If a rectangle is inscribed in an equilateral | vedclass

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(C) Let the side length of the equilateral triangle be $a = 2 \sqrt{2}$.Let the length of the rectangle be $\ell$ and its breadth be $b$.From the simi

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$3$

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(C) Let the side length of the equilateral triangle be $a = 2 \sqrt{2}$.Let the length of the rectangle be $\ell$ and its breadth be $b$.From the similar triangles formed at the top,the ratio of the height of the small triangle to its base is equal to the ratio of the height of the large triangle to its base.The height of the equilateral triangle is $H = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} (2 \sqrt{2}) = \sqrt{6}$.By similar triangles,$\frac{H-b}{H} = \frac{\ell}{a}$.$\frac{\sqrt{6}-b}{\sqrt{6}} = \frac{\ell}{2 \sqrt{2}}$.$b = \sqrt{6} (1 - \frac{\ell}{2 \sqrt{2}}) = \sqrt{6} - \frac{\sqrt{6} \ell}{2 \sqrt{2}} = \sqrt{6} - \frac{\sqrt{3}}{2} \ell$.The area $A = \ell 	imes b = \ell (\sqrt{6} - \frac{\sqrt{3}}{2} \ell) = \sqrt{6} \ell - \frac{\sqrt{3}}{2} \ell^2$.To maximize $A$,set $\frac{dA}{d\ell} = 0$.$\sqrt{6} - \sqrt{3} \ell = 0 \Rightarrow \ell = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}$.The maximum area $A = \sqrt{2} (\sqrt{6} - \frac{\sqrt{3}}{2} \sqrt{2}) = \sqrt{12} - \frac{\sqrt{6}}{2} \sqrt{2} = 2 \sqrt{3} - \sqrt{3} = \sqrt{3}$.The square of the largest area is $A^2 = (\sqrt{3})^2 = 3$.

If a rectangle is inscribed in an equilateral triangle of side length $2 \sqrt{2}$ as shown in the figure,then the square of the largest area of such a rectangle is $....$

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