If a rectangle is inscribed in an equilateral triangle of side...

In $\delta DBF,$

$\tan 60 ° = \dfrac{2b}{2\sqrt{2} - ℓ}$

$⇒ b = \dfrac{\sqrt{3}\left(2\sqrt{2} - ℓ\right)}{2}$

$A =$Area of rectangle

$= ℓ \times b$

$⇒ A = ℓ \times \dfrac{\sqrt{3}}{2}\left(2\sqrt{2} - ℓ\right)$

On differentiating, we get

$\dfrac{dA}{dℓ} = \dfrac{\sqrt{3}}{2}\left(2\sqrt{2} - ℓ\right) - \dfrac{ℓ \cdot \sqrt{3}}{2} = 0$

$⇒ ℓ = \sqrt{2}$

Now, $A = ℓ \times b$

$= \sqrt{2} \times \dfrac{\sqrt{3}}{2}\left(\sqrt{2}\right) = \sqrt{3}$

$⇒ {A}^{2} = 3 .$