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Math Help - please can anybody give the height of the pile

  1. #1
    rcs
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    please can anybody give the height of the pile

    kindly please help on this

    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
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  2. #2
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    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.
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  3. #3
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    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.
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  4. #4
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    Quote Originally Posted by rcs View Post
    kindly please help on this

    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.
    Attached Thumbnails Attached Thumbnails please can anybody give the height of the pile-tetraeder.png  
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  5. #5
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    Hello
    The height of the pile is 1.5rad 3 +3. Obtained using 30-60-90 triangle relationships


    bjh
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  6. #6
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    height of pile

    Hello.

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


    bjh
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  7. #7
    rcs
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    i still dont know what and how to solve it... still super confused how to get the answer
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  8. #8
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    Quote Originally Posted by rcs View Post
    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?
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  9. #9
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    Quote Originally Posted by rcs View Post
    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)
    and finally add on the radius of the top sphere,
    you will have the perpendicular height of the "nucleus".
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