A rectangle is inscribed in an equilateral triangle of side length 2a units. The maximum area of this rectangle can be

Let ${\mathrm{BD}}_1=\mathrm{x}\text{ or }{\mathrm{BC}}_1=(\mathrm{a}-\mathrm{x})$
$\therefore \mathrm{BC}=(\mathrm{a}-\mathrm{x})\tan \frac{\mathrm{\pi }}{3}=\sqrt{3}(\mathrm{a}-\mathrm{x})$
Now, area of rectangle ABCD,
$\begin{array}{l}\mathrm{\Delta }=\left(\mathrm{AB}\right)\left(\mathrm{BC}\right)=2\sqrt{3}\mathrm{x}(\mathrm{a}-\mathrm{x})\\ \left.\le 2\sqrt{3}{\left(\frac{\mathrm{x}+\mathrm{a}-\mathrm{x}}{2}\right)}^2=\frac{\sqrt{3}{\mathrm{a}}^2}{2}\text{ (using A.M. }\ge \text{ G.M. }\right)\end{array}$