There is a cone.The slant height of cone is 6√3 and radius is 3√3. There is a sphere inscribed in the cone(touching its base and curved surface). Find the radius of sphere.
+128460 We can solve this through the method of cross-section
See the following image :
6sqrt (3) = sqrt (108) and 3sqrt (3) = sqrt (27)
Note that triangle ABC is right......and we can find AB as sqrt ( 108 - 27) = sqrt (81) = 9
Draw BE perpendicular to AC
And triangles CEB and CBA are similar....so
BC/ AC = BE / AB
3sqrt (3) / 6sqrt (3) = BE / 9
3/ 6 = BE / 9
1/2 = BE/ 9
BE = 9/2 = 4.5
And triangle AFG is similar to triangle ABE
So
AF / FG = AB / BE
9 - R 9
____ = ______ cross-multiply
R 4.5
4.5 (9 - R) = 9R
40.5 - 4.5R = 9R
40.5 = 13.5R
R = 3 = the radius of the sphere
