Prove the Pythagorean Theorem.

Question

Please discuss various proofs that most students will learn.

Answer 1

Here's a simple proof.

 

Consider a square in which has a side equal to a + b, then draw a square inscribed within the original square in such a way that its vertices intersect each side of the original square to make each side containing the two segments which equal to a and b; let the inscribed square have a side equal to c.

 

Now, the figure contains 4 right triangles containing lengths of a, b, and c.

 

The first step is to find the area of the square in which its side is a+b:

 

A_ab=(a+b)² = a²+2ab+b²

 

Next, figure out the area of each triangle being 1/2*ab.

 

Next, add the areas of the triangles in which is:

 

A_total = 4*(1/2*ab)=2ab

 

Now, subtract the total areas of the triangles from the original area of the square contain the side equal to a+b in which is:

 

A_csqr = (a+b)² - 2ab = a²+2ab+b² - 2ab = a²+b²

 

Hence, this is the area of the inscribed square being equal to c².

 

Therefore,

 

c² = a²+b².

Answer 2

Bob,

 

The theorem only applies to right triangles containing a 90 degree angle.

 

Let c = length of triangle side opposite the 90 degree angle

Let a = the length of the shorter side left

Let b = the remaining side' length

Let T = the angle opposite side a

 

Then prove the theorem:    c² = a² + b²  for any non-zero a, b lengths

 

(1) Proof if you know sin(T) and cos(T) and how to manipulate them

 

Note by definition (b/c) = cos(T)    and  (a/c) = sin(T)

We know that [sin(T)]² + [cos(T)]² = 1  so by substitution

[a/c]² + [b/c]² = 1  then squaring and multiplying both sides of the equality by c²

a² + b² = c²   and we're done

 

(2) Proof by geometrical shapes

 

Draw a square with side length c and area c²

Draw a larger square that goes through the 4 vertices of the smaller one.  It appears rotated somewhat compared to the embedded smaller square or vice versa. Draw it.

 

Each vertex of the smaller embedded square divides the larger squat's length into two parts a & b, where a is the shorter length.  Then by construction, the area of the bigger square (a + b)² is just the sum of area of the smaller square c² plus the aras of the 4 identical right triangles between the squares. Each has area (1/2)ab.

 

So     (a + b)² = c² + 4[(1/2)ab 

a² + 2ab + b² = c² + 2ab     and the ab terms cancel out, leaving

a² + b² = c².  Again, we're done.

 

There are other ways of proving this theorem, but why beat a dead horse?