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'30-60-90 Polypuzzle' printed from https://nrich.maths.org/
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 $r=\sqrt{(1+s)^2+t^2}-n\approx
0.864$.