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: