An equilateral triangle is circumscribed and a square is inscribed in a circle of radius $\text{'}r\text{'}$. The area of the triangle is $T$ and the area of the square is $S$. Then $\frac{T}{S}=$

Given $\Delta ABC$ is an equilateral triangle

$AB=BC=CA$

Let $BD$ is the median

Centroid divides the median ratio is $2:1$

So $BF=FO=OD$

From square $DEFG;FD=2r$

$EF=\sqrt{2}.r$

Area of the square $\left(S\right)=\sqrt{2}.r\times \sqrt{2}.r$

$S=2r^2\cdots \cdots \left(1\right)$

From the fig. $OP=r,BO=2r$

$\begin{array}{l}\Delta BPO\\ B{O}^2=O{P}^2+B{P}^2\\ {\left(2r\right)}^2=r^2+B{P}^2\\ 3r^2=B{P}^2\\ BP=\sqrt{3}.r\end{array}$

From the fig $BP=CP=CD=DA=AQ=QB(\because \text{tangents})$

$\text{So }BC=2\sqrt{3}.r$

Area of the equilateral triangle  $=\frac{\sqrt{3}}{4}.a^2$

$\begin{array}{l}=\frac{\sqrt{3}}{4}{\left(2\sqrt{3}\right)}^2\\ =\frac{\sqrt{3}}{4}\mathrm{.4.3.}{r}^2\\ T=3\sqrt{3}.r^2\cdots \cdots \left(2\right)\\ \text{Then\hspace{0.17em}\hspace{0.17em}}\frac{T}{S}=\frac{3\sqrt{3}{r}^2}{2r^2}\\ \frac{T}{S}=\frac{3\sqrt{3}}{2}\end{array}$