The sum of the squares of the length the sides of a right triangle is equal to the square of the length of the hypotenuse.
Visually, the theorem can be illustrated as follows:
Given a right triangle, with sides a and b and hypotenuse c, (Figure 1)
/ l
/ l
c / l
/ l
/ l b
/ l
/________________ l
a
Figure 1
the hypotenuse is the side opposite the right (90 degree) angle in a right triangle.
Then, c^2 = a^2 + b^2, or c = sqrt(a^2 + b^2).
Certain sets of 3 integers are useful to remember as being Pythagorean triples, that is, they are possible lengths of the sides of a right triangle. For example:
a b c
3 4 5
9 12 15
The PythagoreanTheorem is an important tool in the study of TrigonometricFunctions.
Poser: (3,4,5) is a pythagorean triplet since 3^2 + 4^2 = 5^2. Which positive integers are not part of a pythagorean triplet?
Proof: Draw right triangle with sides a,b,and c as above. Turn an identical triangle 180 degrees and stick its c side to the original triangle's to form a rectangle of sides a and b. Draw an identical rectangle perpendicular to the first with only a corner touching. Draw squares of side a and b to connect the two rectangles into one large square of side (a+b). From this diagram, the area of the large square is (a+b)^2 = a^2 + b^2 + 4(.5ab) by the formulas for the areas of squares and triangles.
Now form a square from four of the triangles by placing the a side of one triangle in line with the b side of another, so that all four sides of the figure are (a+b). Note that within the large square is a square of side c. From this diagram, the area of the large square is (a+b)^2 = 4(.5ab) + c^2.
Since (a+b)^2=itself, a^2 + b^2 + 4(.5ab) = 4(.5ab) + c^2.
Therefore, a^2 + b^2 = c^2.
I think a visual of the triangles involved in this proof would be very helpful- even if they are really crude like mine.
Actually, the sheer volume of distinct known proofs of this theorem is staggering. See [Pythagorean Theorem Proofs] for just a sampling.