http://www.twelvearound1.com/diagp.html

The picture below shows nine small squares arranged into a 3 x 3 square; within it is
another (red) square made by the diagonals of four 2 x 1 rectangles.
The four red diagonals cut their 2 x 1 rectangles in half, each leaving an area equal to the
area of one of the small squares which is in the 3 x 3 square but out of the red square.
The inner square must have an area of (nine minus four) five squares so its sides, the 2
by 1 rectangles' diagonals, have a length of root five times the side of one square.

This argument can easily be extended to make a general proof
that the square drawn on the longest side (called the
hypotenuse (Greek for stretched under)) of a right angled
triangle has an area equal to the areas of the squares drawn
on the other two sides added together.

Lets start with an oblong. We'll say it has a short side and a
long side.

If we draw in a diagonal this
cuts the oblong in half.

This is the triangle that is half
of the oblong. Any right angled
triangle can be thought of as
half of some oblong or square.

Here I have arranged four of the right
angled triangles so that their right angles
make the corners of a square, by joining
the sides which correspond to the short
sides of the oblong to the sides which
correspond to the long sides making
sides of the square which equal the
short and long sides added together.
So the area of this outer square is the
area of the square drawn on the longest
side of the triangles plus the area of the
four triangles, which you may remember
is half of the oblong each, or the area of
two such oblongs.

This also has sides equal to the short
and long sides added together and so is
an equal square, but you will see I have
made it this time out of a square with
sides equal to the long sides of the
oblong (in green), a square with sides
equal to the short sides of the oblong
(also green) and two oblongs which are
equal to the one I started with.
The two squares are equal, the first is
equal in area to twice the oblong plus
the square drawn on the long side of the
triangle, and the second is equal to two
oblongs and the squares drawn on the
other two sides.
So the square drawn on the long side of
the triangle equals in area the squares
drawn on the other two sides.

This will be true for any triangle which
can be drawn as half an oblong or
square i.e. any right angled triangle.

Back to golden rec tangle

Back to sacred geometr y

You may prefer to think of the four half rectangles left out of the red square as adding
together to make two whole 2 x 1 rectangles which clearly have an area of two squares
each leaving five to be contained in the red square.

Or perhaps even that the red square is made up of four half rectangles round a single
small square.

Anyhow, the area of the red square is that of five small squares so its sides, the diagonals
have a length of the square root of five times the length of the small squares' sides.