Question

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_______.

Answer 1

In $\triangle D B F$
$\tan 60^{\circ}=\frac{2 b}{2 \sqrt{2}-l}$
$\Rightarrow b=\frac{\sqrt{3}(2 \sqrt{2}-l)}{2}$
$A=$ Area of rectangle $=l \times b$
$A=l \times \frac{\sqrt{3}}{2}(2 \sqrt{2}-l)$
$\frac{d A}{d l}=\frac{\sqrt{3}}{2}(2 \sqrt{2}-l)-\frac{l \cdot \sqrt{3}}{2}=0$
$l=\sqrt{2}$
$A=l \times b=\sqrt{2} \times \frac{\sqrt{3}}{2}(\sqrt{2})=\sqrt{3}$
$\Rightarrow A^{2}=3$