Copyright © University of Cambridge. All rights reserved.

## 'Pythagoras Proofs' printed from http://nrich.maths.org/

Herschel of the European School of Varese,
Italy and Patrick of Woodbridge School both sent us in their ideas
for the first proof.
We rearrange the construction on the left into the L shape on the
right as shown above. Each of the four triangles are congruent and
have side lengths $a$ and $b$ and hypotenuse $c$. Clearly the area
of the original square is $c^2$. After rearranging the shapes we
find that the dark blue square has sides of length $b$ and the
light blue square has sides of length $a$. The total area of the L
shape must therefore be $a^2 + b^2$. Hence $a^2 + b^2 = c^2$.

This is a good start and explains the ideas
nicely. However there are a few details which you may want to
clarify. How do we know the triangles are right angled triangles?
How do we know that the L shape on the right is definitely made up
of two squares i.e. how do we know it isn't made up of two
rectangles?
Herschel and Patrick then continued with
the second proof:
$$\eqalign{

\mbox{Area of Square} &= (a+b)^2\cr

&= {{a^2 + 2ab + b^2}}}$$ Area of the Trapezium = Area of
square divided by 2 (rotational symmetry) $$\mbox{Area of
Trapezium} = {{a^2 + 2ab + b^2}\over{2}}$$ We can also think of the
area of the Trapezium as the sum of areas of the three
triangles. So: $$\eqalign{

\mbox{Area of Trapezium} &= {{ab\over{2}}+ {ab\over{2}}+
{c^2\over{2}}}\cr

&={{ab + ab + c^2\over{2}}}\cr

&= {{2ab + c^2\over{2}} }}$$

Hence$$\eqalign{

{{2ab + c^2\over{2}} } &= {{a^2 + 2ab + b^2}\over{2}}\cr

2ab+c^2 &= a^2+2ab+c^2\cr

c^2 &= a^2 + b^2}$$

Herschel: "This proof was the
simplest"

Patrick: "I find this proof the most
interesting, and probably easier to explain. I was quite surprised
to find Pythagoras's Theorem emerging from the formulae."

Andrew from Island School sent us his work
on the third proof.
For the red triangle, $DA = a^2$, $DB=ac$ and $AB=ab$.

For the blue triangle, $AB=ab$, $AC=b^2$ and $BC=bc$.

For the combined triangle, $DB=ac$ and $BC=bc$.

To get the missing length which is $DC$, the enlargement has
the scale factor of $c$. Then the sides of the triangle with the
scale factor of $c$ would be $ca$, $cb$ and $c^2$.

But $DC=DA+AC$, so $a^2+b^2=c^2$.

Good work, Andrew!