In a triangle PQR, let∠PQR = 30 ^ o and the sides PQandQR have...

$cos {30}^{o} = \dfrac{{\left(10\sqrt{3}\right)}^{2} + {\left(10\right)}^{2} - {\left(PR\right)}^{2}}{2 \times 10\sqrt{3} \times 10}$
$⇒ PR = 10$
$∵ QR = PR ⇒ ∠PQR = ∠QPR$
$∠QPR = {30}^{o}$
\left(B\right)$Area of$∆ PQR = \dfrac{1}{2} \times 10\sqrt{3} \times 10 \times sin {30}^{o} = \dfrac{1}{2} \times 10 \times 10\sqrt{3} \times \dfrac{1}{2}$<br >$= 25\sqrt{3}$<br >$∠QRP = {180}^{o} - \left({30}^{o} + {30}^{o}\right) = {120}^{o}$<br >\left(C\right) r = \dfrac{∆}{S} = \dfrac{25\sqrt{3}}{\left(\dfrac{10 + 10 + 10\sqrt{3}}{2}\right)} = \dfrac{25\sqrt{3}}{10 + 5\sqrt{3}}$
$= 5\sqrt{3} . \left(2 - \sqrt{3}\right) = 10\sqrt{3} - 15$
\left(D\right) R = \dfrac{a}{2sin A} = \dfrac{10}{2sin {30}^{o}} = 10$<br >$∴ Area = \pi{R}^{2} = 100\pi