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
∴ \angle QPR=30^{0} and \angle QRP=120^{0}
Area of ΔPQR = \frac{1}{2} PQ.QR \sin 30^{0}
= \frac{1}{2}\times10\sqrt{3}\times 10\times\frac{1}{2}=25\sqrt{3}
Radius of incircle of
Δ PQR= \frac{Area of \triangle PQR }{Semi-perimetre of \triangle PQR}
i.e r=\frac{\triangle }{s}= \frac{25\sqrt{3}}{\frac{10\sqrt{3}+10+10}{2}}=\frac{25\sqrt{3}}{5(\sqrt{3}+2)}
\Rightarrow r= 5\sqrt{3}(2-\sqrt{3})=10\sqrt{3}-15
and radius of the circumcircle
(R)=\frac{abc}{4\triangle}= \frac{10\sqrt{3}\times10\times10}{4\times25\sqrt{3}}=10
\therefore Area of circumcircle of Δ PQR= \pi R^{2}=100\pi
Hence, option (b), (c) and (d) are correct answer.