Question 36023
        A sphere is inscribed in a cone with radius 6 and height 8. Find the volume of the sphere.
DRAW A SECTION...TRIANGLE ABC REPRESENTS THE SECTION OF THE CONE,WITH A AS VERTEX AND BC AS BASE.HENCE BC =6+6=12
DRAW AD PERPENDICULAR FROM A TO BC.AD IS HEIGHT OF CONE =8
DRAW A CIRCLE WITH CENTRE AT O ON AD 

{{{drawing(600,600,-5,15,-5,15,green( line( 0,0, 12,0 ) ),green(line(6,0,6,8)),locate( 0, 0, B ),locate(6,0,D),green( line( 12,0,6,8 ) ),locate(6,3,O),locate(6,8,A),locate(8.5,4.5,E),locate(3.5,4.5,F),green(line(6,3,8.5,4.5)),locate(12,0,C),circle(6,3,3),green(line(0,0,6,8)))}}}

AS SECTION OF SPHERE TOUCHING BC,CA,AB AT D,E,F.
HENCE OD=OE=OF=RADIUS OF SPHERE=R SAY
NOW TRIANGLES ADC,AND AEO ARE RIGHT ANGLED
ANGLE ADC=90=ANGLE AEO...AS AEC IS TANGENT..SPHERE TOUCHING CONE.
ANGLE DAC=ANGLE OAE=SAME ANGLE
HENCE THE 2 TRIANGLES ADC AND AEO ARE SIMILAR.
   HENCE DC/EO=AD/AE
DC=6.....AD=8......EO=R=RADIUS OF SPHERE.....AE=SQRT(AO^2-OE^2)
=SQRT{(AD-OD)^2-R^2}
=SQRT{(8-R)^2-R^2}=SQRT{64+R^2-16R-R^2}
=SQRT(64-16R)
HENCE WE HAVE 
6/R=8/SQRT(64-16R)
6SQRT(64-16R)=8R
6*4SQRT(4-R)=8R
3SQRT(4-R)=R....SQUARING....
9(4-R)=R^2
R^2+9R-36=0
(R-3)(R+12)=0
R=3