In a non-right angled triangle ΔPQR  let p, q, r denote the lengths of the sides opposite to the angles at P, Q, R respectively. The median from R meets the side PQ at S, the perpendicular from P meets the side QR at E, and RS and PE intersect at O. If p=3,q=1  and the radius of the circumcircle of the ΔPQR  equals 1, then which of the following options is/are correct?

From sine rule ,3sinp=2⇒sinp=32 so ¯¯P=60°,120° again  from sine rule1sinQ=2⇒sinQ=12 so Q=30°,150° Clearly P=60° not possible.if P=60 ,Q=30 then R =90 ,not possible   So ¯¯P=120°,Q=30°,R=30°PQ=PR=1,QE=RE=32 O must be centroid so OE=13·PE=131-34=16 option 1 is correct  Area of ΔPQR ∆=12×3×12=34,  Semiperimeter s=(3+1+1)/2=3+22 So in radius of ΔPQR  r=343+22=3(2-3)2 option 2 is correct   since r=∆s∣RS=PR2+QR2-2PS22 =1+3-2×142=72 option 4 is correct  Area of ΔSOE=13. Area of ΔSPE =13×14 area of ΔPQR=112×12×3×12=348 option 3 is incorrect