A circle with radius $6$ is externally tangent to a circle with radius $24$. Find the area of the triangular region bounded by the three common tangent lines of these two circles.
$r_1 = O_1A = 24$, $r_2 = O_2B = 6$, $AG = BO_2 = r_2 = 6$, $O_1G = r_1 - r_2 = 24 - 6 = 18$, $O_1O_2 = r_1 + r_2 = 30$
$\triangle O_2BD \sim \triangle O_1GO_2$, $\frac{O_2D}{O_1O_2} = \frac{BO_2}{GO_1}$,  $\frac{O_2D}{30} = \frac{6}{18}$, $O_2D = 10$
$CD = O_2D + r_2 = 10 + 6 = 16$,
$EF = 2EC = EA + EB = AB = GO_2 = \sqrt{(O_1O_2)^2-(O_1G)^2} = \sqrt{30^2-18^2} = 24$
$DEF = \frac12 \cdot EF \cdot CD = \frac12 \cdot 24 \cdot 16 = \boxed{\textbf{192}}$
~[isabelchen](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Isabelchen)

Alternative Finish
Note that $\triangle{O_1 O_2 G} \sim \triangle{O_1 D A}$ by AA similarity. Setting up the ratio $\frac{18}{24}=\frac{24}{AB+8}$ and then substituting $AB$ with our known values will lead us to the same solution.
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