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.
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 $\boxed{12}.$
The height, $h$, of the cone can be found by using the Pythagorean Theorem. We can consider one of the radii of the base as the hypotenuse of a right triangle, with the height of the cone as one leg, and the distance from the center of the base to the edge of the sphere as the other leg. The distance from the center of the base to the edge of the sphere is half the diameter of the sphere, or $4$ inches. Since the radius of the base is $8$ inches, then this distance is also the height.
[asy]
unitsize(0.4 cm);
draw((0,0)--(24,0)--(8*sqrt(3),24)--cycle);
draw((12,sqrt(48))--(12,0));
label("$8$", (0,0)--(8*sqrt(3),24), NW);
label("$4$", (8*sqrt(3)/2,24/2)--(8*sqrt(3),24)/2, NE);
label("$h$", (0,0)--(8*sqrt(3)/2,24/2), W);
label("$4$", (12,sqrt(48))/2,(0,-1.5));
dot((12,sqrt(48))/2);
[/asy]
By the Pythagorean Theorem, we have that
\[h^2 + 4^2 = 8^2.\]Solving, we find $h = \sqrt{8^2 - 4^2} = \sqrt{64 - 16} = \sqrt{48} \approx \boxed{6.9}$ inches.