30-60-90 Polypuzzle
Can you re-arrange the pieces of the puzzle to form a rectangle by sliding the pieces without rotating them? Now can you re-arrange the pieces to form an equilateral triangle by flipping the pieces numbered $2$ and $5$ and moving them into new positions?
You can assume that pieces $1$ and $5$ each have a side of length one unit, that the pieces as shown form a perfect square of area one square unit and that they do fit together to form a perfect equilateral triangle of the same area.
Calculate the length $t$ of the edge of piece $3$ and then calculate the lengths of all the other edges giving answers correct to $3$ significant figures. You can use the interactivity below to explore how the pieces fit together.
Finding the lengths depends on using the ratio for the sides of a 30-60-90 triangle as the name of the problem suggests. Below the diagrams show how to take the pieces which make a square of unit area and fit the pieces together to make an equilateral triangle of the same area with side $2t$ and knowing this you can calculate $t$. The way pieces fit together gives you that $p=t$ and the rest is up to you!
You can calculate the length '$t$' knowing the area of the equilateral triangle. Pythagoras theorem and the sine rule can be used in finding the other lengths.
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 $r=\sqrt{(1+s)^2+t^2}-n\approx 0.864$.