What is the sum of the lengths of the $\textbf{altitudes}$ of a triangle whose side lengths are $10,$ $10,$ and $12$? Express your answer as a decimal to the nearest tenth.
Let us draw our triangle and altitudes and label our points of interest: [asy]
pair A, B, C, D, E, F;
A = (0, 8);
B = (-6, 0);
C = (6, 0);
D = (0, 0);
E = foot(B,A,C);
F = foot(C,A,B);
draw(A--B--C--cycle);
draw(A--D);
draw(B--E);
draw(C--F);
draw(rightanglemark(B,E,C,10));
draw(rightanglemark(C,F,B,10));
draw(rightanglemark(A,D,C,10));
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
label("$F$", F, NW);
[/asy] We have made $AB = AC = 10$ and $BC = 12.$ We can notice a few useful things. Since $ABC$ is isosceles, it follows that $AD$ is a median as well as an altitude, which is useful since that means $BD = DC = \frac{1}{2} \cdot BC = 6.$ Now, since $DC = 6$ and $AC = 10,$ we have a $3:4:5$ Pythagorean triple and $AD = 8$. Now we can find the area of $ABC$ by finding $\frac{1}{2} \cdot AD \cdot BC = \frac{1}{2} \cdot 8 \cdot 12 = 48.$

Now, we can find $BE$ by using the area we just found: $\frac{1}{2} \cdot AC \cdot BE = 48.$ Since $AC = 10,$ that means $BE = 9.6$ By symmetry, $CF$ is also $9.6.$ Our answer is: $9.6 + 9.6 + 8 = \boxed{27.2}.$