In a $\mathrm{△}PQR,\text{ let }\mathrm{\angle }PQR={30}^{\circ }$ and the sides PQ and QR have lengths $10\sqrt{3}\text{ and }10$, respectively. Then, which of the following statement(s) is (are) 'TRUE?

We have,

In $\mathrm{△}PQR$

                          $\begin{array}{l}\mathrm{\angle }PQR={30}^{\circ }\\ PQ=10\sqrt{3}\\ QR=10\end{array}$

By cosine rule 

$\begin{array}{l}\cos {30}^{\circ }=\frac{P{Q}^2+Q{R}^2-P{R}^2}{2PQ\cdot QR}\\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{300+100-P{R}^2}{200\sqrt{3}}\\ \Rightarrow 300=300+100-P{R}^2\\ \Rightarrow PR=10\end{array}$

Since, $PR=QR=10$

$\therefore \mathrm{\angle }QPR={30}^{\circ }$

and   $\mathrm{\angle }QRP={120}^{\circ }$

are of $\mathrm{△}PQR=\frac{1}{2}PQ\cdot QR\cdot \sin {30}^{\circ }$

$=\frac{1}{2}\times 10\sqrt{3}\times 10\times \frac{1}{2}=25\sqrt{3}$

Radius of incircle of 

$\mathrm{\Delta }PQR=\frac{\text{ Area of }\mathrm{△}PQR}{\text{ Semi-perimetre of }\mathrm{\Delta }PQR}$

i.e.  $r=\frac{\mathrm{\Delta }}{s}=\frac{25\sqrt{3}}{\frac{10\sqrt{3}+10+10}{2}}$

$\begin{array}{l}\Rightarrow =\frac{25\sqrt{3}}{5(\sqrt{3}+2)}\\ r=5\sqrt{3}(2-\sqrt{3})\\ =10\sqrt{3}-15\end{array}$

and radius of circumcircle 

$\left(R\right)=\frac{abc}{4\mathrm{\Delta }}=\frac{10\sqrt{3}\times 10\times 10}{4\times 25\sqrt{3}}=10$

Area of circumcircle of 

$\mathrm{\Delta }PQR=\pi R^2=100\pi$

Hence, option (b), (c) and (d) are correct answer.