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(D) Let the centers of the circles be $C_1, C_2, C_3$ with radii $r_1=3, r_2=3, r_3=1$. The sides of the triangle $C_1C_2C_3$ are $C_1C_2 = 3+3=6$,$C_

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(D) Let the centers of the circles be $C_1, C_2, C_3$ with radii $r_1=3, r_2=3, r_3=1$. The sides of the triangle $C_1C_2C_3$ are $C_1C_2 = 3+3=6$,$C_1C_3 = 3+1=4$,and $C_2C_3 = 3+1=4$.This is an isosceles triangle. Let $M$ be the midpoint of $C_1C_2$. The height $h = C_3M = \sqrt{4^2 - 3^2} = \sqrt{7}$.Points $A$ and $B$ lie on $C_1C_3$ and $C_2C_3$ respectively. By similar triangles,the side length $AB$ of triangle $ABC$ is $\frac{AB}{C_1C_2} = \frac{C_3A}{C_3C_1} = \frac{4-3}{4} = \frac{1}{4}$. Thus,$AB = \frac{1}{4} 	imes 6 = \frac{3}{2}$.The height of triangle $ABC$ from $C$ to $AB$ is $h' = h - (	ext{distance of } C 	ext{ from } AB) = \sqrt{7} - \frac{3}{4}\sqrt{7} = \frac{\sqrt{7}}{4}$.Area of $	riangle ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{3}{2} 	imes \frac{\sqrt{7}}{4} = \frac{3\sqrt{7}}{16}$.

Three circles lie on a plane such that each of them externally touches the other two. Two of them have radius $3$,and the third has radius $1$. If $A, B$,and $C$ are the points of tangency of the circles,then the area of the triangle $ABC$ is

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