A sphere is inscribed in a cone with height $3$ and base radius $3$. What is the ratio of the volume of the sphere to the volume of the cone?
The formula for the volume of a sphere is given by:
Volume of Sphere = (4/3)πr^3
= (4/3)π(3^3) = 36π cubic units.
The formula for the volume of a cone is given by:
Volume of Cone = (1/3)πr^2h
= (1/3)π(3^2)(3) = 27π cubic units
Ratio = (Volume of Sphere) / (Volume of Cone)
= (36π) / (27π) = 4/3
Through the use of similar triangles, we can find the radius, r, of the sphere thusly :
Let one right triangle CDB be composed of two legs.....DB = the base radius of the cone = 3 and the other DC = the height of the cone = 3
A simiar second right riangle CFE can be formed (see the illustration) with legs EF = r and FC = r
The hypotenuse of this triangle = CE = sqrt [ r^2 + (r^2) ] = r sqrt 2
And we can find the radius of the sphere thusly :
CD = 3
ED + CE = 3
r + r sqrt 2 = 3
r ( 1 + sqrt 2) = 3
r = 3 / ( 1 + sqrt 2)
Volume of sphere to volume of the cone =
(4/3) pi ( 3 / (1+sqrt 2) )^3 4
_____________________ = ____________ ≈ .284
(1/3) pi ( 3)^2 * 3 (1 + sqrt 2)^3