Take an A4 piece of paper and halve it by drawing a line across
the middle, parallel to the shorter side. Each half is called an A5
rectangle.

Halve the bottom half by drawing a vertical line down the middle.
This creates two A6 rectangles.

Halve the right hand one by drawing a horizontal line across its
middle. This creates two A7 rectangles.

Halve the bottom one by drawing a vertical line down the middle.
This creates two A8 rectangles.

You should have something like this:

Halve the right hand A8 shape by drawing a horizontal line
across its middle. This creates two A9 rectangles.

And so on ... Keep going until the rectangles are too small to be
seen clearly.

Now draw the diagonal of the A4 piece of paper from the top left corner to the bottom right corner. This creates a sequence of triangles. The first two are numbered in the diagram below but you will have many more drawn on your sheet:

What is the total area of the first two triangles
as a fraction of the original A4 rectangle?

What is the total area of the first three triangles as a fraction
of the original A4 rectangle?

If you could you go on adding all the triangles' areas, what do you
think the total would be as a fraction of the original A4
rectangle?

Many thanks to Professor Michael Sewell
for providing us with this idea.

It has a connection with Zeno's paradox. Can you find out about
that paradox?

What is the connection between the method used to find all the
triangles' areas and the method used to explain Zeno's paradox?