In a non right triangle $PQR$, the median from $R$ meets the sides $PQ$ at $S$, the perpendicular from $P$ meets sides $QR$ at $E$ and $RS$ and $PE$ intersect at $O$. $p=\sqrt 3, q=1$ and circumradius of $PQR$ is $1$.
I don’t want the complete answer for this. In the solution, there was point which mentioned area of $\Delta OQR =\frac 13 \Delta PQR$
Why is that the case?
If you apply the sine rule on triangle $PQR$, which states that $$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$$ where $a,b,c$ are sides of the triangle opposite angles $A,B,C$ and $R$ is the circumradius, you get $$\dfrac{\sqrt{3}}{\sin P}=\dfrac{1}{\sin Q}=\dfrac{r}{\sin R}=2$$ so that $P=120^{\circ}$ and $Q=30^{\circ}$ which implies $R=30^{\circ}$ which implies $r=1$. Now, I hope you can continue from here.
Note that I have assumed that $p=QR$ and $q=PR$ which is not mentioned in the question.
Please note that $PE$ is a median and hence $PE = 3OE$. So $\triangle PQR = 3 \times \triangle OQR$.
The question states that $\triangle PQR$ is not a right angled triangle otherwise $\angle R = 90^0$ is another obvious configuration for the given sides and circumradius.
As $p = \sqrt3 \gt q = 1, \angle Q \, $ is acute. That leads to only other possible configuration of $\angle P \,$ being obtuse.
If $C$ is the circumcenter, you can see why $\angle P = 120^0 \, PQ = PR$ and $E$ is the midpoint of $QR$.
Let $a=1$ be the circumradius and, per the sine rule $$ \sin Q=\frac q{2a}=\frac12,\>\>\>\>\> \sin P=\frac p{2a}=\frac{\sqrt3}2$$ Given the non-right triangle, $Q=30^\circ$ and $P=120^\circ$. Thus, the triangle $PRQ$ is isosceles with $PE$ as a median as well. So, $O$ is the centroid point, hence the area ratio $\frac13$.
