Square inscribed in a square with 30 and 60 degree angles

Text

Inside a square $ABCD$ another square $EFGH$ is inscribed, so that each vertex $E,F,G,H$ lies on a different side of $ABCD$, and so that the sides of $EFGH$ form angles of $30^\circ$ and $60^\circ$ with those of $ABCD$. You know that the diagonal of $EFGH$ is $1$. Determine the area of $ABCD$.

Solution

Let $s$ be the side of the inner square. From the diagonal $s\sqrt2=1$ we get $s=\dfrac{1}{\sqrt2}$. Each side of $ABCD$ is ”covered” by the projections of two consecutive sides of $EFGH$, tilted by $30^\circ$: with $L$ the side of $ABCD$, $L=s\cos30^\circ+s\sin30^\circ=s\,\frac{\sqrt3+1}{2}.$ Substituting $s=\dfrac{1}{\sqrt2}$: $L=\frac{\sqrt3+1}{2\sqrt2}\ \Longrightarrow\ \text{Area}_{ABCD}=L^2=\frac{4+2\sqrt3}{8}=\frac{2+\sqrt3}{4}.$ $\boxed{\text{Area}_{ABCD}=\frac{2+\sqrt3}{4}\approx0{,}933.}$