In adjacent figure three circles each of radius 1 unit inscribed in an equilateral triangle then area of triangle in sq.units.

Given that three circles each of radius 1 inscribed in an equilateral triangle. Here ${\mathrm{C}}_2\mathrm{D}=1$

Side of the equilateral triangle

 $\begin{array}{rcl}\mathrm{BC}& =& \mathrm{BD}+DE+\mathrm{EC}\\ & =& \mathrm{BD}+2\mathrm{r}+\mathrm{BD} (\because \mathrm{DE}=2\mathrm{r})\\ & =& 2\mathrm{BD}+2\mathrm{r}\\ & =& 2.{\mathrm{C}}_2\mathrm{D} cot{30}^0+2\mathrm{r} \left(\because \mathrm{from} {∆}^{\mathrm{le}} {\mathrm{BC}}_2\mathrm{D}, \tan {30}^0=\frac{{\mathrm{C}}_2\mathrm{D}}{\mathrm{BD}}\right)\\ & =& 2\mathrm{r}\left(\sqrt{3}\right)+2\mathrm{r}\\ & =& 2\sqrt{3} \left(1\right)+2\left(1\right)\\ \mathrm{BC}& =& 2\left(\sqrt{3}+1\right)\end{array}$

Now Area of $\begin{array}{rcl}{∆}^{\mathrm{le}} \mathrm{ABC}& =& \frac{\sqrt{3}}{4} {\mathrm{BC}}^2\end{array}$

                                         $\begin{array}{rcl} & =& \frac{\sqrt{3}}{4} {\left(2\left(\sqrt{3}+1\right)\right)}^2\\ & =& \frac{\sqrt{3}}{4} \left(4\right) (3+1+2\sqrt{3})\\ & =& \sqrt{3} (4+2\sqrt{3})\\ & =& 6+4\sqrt{3}\end{array}$