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

Answer: 3

In $△DBF$
$\tan 60^{\circ }=\frac{2b}{2\sqrt{2}-l}$
$\Rightarrow b=\frac{\sqrt{3}(2\sqrt{2}-l)}{2}$
$A=$ Area of rectangle $=l\times b$
$A=l\times \frac{\sqrt{3}}{2}(2\sqrt{2}-l)$
$\frac{dA}{dl}=\frac{\sqrt{3}}{2}(2\sqrt{2}-l)-\frac{l\cdot \sqrt{3}}{2}=0$
$l=\sqrt{2}$
$A=l\times b=\sqrt{2}\times \frac{\sqrt{3}}{2}(\sqrt{2})=\sqrt{3}$
$\Rightarrow A^2=3$