In triangle pqr, m is the midpoint of pq. Let x be the point on qr such that px bisects angle qpr and let the perpendicular bisector of pq

Let's start by drawing a diagram:

 

We see that $\triangle PQR$ is a right triangle with $\angle PQR = 90^\circ$ and $PR$ as the hypotenuse. Therefore, $PR^2 = PQ^2 + QR^2$, or $22^2 = 36^2 + 26^2$, so $PR = \sqrt{36^2 + 26^2} = 10\sqrt{17}$.

Since $M$ is the midpoint of $PQ$, $PM = QM = 18$, so $PX = 18 - MY = 10$. Therefore, $\triangle PQX$ is a $9$-$10$-$17$ right triangle, so $\angle QPX = \arctan \frac{9}{10}$ and $\angle RPX = \arctan \frac{17}{10}$.

Since $PX$ bisects $\angle QPR$, we have $\angle QPA = \angle APR$, so $\triangle PQA \cong \triangle APR$ by ASA. Therefore, $PA = PR = 10\sqrt{17}$.

Since $Y$ lies on the perpendicular bisector of $PQ$, we have $PY = QY = 18$, so $AY = AP - PY = 10\sqrt{17} - 18$.

Since $\triangle APX$ and $\triangle AQY$ are similar, we have $\frac{AY}{AX} = \frac{QY}{PX}$, or $\frac{10\sqrt{17} - 18}{AX} = \frac{18}{10}$. Solving for $AX$, we get $AX = \frac{9(10\sqrt{17} - 18)}{5} = 18\sqrt{17} - 36$.

Therefore, the altitude from $P$ to $YR$ has length $MY + AY = 8 + (10\sqrt{17} - 18) = 10\sqrt{17} - 10$. Thus, the area of $\triangle PYR$ is $\frac{1}{2} \cdot PQ \cdot PR = \frac{1}{2} \cdot 36 \cdot 10\sqrt{17} = 180 \sqrt{17}$.