Pythagoras' Theorem
Pythagoras' theorem  proving
Pythagoras' theorem
Pythagoras' Theorem


Pythagoras' theorem states that the square of the hypotenuse
will equal the sum of the squares of the other two sides. (It doesn't
matter which way around a and b are.)
a^{2} + b^{2} = c^{2}
Two easy to remember right triangles which fulfill this condition
with whole numbers are the 3, 4, 5 triangle and the 5, 12, 13 triangle.
3^{2} + 4^{2} = 5^{2}
9 + 16 = 25
5^{2} + 12^{2} = 13^{2}
25 + 144 = 169
These three numbers are together called Pythagorean
triples. 
Proving Pythagoras' theorem
One way of proving Pythagoras' theorem is by algebra.
We can arrange four triangle so that they bound a square area
(light blue). We know it's a square because the angles in a triangle
add to 180° and so do angles in a straight line.
The area of a triangle is given by ^{1}/_{2} base
x height, and the area of a square is given by (length)^{2}.
So the area of the whole arrangement in the diagram will be given
by the area of 4 triangles plus the large blue square.
4 x (^{1}/_{2} a x b ) + c^{2}
But the arrangement is a square  there is a right angle at
each corner and each side has a length of a + b. So the total
area can also be given by:
( a + b )^{2}
Of course, these areas are the same, so we equate the two then
simplify.
( a + b )^{2} = 4( ^{1}/_{2} ab
) + c^{2}
( a + b )( a + b ) = ^{4}/_{2} ab
+ c^{2}
a^{2} + ab + ba + b^{2} = 2ab
+ c^{2}
a^{2} + 2ab + b^{2} = 2ab +
c^{2}
a^{2} + b^{2} = c^{2} 


It's also possible to prove by rearrangement.
What happens if you double the size of one of the examples given,
say 3, 4, 5 becomes 6, 8, 10? Will the triangle still be a right
angle triangle? Do the numbers still work out for Pythagoras' theorem?
For information see the Pythagoreas Triples page.
