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

$\begin{array}{l}\left(A\right)\cos {30}^{\circ }=\frac{P{Q}^2+Q{R}^2-P{R}^2}{2PQ\cdot QR}\\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{(10\sqrt{3}{)}^2+{10}^2-P{R}^2}{2\times 10\sqrt{3}\times 10}\\ \Rightarrow P{R}^2=100\text{ or }PR=10\\ \therefore \mathrm{\angle }P=\mathrm{\angle }Q={30}^{\circ }\end{array}$

$\therefore \left(A\right)\text{ is false }$

$\text{ (B) Area of }\mathrm{△}PQR=\frac{1}{2}PQ\times QR\times \sin {30}^{\circ }$

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

$\mathrm{\angle }R={180}^{\circ }-{30}^{\circ }-{30}^{\circ }={120}^{\circ }$

$\therefore \left(B\right)\text{ is true }$

$\begin{array}{l}\text{ (C) }r=\frac{\mathrm{\Delta }}{s}=\frac{25\sqrt{3}}{\left(\frac{10\sqrt{3}+10+10}{2}\right)}=\frac{25\sqrt{3}}{10+5\sqrt{3}}=\frac{5\sqrt{3}}{2+\sqrt{3}}\\ =5\sqrt{3}(2-\sqrt{3})=10\sqrt{3}-15\end{array}$

$\therefore \left(\mathrm{C}\right)\text{ is true. }$

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

$\therefore \text{ area of circumcircle }=\pi R^2=100\pi$

$\therefore \left(D\right)\text{ is true }$