If the shorter edge is 1, then by Pythagoras’ theorem 
the length of a must be V2 .
It is clear that the length b is V2 - 1 and this is also 
the length of the equal sides of the folded triangle so 
length c can be found using Pythagoras’ theorem.
C2 = (V2 — 1 )2 + (V2~— 1 )2 
c2 = (3 -V2) + (3 - V2)
c2 = 6 - 2V2
To work out the length d, we said that if the original 
width was 1 and the folded part was
V2 - 1 then d = 1 - (V2 - 1) so d = 2 - V2
Therefore d2 = (2 - V2)2 = 6 - 4V2"
Hence c and d must be of equal length and so the 
shape is a kite.
We discussed how this was good practice of surd 
manipulation and also how we could extend this 
to find areas and perimeters and to prove that the 
diagonals meet at right angles.
The fact that A-size paper has its sides in the ratio 
V2 : 1 can be proved using similarity and algebra, 
but we were also shown a demonstration using just 
two folds: the square fold to get the V2 diagonal, and 
a second fold from the same corner to match the 
rectangle’s longer side to the square’s diagonal. This 
raised the question of what constitutes a ‘proof.
Another activity used ninja stars, 
known as ‘shuriken’ in Japanese. 
Instructions on how to make shuriken 
can be found on the internet. We 
discussed the fact that they had 
rotational symmetry of order 4, but had no 
reflectional symmetry; although we suspected that 
some students would want to add lines of symmetry.
Then folding down a second corner: