Question

$\cos {30}^o=\frac{{\left(10\sqrt{3}\right)}^2+{\left(10\right)}^2-{\left(PR\right)}^2}{2\times 10\sqrt{3}\times 10}$                     
$\Rightarrow$PR=10
$\therefore$QR = PR    $$\begin{gathered} \Rightarrow \angle PQR=\angle QPR \\ \angle QPR={30}^o \end{gathered}$$
(B) area of $∆$PQR= $$\begin{gathered} \frac{1}{2}\times 10\sqrt{3}\times 10\times \mathrm{si}n{30}^o=\frac{{\displaystyle 1}}{{\displaystyle 2}}\times 10\times 10\sqrt{3}\times \frac{{\displaystyle 1}}{{\displaystyle 2}} \\ =25\sqrt{3} \\ \angle QPR={180}^o-({30}^o+{30}^o)={120}^o \end{gathered}$$
QRP = 180° – (30° + 30°) = 120°
$$\begin{gathered} \left(c\right) r=\frac{∆}{s}=\frac{25\sqrt{3}}{\left({\displaystyle \frac{10+10+10\sqrt{3}}{2}}\right)}=\frac{25\sqrt{3}}{10+5\sqrt{3}} \\ =5\sqrt{3}.(2-\sqrt{3})=10\sqrt{3}-15 \\ \left(D\right)R=\frac{a}{2 \sin A}=\frac{10}{2\sin {30}^o}=10 \\ \therefore Area={\mathrm{\pi R}}^2=100\mathrm{\pi } \end{gathered}$$