+743 I keep trying this problem but i'm not getting anywhere
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?
+129845 See the following :
We can find the inradius of the triangle shown
The triangle is isosceles with equal sides of sqrt (3^2 + 3^2) = sqrt (18) = 3sqrt (2) and the remaining side of 6
Area of triangle = (1/2) (6) (3) = 9
Semiperimeter = ( 2 * 3 sqrt 2 + 6) / 2 = 3sqrt (2) + 3
Inradius = Area / semi-perimeter = 9 / ( 3sqrt (2) + 3) = 3 / ( sqrt (2) + 1) = radius of sphere
Volume of sphere / Volume of cone = (4/3) pi * ( 3 / [sqrt (2) + 1])^3 / [ (1/3) pi (3^2) * 3 ] =
4 / [ sqrt (2) + 1 ] ^3 ≈ .284
