2016 AMC 12A Problem 15

The centers lie at heights $1,$ $2,$ and $3$ above line $l.$ Since circle $Q$ is externally tangent to circle $P,$ we have $PQ=3,$ so the horizontal distance is $P'Q'=\sqrt{3^2-1^2}=\sqrt{8}.$ Since circle $Q$ is tangent to circle $R,$ we have $QR=5,$ so $Q'R'=\sqrt{5^2-1^2}=\sqrt{24}.$ Place $P=(0,1),$ $Q=(\sqrt8,2),$ and $R=(\sqrt8+\sqrt{24},3).$ By the shoelace formula, the area is $$\begin{gathered} \small \dfrac12\left|\sqrt8(3-1)+(\sqrt8+\sqrt{24})(1-2)\right|\\ =\dfrac12\left(\sqrt{24}-\sqrt8\right)\\ =\sqrt6-\sqrt2. \end{gathered}$$ Thus, the correct answer is D.