A triangle of maximum area is inscribed in a circle. If a side of the triangle is  $20\sqrt{3}$   then the radius of the circle is

$$\begin{gathered} \mathrm{PS}\perp \mathrm{QR}, From ∆OSQ \\ {\left(OQ\right)}^2={\left(QS\right)}^2+{\left(OS\right)}^2 \\ {\left(QS\right)}^2={\left(OQ\right)}^2-{\left(OS\right)}^2 \\ \mathrm{QS}=\sqrt{{\mathrm{r}}^2-{\mathrm{x}}^2} \end{gathered}$$
 

$$\begin{gathered} \text{ Area of }\mathrm{\Delta PQR} \mathrm{is}=\frac{1}{2}QR\times PS \\ A=\frac{1}{2}\times 2\times \sqrt{r^2-x^2}(r+x) \end{gathered}$$
$\begin{array}{l}\frac{\mathrm{dA}}{\mathrm{dx}}=\sqrt{{\mathrm{r}}^2-{\mathrm{x}}^2}\left(1\right)+(\mathrm{r}+\mathrm{x})\cdot \frac{-2\mathrm{x}}{2\sqrt{{\mathrm{r}}^2-{\mathrm{x}}^2}}\Rightarrow \frac{dA}{dx}=\frac{{\mathrm{r}}^2-{\mathrm{x}}^2-\mathrm{rx}-{\mathrm{x}}^2}{\sqrt{{\mathrm{r}}^2-{\mathrm{x}}^2}}\\ For max or min \frac{dA}{dx}=0 then\\ \Rightarrow (\mathrm{r}+\mathrm{x})(\mathrm{r}-2\mathrm{x})=0\\ \mathrm{x}=-\mathrm{r} \mathrm{and} \mathrm{x}=\frac{\mathrm{r}}{2}\\ Maximum at \mathrm{x}=\frac{\mathrm{r}}{2}\\ \frac{{\mathrm{d}}^2\mathrm{A}}{{\mathrm{dx}}^2}=\frac{(-4\mathrm{x}-\mathrm{r})\left(\sqrt{{\mathrm{r}}^2-{\mathrm{x}}^2}\right)}{\left({\mathrm{r}}^2-{\mathrm{x}}^2\right)}-\left({\mathrm{r}}^2-\mathrm{rx}-2{\mathrm{x}}^2\right)\cdot \frac{-2\mathrm{x}}{2\sqrt{{\mathrm{r}}^2-{\mathrm{x}}^2}}<0\end{array}$
$\begin{array}{l}Given that length of a side of triangle is 20\sqrt{3} so, \\ \mathrm{QR}=20\sqrt{3},\mathrm{QS}=10\sqrt{3}\\ From the triangle ∆OSQ \\ {\mathrm{r}}^2={\mathrm{x}}^2+300\Rightarrow {\mathrm{r}}^2=\frac{{\mathrm{r}}^2}{4}+300\\ \frac{3{\mathrm{r}}^2}{4}=300\Rightarrow {\mathrm{r}}^2=400\Rightarrow \mathrm{r}=20\end{array}$