$\Gamma$ Circle

$Chord~ XY~is~||~PQ~since~RX/RP=RY/RQ=2/3.\\ \therefore~XY=20.\\ \implies~~\bot~distance~between~XY~and~PQ~=\sqrt{15^2-10^2}=5\sqrt5.\\ So~altitude~from~R~to~PQ=3*5\sqrt5.\\ Area~\Delta~RPQ=\frac 1 2*30*15\sqrt5=225\sqrt5=a\sqrt b.\\ a+b=\large~~\color{#D61F06}{230}.$

First, we show that $PR = QR.$ By the problem condition, we have $RX = \dfrac{2}{3}PR$ and $RY = \dfrac{2}{3} RQ.$ Applying Power of a Point to $R$ yields

$\begin{aligned} RX(RP) &= RY(RQ) \\ \dfrac{2}{3}RP^2 &= \dfrac{2}{3}RQ^2 \\ RP &= RQ. \end{aligned}$

Thus, $\triangle PQR$ is isosceles. Let $PX = QY = a,$ so that $XR = YR = 2a.$ Draw $XQ,$ and note that $\angle PXQ = 90^{\circ},$ since it subtends a semicircle arc. By applying Pythagorean Theorem on right triangle $RXQ,$ we can find $XQ$ in terms of $a$ :

$\begin{aligned} RX^2 + XQ^2 &= QR^2 \\ (2x)^2 + XQ^2 &= (3x)^2 \\ XQ &= x \sqrt{5}. \end{aligned}$

Doing the same for $\triangle PXQ$ gives us a value for $a^2$ :

$\begin{aligned} PX^2 + XQ^2 &= PQ^2 \\ a^2 + (a \sqrt{5})^2 &= 30^2 \\ 6a^2 &= 900 \\ a^2 &= 150. \end{aligned}$

Therefore, the area of $\triangle PQR$ is

$\dfrac{(PR)(XQ)}{2} = \dfrac{3a^2 \sqrt{5}}{2} = \dfrac{3(150) \sqrt{5}}{2} = 225 \sqrt{5},$

and $m + n = 225 + 5 = \boxed{230}.$

notice by symmetry that $\Delta PQR$ is iscoceles. let O be the center of the circle, then $OR=\sqrt{(3x)^2-15^2}=3\sqrt{x^2-5^2}$ is the altitude. applying power of a point to R, $2x(3x)=(3\sqrt{x^2-5^2}-15)(3\sqrt{x^2-5^2}+15)\to 6x^2=9x^2-225-225 \\ \to x=\sqrt{150}$ we get the area $[\Delta PQR]=15\times 3\sqrt{x^2-5^2}=225\sqrt{5}$