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Help would be greatly appreciated, I can't visualize the problem and I don't understand the problem very well. An explanation to how the problem was solved would be very helpful, thank you.

 

A sphere with radius 3 is inscribed in a conical frustum of slant height 10. (The sphere is tangent to both bases and the side of the frustum.) Find the volume of the frustum.

Guest Nov 6, 2018
edited by Guest  Nov 6, 2018
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There may be a much easier way to do this, but we can find the volume of the frustum by creating a cross-sectional kite .......the sphere will  have a circular cross-section and the frustum will have a cross-sectional trapezoid 

 

 

First......since the sphere is tangent to the frustum at both the top and bottom bases, the frustum height  is  =  6

 

So.....we can create a 6-8-10  Pythagorean right triangle DGF  as shown in the illustration - with one leg of 6 - (GF)  - one leg of 8 - (DG) -  and the slant height - (DF)  -forming the hypotenuse........the cosine of  angle FDG  ( at the bottom right vertex of the frustum ) =   8/10   =  4/5

 

Now......the radii of the circle will form kite  ABCD  

 

Connect AC......angle ABC will be supplementary to angle FDG......so its cosine  = -4/5

And because DA and DC are tangents drawn from the same point D, they are equal

And we can use the Law of Cosines to find AC

Note: AB  and BC are radii of the circle

 

AC^2 =  AB^2 + BC^2  - 2(AB)(BC)(cos ABC)

AC^2  =  3^2 + 3^2  - 2(3)(3) (-4/5)

AC^2 = 18  + 72/5

AC^2  =  [ 90 + 72] / 5

AC^2  =  [ 162 / 5 ]

 

 

And Using the Law of Cosines once more, we can find the lenths of DA and DC.....let their lengths = x......and we have

 

AC^2  =   x^2 + x^2  -  2 (x^2)cos(CDA)

[ 162 / 5 ]  =  2x^2 - 2x^2 [ 4/5 ]

[162/ 5 ]  = 2x^2 [ 1 - 4/5]

[ 162 / 5 ] = 2x^2 [ 1/5]

162 = 2x^2

x^2  = 81

x = 9  =   DA  =  DC

 

But DA  =  radius  of the bottom base of the frustum

And since DC = 9.....then CF  = FD - DC  =  10 - 9   =  1

But CF = FE since they are tangents drawn from the point F

And FE  =  radius of the top base of the frustum

 

So.....using the fomula for the volume of a frustum, we have

 

V  = pi * GF * ( DA^2 + DA*FE^2 + FE^2 ) / 3  

 

Where 

GF  = frustum height

DA =  radius of bottom base

FE = radius of top base

 

So we have

 

 

pi * ( 6) *  ( 9^2  + 9 * 1 + 1^2 ) / 3  =

 

pi (6)  ( 81 + 9 + 1 ) / 3    =

 

pi (6) (91) / 3  =

 

2  * 91 pi  =

 

182 pi  units^3

 

 

cool cool cool

CPhill  Nov 6, 2018
edited by CPhill  Nov 6, 2018
edited by CPhill  Nov 6, 2018

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