Ratio of inscribed areas

Let $h$ be the height of both the trapezoid and the triangle. Since the triangle is equilateral, we easily find that its area is $\mathcal{A}_{1}=\frac{\sqrt{3}}{3}h$ .

To find the area of the trapezoid, we consider the red and purple triangles in the diagram on the right. They share a diagonal as a base, while the sum of their heights gives the diagonal again, so that the area of the trapezoid is $\mathcal{A}_2=\frac{d^2}{2}$ , and since the diagonal forms a $45°$ angle with the vertical direction we have $d=\sqrt{2}h$ and therefore $\mathcal{A}_2=h^2$ .

It is now proved that the ratio between $\mathcal{A}_2$ and $\mathcal{A}_1$ is $\sqrt{3}$ .