A piece of A0-sized paper has a total surface area of 1 m^2. The length and breadth of the A0-sized paper are x m and y m respectively. An A1 paper is an A0 paper folded in half, length-wise. An A2 paper is an A1 paper folded in half, so on and so forth.

1. Relate the rectangular area to the given total surface area

Area of A0 paper $= x \times y$.

Given area $=1\text{ m}^2$,

$xy=1.$

1. Set up the ratio of corresponding sides using similarity of A0 and A1 papers

A0 has dimensions $x$ by $y$.

A1 has dimensions $y$ by $\dfrac{x}{2}$.

Since A0 and A1 are geometrically similar,
$\frac{x}{y}=\frac{y}{x/2}.$

2. Solve the proportion to obtain the required ratio

$\frac{x}{y}=\frac{2y}{x}$
$x^2=2y^2$
$\left(\frac{x}{y}\right)^2=2$
Since lengths are positive,
$\frac{x}{y}=\sqrt{2}.$

1. Use the ratio from part (ii) to express one variable in terms of the other

From part (ii),
$\frac{x}{y}=\sqrt{2}$
so
$x=\sqrt{2}y.$

2. Substitute into the area relationship from part (i)

From part (i),
$xy=1.$
Substitute $x=\sqrt{2}y$:
$(\sqrt{2}y)y=1$
$\sqrt{2}y^2=1.$

3. Solve for y

$y^2=\frac{1}{\sqrt{2}}$
Since $y$ is a length, $y>0$.
$y=\sqrt{\frac{1}{\sqrt{2}}}=\frac{1}{\sqrt[4]{2}}.$

4. Find x using the ratio relationship

$x=\sqrt{2}y$
$x=\sqrt{2}\left(\frac{1}{\sqrt[4]{2}}\right)=\sqrt[4]{2}.$

1. Find the scale factor from A0 to A4 for each side

From A0 to A4, the area is halved 4 times, so the dimensions are reduced by a linear factor of $\dfrac{1}{4}$.

Hence A4 dimensions are
$\frac{x}{4}\text{ m} \times \frac{y}{4}\text{ m}.$

2. Substitute the A0 dimensions and convert to millimetres

Using $x=\sqrt[4]{2}$ and $y=\dfrac{1}{\sqrt[4]{2}}$,
$\frac{x}{4}=\frac{\sqrt[4]{2}}{4}\text{ m}=\frac{1000\sqrt[4]{2}}{4}\text{ mm}\approx 297\text{ mm},$
$\frac{y}{4}=\frac{1}{4\sqrt[4]{2}}\text{ m}=\frac{1000}{4\sqrt[4]{2}}\text{ mm}\approx 210\text{ mm}.$

1. Recognise the doubling pattern for the number of pieces after each fold

A0 gives $1=2^0$ piece.

A1 gives $2=2^1$ pieces.

A2 gives $4=2^2$ pieces.

Therefore, AN gives $2^N$ pieces.