22. A square with perimeter $4 \mathrm{~cm}$ can be cut into two congruent rightangled triangles and two congruent trapezia as shown in the first diagram in such a way that the four pieces can be rearranged to form the rectangle shown in the second diagram.

What is the perimeter, in centimetres, of this rectangle?
A $2 \sqrt{5}$
B $4 \sqrt{2}$
C 5
D $4 \sqrt{3}$
E $3 \sqrt{7}$
Solution
A
The square has perimeter $4 \mathrm{~cm}$ and hence its sides each have length $1 \mathrm{~cm}$.
Let the vertices in the diagram on the left be labelled as shown.
We let the length of $S U$ be $x \mathrm{~cm}$.
For the pieces to fit together as shown on the right, $T V$ and $W R$ must also have length $x \mathrm{~cm}$. It follows that the other lengths, in $\mathrm{cm}$, are as shown.

Method 1
For the pieces to fit together the angles $\phi$ and $\theta$ must be equal, as shown on the right of the above diagram. Therefore $\tan \phi=\tan \theta$. That is,
$$
\frac{1-2 x}{1-x}=\frac{x}{1} \text {. }
$$

Hence
$$
1-2 x=x(1-x)
$$

This last equation may be rearranged to give
$$
x^{2}-3 x+1=0 \text {. }
$$

From the standard formula $\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ for the solutions of a quadratic equation, we can deduce that
$$
x=\frac{3 \pm \sqrt{5}}{2} .
$$

Since $x<1$ it follows that $x=\frac{3-\sqrt{5}}{2}$.
From the diagram on the right above we see that the perimeter of the rectangle is given by
$$
(1-x)+(2-x)+(1-x)+(2-x)=6-4 x=6-4\left(\frac{3+\sqrt{5}}{2}\right)=2 \sqrt{5} \text {. }
$$

Method 2
Since they are made up from the same two triangles and two trapezia, the area of the rectangle equals the area of the square.
Therefore, from the diagram above, we have
$$
(1+(1-x)) \times(1-x)=1 \times 1,
$$
that is,
$$
(2-x)(1-x)=1
$$

Hence
$$
2-3 x+x^{2}=1
$$
and therefore
$$
x^{2}-3 x+1=0 \text {. }
$$

Thus, as in Method $1, x=\frac{3-\sqrt{5}}{2}$, and the perimeter of the rectangle is $2 \sqrt{5}$.
FOR INVESTIGATION
The diagram above shows on the left an $8 \times 8$ square divided into four pieces which are rearranged to make the $13 \times 5$ rectangle shown on the right.
However $8 \times 8<13 \times 5$.
Where has the additional area come from?
Note: This puzzle has been attributed to William Hooper in his book Rational Recreations of 1774 .