Circles with centers $P, Q$ and $R$, having radii $1, 3$ and $5$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQR$?

1. For two circles of radii $r_1,r_2$ that are externally tangent and both tangent to a line, the distance between the points of tangency on the line is $2\sqrt{r_1r_2}$. Hence $P'Q'=2\sqrt{1\cdot3}=2\sqrt3$ and $Q'R'=2\sqrt{3\cdot5}=2\sqrt{15}$.
2. Using the trapezoids $PP'Q'Q$, $QQ'R'R$ and $PP'R'R$ (all with parallel sides perpendicular to $l$): $$[PQR]=[PP'Q'Q]+[QQ'R'R]-[PP'R'R].$$
3. $[PP'Q'Q]=\frac{1+3}2\cdot2\sqrt3=4\sqrt3$, $[QQ'R'R]=\frac{3+5}2\cdot2\sqrt{15}=8\sqrt{15}$, and $[PP'R'R]=\frac{1+5}2\left(2\sqrt3+2\sqrt{15}\right)=6\sqrt3+6\sqrt{15}$.
4. Therefore $[PQR]=4\sqrt3+8\sqrt{15}-6\sqrt3-6\sqrt{15}=\boxed{2\sqrt{15}-2\sqrt3}\approx4.28$.