Question

A sphere is inscribed in a cone with height $3$ and base radius $4$. What is the ratio of the volume of the sphere to the volume of the cone?

Answer 1

Consider the cross section of the cone and sphere. The diagram below shows the cross section when we cut the cone and sphere vertically.

[asy]
real h=3;
draw(Circle((0,0), 4));
draw((-sqrt(7),1.2)--(sqrt(7),1.2));
draw((sqrt(7),1.2)--(0,7*h/8));
draw((0,7*h/8)--(-sqrt(7),1.2));
draw((0,0)--(0,7*h/8));
label("$4$", (sqrt(7),1.2), E);
label("$4$", (-sqrt(7),1.2), W);
label("$3$", (0,7*h/8), N);
[/asy]

Let $x$ be half the length of a side of the equilateral triangle formed by the circle's cross section. Since the altitude of an equilateral triangle divides the base into two $30^\circ$-$60^\circ$-$90^\circ$ right triangles, $x = 4 \tan 60^\circ = 4 \sqrt{3}.$ Hence, the radius of the circle is $2x = 8 \sqrt{3},$ and the volume of the sphere is
\[\frac{4}{3} \pi (8 \sqrt{3})^3 = \frac{4 \pi \cdot 8^3 \cdot 3^2}{3^4} = 256 \pi.\]The volume of the cone is
\[\frac{1}{3} \cdot \pi \cdot 4^2 \cdot 3 = 16 \pi.\]Therefore, the ratio of the volume of the sphere to the volume of the cone is
\[\frac{256 \pi}{16 \pi} = \boxed{16}.\]