Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $26$. Circle $\Omega$ is tangent to $\overline{AB}$ at $P$ and meets $\Gamma$ at $Q,R$. If $QR=7\sqrt2$ and $\angle QPR=45^\circ$, the area of $\triangle PQR$ is $\frac ab$ in lowest terms. What is $a+b$?

1. By the inscribed angle theorem in $\Omega$, chord $QR=7\sqrt2=2r\sin45^\circ$, so $\Omega$ has radius $r=7$. Let $O$ be the midpoint of $AB$ (the center of $\Gamma$, radius $13$) and let $D$ be the midpoint of $QR$. Then $OD=\sqrt{13^2-\left(\frac{7\sqrt2}2\right)^2}=\frac{17}{\sqrt2}$, and the center $C$ of $\Omega$ has distance $\sqrt{7^2-\left(\frac{7\sqrt2}2\right)^2}=\frac7{\sqrt2}$ from $QR$, so $OC=\frac{17}{\sqrt2}-\frac7{\sqrt2}=5\sqrt2$. Since $CP\perp AB$ and $CP=7$, $OP=\sqrt{OC^2-CP^2}=\sqrt{50-49}=1$.
2. Put $O=(0,0)$ with $AB$ on the $x$-axis, $P=(1,0)$ and $C=(1,7)$. The circles are $x^2+y^2=169$ and $(x-1)^2+(y-7)^2=49$; subtracting gives $x+7y=85$. Substituting, $25y^2-595y+3528=0$, so $y=\frac{63}5$ or $y=\frac{56}5$, giving $Q=\left(-\frac{16}5,\frac{63}5\right)$ and $R=\left(\frac{33}5,\frac{56}5\right)$ (one checks $QR=7\sqrt2$). Then $PQ^2=\frac{882}5$, $PR^2=\frac{784}5$, so $PQ\cdot PR=\frac{588\sqrt2}5$.
3. $[PQR]=\frac12RP\cdot PQ\sin45^\circ=\frac{294}5$; $a+b=294+5=\boxed{299}$.