30-60-90 Polypuzzle

Teresa solved this problem, using this hints we gave:

The area of the equilateral triangle is $1=\frac{1}{2}\times(2t)^2\times\sin 60^{\circ}$ so $t^2=\frac{1}{\sqrt{3}}$, so $t=\frac{1}{\sqrt[4]{3}}\approx 0.760$.

We know that $p=t$ and $q=1-t$, so $p\approx 0.760$ and $q\approx 0.240$.

The height of the equilateral triangle is $1+s=t\sqrt{3}$ so $s\approx 0.316$.

We can use Pythagoras' Theorem to find $m$: $m^2=s^2+(1-t)^2\approx 0.156$, so $m\approx 0.397$.

Now we can work out $\theta$: $\tan\theta=\frac{1-t}{s}\approx 0.760$, so $\theta\approx 37.2^{\circ}$.

From the sine rule, we have $\frac{u}{\sin 30^{\circ}}=\frac{1}{\sin(150-\theta)}$, so $u=\frac{1}{2\sin(150-\theta)}\approx 0.542$.

Also from the sine rule, $n=\frac{\sin\theta}{\sin(150-\theta)}\approx 0.656$.

By looking at the longest line across the square, we have $\frac{1}{m+u+v}=\cos\theta$, so $v=\frac{1}{\cos\theta}-m-u\approx 0.317$.

By looking at the rectangle and considering the diagonal, we see that $(r+n)^2=(1+s)^2+t^2$, so $r=\sqrt{(1+s)^2+t^2}-n\approx 0.864$.