Answer:
Option B,C,D
Explanation:
In Δ PQR
∠PQR=300
PQ = 10√3 , QR=10

By cosine rule
cos300=PQ2+QR2−PR22PQ.QR
⇒√32=300+100−PR2200√3
⇒ 300=300+100-PR2
⇒ PR=10
Since , PR=QR= 10
∴ ∠QPR=300 and ∠QRP=1200
Area of ΔPQR = 12 PQ.QR sin300
= 12×10√3×10×12=25√3
Radius of incircle of
Δ PQR= Areaof△PQRSemi−perimetreof△PQR
i.e r=△s=25√310√3+10+102=25√35(√3+2)
⇒ r= 5√3(2−√3)=10√3−15
and radius of the circumcircle
(R)=abc4△=10√3×10×104×25√3=10
∴ Area of circumcircle of Δ PQR= πR2=100π
Hence, option (b), (c) and (d) are correct answer.