A sphere of radius $4$ inches is inscribed in a cone with a base of radius $8$ inches. In inches, what is the height of the cone? Express your answer as a decimal to the nearest tenth.

Answer 1

Consider a cross section of the cone, which is a triangle with height $h$ and base $16$ inches. Let $x$ represent the height of the triangle dropped from the apex of the cone to the base of the cone, as shown.

[asy]
unitsize(0.5 cm);

pair A, B, C, D, F;

A = (0,0);
B = (16,0);
C = (8,20);
D = (8,0);
F = (8,8);

draw(A--C--B--cycle);
draw(F--D);

label("$h$", (D + C)/2, NW);
label("$8$", (D + F)/2, N);
label("$8$", (F + B)/2, NE);
label("$4$", (D + A)/2, S);
[/asy]

By the Pythagorean Theorem, $(8 - x)^2 + 4^2 = h^2.$ Expanding, we get
\[64 - 16x + x^2 + 16 = h^2.\]Since $\triangle DAF$ is similar to $\triangle DFC,$ we have that
\[\frac{h - 4}{8} = \frac{h}{8 - x}.\]Cross-multiplying gives $h(8 - x) = 8(h - 4),$ so $8h - hx = 8h - 32,$ or $hx = 32.$

Substituting into the equation $64 - 16x + x^2 + 16 = h^2,$ we get
\[x^2 - 16x + 48 = 0.\]This factors as $(x - 4)(x - 12) = 0,$ so $x = 4$ or $x = 12.$ We can check that $h = \frac{32}{x}$ has a solution for $x = 12,$ so the answer is $\boxed{12}.$