 2011, 12:53 AM
rcs
i still dont know what and how to solve it... still super confused how to get the answer
• Jan 23rd 2011, 01:22 AM
Prove It
Quote:

Originally Posted by rcs
i still dont know what and how to solve it... still super confused how to get the answer

Have you at least drawn the tetrahedron?
• Jan 23rd 2011, 12:28 PM
Quote:

Originally Posted by rcs
i still dont know what and how to solve it... still super confused how to get the answer

Place 3 of the 4 spheres on a flat surface, so that they all make contact.

Lines from centre to centre form an equilateral triangle with side lengths 2(1.5)=3 units.

First, be clear on the above.

Next, place the 4th sphere on top of the other 3.
It will slide into place.

Taking the top sphere and any 2 from the 3 at the base,
what shape is formed between their centres ?

Again, only an equilateral triangle can be formed because the spheres are in contact.
Hence, 3 equilateral triangles are meeting at the centre of the top sphere,
each apex being the centre of each sphere.

The shape formed between the centres is constructed from 4 equilateral triangles
of sides 3 units in length.

If you find the perpendicular height of this 3-sided pyramid,
then add it to the radius of the base spheres (since the base of the "pyramid" is 1.5 units off the floor)