	PYTHAGOREAN THEOREMS		Contact Us At Contra Costa College	Math Dept. Home Page

What's this? See below.

Pythagorean TheoremS

Some "Not So Familiar" Implications

Traditional The Pythagorean Theorem states that for a right triangle with legs a and b and hypotenuse c, we have the Pythagorean Equation:
a² + b² = c²
If we think of this geometrically, we are saying that the sum of the areas of the squares drawn on the two legs, as shown to the left, is equal to the area of the square drawn on the hypotenuse. This is the traditional Pythagorean result, both algebraically and geometrically.
(Note: There are many different proofs of the Pythagorean Theorem, including one by U. S. President J.A. Garfield in 1876. There is a link at the bottom of this page to more information and some the many proofs of this theorem.)

Pytagorean Hexagons
True for Regular Hexagons.
However, still thinking in geometric terms, we can construct any set of 3 regular polygons (triangles, pentagons, hexagons, etc.) using a and b and hypotenuse c, respectively, as the common sides of the three polygons. And still, the sum of the areas of the figures drawn on the two legs is equal to the area of the figure drawn on the hypotenuse.
(You can show this algebraically for equilateral triangles, by multiplying both sides of the Pythagorean Equation by 3/4, the area of an equilateral triangle being 3/4 times the square of the common side. For the other regular polygons, you can use the basic formula for the area of an isosceles triangle multiplied by n, where n is the number of sides of the polygon - see Area of the Pentagram & Regular Polygons )

Pytagorean Triangles
True for Pythagorean Similar Triangles. Further, the Pythagorean result is valid for similar figures so constructed on the legs and the hypotenuse. They don't have to be only regular polygons. However, the two controlling dimensions of of the figure that determine its area, must satisfy what we will call the Pythagorean Similarity Condition. That is, if we consider the leg a, then the two controlling dimensions of the figure constructed on leg a might be xa and ya, some real multiples of a. The Similarity Condition is that these same multiples of b and c must also exist in the controlling dimensions of those figures, xb, yb, xc and yc. Thus, the areas of the three figures are kxaya, kxbyb and kxcyc. k is just another factor, it could be 1 or pi or any other real number. The result is that the Pythagorean Equation is just multiplied by kxy:
(kxy)a² + (kxy)b² = (kxy)c²

Sum of Stars True for similar Pentagrams.
The three pentagrams at the top of this page are Pythagorean Similar.
Hence, the sum of the areas of the pentagrams drawn on the two legs is equal to the area of the pentagram drawn on the hypotenuse.

Pythagorean Circles
True for Circles & Semi-circles. The sum of the areas of the circles drawn on the two legs is equal to the area of the circle drawn on the hypotenuse.
(Can you write the appropriate equation?)

Also, related to these Pythagorean circles,
see Problem 4 below in the Problem Section.

Pythagorean Buffalos
What about Buffalos? These buffalos are of the Buffalo Nickel fame. However, these three are also Pythagorean Buffalos in that they are Pythagorean Similar.
Click below to see the Pythagorean relationship and a gallery of other familiar figures presented as Pythagorean Similar.
Pythagorean Gallery

Pythagorean Boxes
True for Boxes.

Suppose we extend the Pythagorean figure into the third dimension. Then we have
ma² + mb² = mc²
and that is neat!
The sum of the volumes of the two smaller boxes equals the volume of the larger box, mc-squared.

Question: Does the sum of the areas of the two smaller white triangles surrounding the blue right triangle (figure at the top of the page) equal the area of the larger white triangle? (This is a question for you, you don't need to send me your answer.)

PROBLEM SECTION:

Problem 1: Binomial Expansions & The Pythagorean Theorem
Try Problem 1.

(for more about this topic see the link : Pentagons & The Pythagorean Theorem
near the bottom of this page )

Problem 2: Constructing a Square Equal To The Sum of Two Given Squares
Two Squares

Try Problem 2.

The Following Problems Are Related To Problem 2:

SQUARING POLYGONS

                                     Problem 2a: Constructing a Square Equal To The Area of a Given Rectangle
                                     Makes use of the geometric mean of two numbers.

Try Problem 2a.

Problem 2b: Constructing a Square Equal To The Area of a Given Triangle

Try Problem 2b.

Problem 2c: Constructing a Square Equal To The Area of a Given Pentagon

Try Problem 2c.

Historical Note (ckick here):

Problem 3: Sectional Pythagorean Theorem
Pythagorean Section

Try Problem 3.

Problem 4: Pythagorean Similar Rectangles
Pythagorean Rectangles

Try Problem 4.

Problem 5: Circle Pythagorean Theorem (Lunes)
Pythagorean Circles

Try Problem 5.

Problem 6: A Trapezoid Pythagorean Theorem
Sum Of Squares of the Sides of a Trapezoid

Try Problem 6.

Problem 7: A Pythagoragram

Try Problem 7.

THE UNIT CIRCLE
TRIGONOMETRY AND THE PYTHAGOREAN THEOREM

When we examine the unit circle, the x and y coordinates of points on that circle are equal to the sine and cosine of the angle the radius makes with the positive x-axis.

x = COS( q )
y = SIN( q ) = COS( b )

Therefore it follows that

x² + y² = 1 ® COS²( q ) + SIN²( q ) = 1

The latter equation is known as the Pythagorean Identity in trigonometry.

Also, by substitution, COS²( q ) + COS²( b ) = 1

(Notice that b is also the angle that the radius line makes with the y-axis. For future reference, these angles are called the direction angles of the radius line and COS( q ) and COS( b ) are called the direction cosines of the radius line.)

Now we can multiply the above equations by r²

or by examining the non-unit circle, circle with radius r, we have
x = r COS( q )
y = r SIN( q )
and thus, for any circle we easily get

(r COS(q ))² + (r SIN(q ))² = r²

and
x² + y² = r²

the equation for a circle in Cartesian coordinates as a direct result of the Pythagorean Theorem.

(See below, the discussion of the Unit Sphere for more on direction angles and direction cosines. )

NOT RIGHT? PARALLELOGRAMS IN PLACE OF SQUARES

In special circumstances, when the triangle is not a right triangle, we can talk about parallelograms and rhombuses instead of squares and create a special 'Pythagorean Relationship'. See the figures below:

In the second figure above (stretched), we no longer have squares on the sides of the triangle. We have parallelograms (rectangles, in this case) and a rhombus. Still, the Pythagorean relationship holds, the sum of the areas of the rectangles drawn on the two legs is equal to the area of the rhombus drawn on the hypotenuse of the right triangle. But the triangle may not be a right triangle under other transformations. See the next two figures:

In these figures the Pythagorean relationship still holds, the sum of the areas of the parallelograms drawn on the two sides is equal to the area of the parallelogram drawn on the third side of the triangle.

In these special circumstances the area of the "red" figure is equal to twice the area of one of the "yellow" figures (the yellow figures have equal areas) because of the 3 by 3 grid we produced in the original set up.

Try this:
Draw an equilateral triangle, set up a 3 by 3 grid work around it, and show that the areas of rhombuses drawn on two of the sides of the triangle is equal to the area of the larger parallelogram drawn on the third side.

NOT RIGHT? BUY A COS!

If the triangle is not a right triangle, then the Pythagorean Equation no longer holds. But, it can be brought back into balance by including one more term in the equation, 2ab*Cos( g) ,

a² + b² = c² + 2ab*Cos( g )

where g is the angle in the triangle opposite side c.
This equation is known as (one of the) Law of Cosines. The other equations in the Law of Cosines are variations wherein the side and the angle opposite it are chosen from any of the three possibilities for the triangle.

So for any triangle, the sum of the areas of the squares
on two of the sides of the triangle is equal to the area
  of the square on the third side plus (or minus) some
 percentage of twice the area of a rectangle, that is 2ab, 
with sides equal to the first two sides of the triangle!

(The percentage mentioned above being the value given by COS(g) which varies from 0 to ±100%
depending on the angle g.)

The Law of Cosines would work even if the triangle was a right triangle. Why is that?

That's ok if g = p/2, but we can choose any side and the angle opposite it.
What if the angle chosen wasn't p/2, yet p/2 was one of the angles of the triangle?
That's something to think about. Consider an isosceles right triangle and choose g = p/4 and see
what happens when you substitute all of the appropriate values into the Law of Cosines equation.

Ok, now try this, how would you interpret the situation if the angle opposite side c is 0? ®

Pythagorean Boxes
Extended To Boxes. Pythagorean Theorem Extended to 3 Dimensions

In the same way we might cut off the corner of a rectangle, making a right triangle, we might cut off the corner of a box, making a right tetrahedron . And just as
a² + b² = c²
is true for the right triangle, we have
A² + B² + C² = D²

is true for the right tetrahedron, where A, B and C are the areas of the three right triangular faces of the tetrahedron, and D is the area of the white triangle (see fig. at left) created by the cut (the base of the tetrahedron)
and that is really neat!

(One easy check up on this result is to make the cut so that an equilateral triangle is created. Then using some algebra it's easy to check that
A² + B² + C² = D² .
But it's true for any planar cut that removes a corner from the box!)

There is no easy geometric interpretation for this result. The faces whose areas are A, B, C and D are two dimensional and squaring them essentially produces a 4-dimensional result.

There are, however, some interesting numerical results, such as, Pythagorean quadruples. Similar to Pythagorean Triples, quadruples are sets of 4 integers, such that the sum of the squares of the smaller three equals the square of the fourth larger integer.
One such quadruple is [ 3, 4, 12, 13 ].
Check this out: 3² + 4² + 12² = 13²
Can you find others?


A Pythagorean Quadruple Problem.	A Pythagorean Quadruple Problem Try the Problem.

Pythagorean Box
Diagonal of Box. Pythagorean Theorem For Diagonal of a Box

Given the dimensions of a box are a, b, and c. Since
a² + b² = e²
and
e² + c² = d²
we have by substitution
a² + b² + c² = d²
Thus, the square of the diagonal of a box is equal to the sum of the squares of the dimensions of the box.
Note: This and the equation of the previous discussion are precisely the same, but the geometrical interpretation is quite different.

Sum Of Squares of the Sides of the Hexagon Check the problem section above for "Sum Of The Squares Of The Sides Of The Trapezoid ... Problem" as it relates to this figure. Problem Section

Note on the Unit Sphere:
Just as in the case of the Unit Circle (See above, the discussion of the Unit Circle )
a radius line in 3 dimensions to the surface of a Unit Sphere, makes angles with each of the positive coordinate axes, x, y, and z.
Call these angles a, b, g, respectively. Then the Pythagorean relationship is

COS²( a ) + COS²( b ) + COS²( g ) = 1

The angles are called the direction angles of the radius line and COS( a ), COS( b ), and COS( g ) are called the direction cosines of the radius line.
If you draw a picture of Unit Sphere and the radius to the point (x, y, z) on the surface, the direction cosines are equal to x, y and z, respectively. Furthermore, they are the edges of a box inside the sphere and the radius line is its diagonal.
And consequently,

x² + y² + z² = r²

is the equation for a sphere in Cartesian coordinates, a direct result of the Pythagorean Theorem.

Can you find the length of the diagonal of a unit cube? a unit square?

PYTHAGOREAN MOSAIC

Why is this called a Pythagorean Mosaic? Click below to find out.

Pythagorean Mosaic

Click below for a diagrammatic proof of the Pythagorean Theorem involving pentagons
(and also a discussion of the relationship of the square of the binomial ( a + b ) and the Pythagorean Theorem. Also see Problem 1 in the Problem Section above).

Pentagons & The Pythagorean Theorem

For more information on the Pythagorean Theorem:
OnLine you can follow the links below:

®  For more about Pythagorean Triples & Quadruples:

Eric W. Weisstein. "Pythagorean Triple."
        and
Eric W. Weisstein. "Pythagorean Quadruple."
From MathWorld--A Wolfram Web Resource.

http://mathworld.wolfram.com/PythagoreanTriple.html

http://mathworld.wolfram.com/PythagoreanQuadruple.html

¬ This plot shows points
with integer coordinates (a, b),
plotted in the xy-plane where
[a, b, Ö(a² + b²)] is a Pythagorean Triple.

      And
®  For more about Pythagorean Spirals:

Mike Patterson, e-zgeometry.com webmaster & designer
    http://www.e-zgeometry.com/projects/spiral/patterson/spiral.htm

      And
®  click here for a manipulative proof of Pythagorean Theorem
       Java Applet Manipulative Proof of Pythagorean Theorem
http://www.ies.co.jp/math/java/geo/pythagoras.html

¬   Bill Casselman's feature article
On the dissecting table
+plus magazine , writes of Henry Perigal, nineteenth century British amateur mathematician, who is now known principally for an elegant cut-and-shift proof - what is now called a dissection proof - of Pythagoras' Theorem (1830).
    http://plus.maths.org/issue16/features/perigal/

      And
®  For more about (some of) The Many Proofs of the Pythagorean Theorem:

www.cut-the-knot.org
Copyright � 1996-2009 Alexander Bogomolny
    http://www.cut-the-knot.org/pythagoras/index.shtml

      And
®   more about Right Tetrahedrons:
    http://www.cut-the-knot.org/Generalization/pythagoras.shtml

send comments to Thomas M. Green c/o CCC Math Dept. hwalters@contracosta.cc.ca.us

True for Regular Hexagons.	However, still thinking in geometric terms, we can construct any set of 3 regular polygons (triangles, pentagons, hexagons, etc.) using a and b and hypotenuse c, respectively, as the common sides of the three polygons. And still, the sum of the areas of the figures drawn on the two legs is equal to the area of the figure drawn on the hypotenuse. (You can show this algebraically for equilateral triangles, by multiplying both sides of the Pythagorean Equation by 3/4, the area of an equilateral triangle being 3/4 times the square of the common side. For the other regular polygons, you can use the basic formula for the area of an isosceles triangle multiplied by n, where n is the number of sides of the polygon - see Area of the Pentagram & Regular Polygons )
True for Pythagorean Similar Triangles.	Further, the Pythagorean result is valid for similar figures so constructed on the legs and the hypotenuse. They don't have to be only regular polygons. However, the two controlling dimensions of of the figure that determine its area, must satisfy what we will call the Pythagorean Similarity Condition. That is, if we consider the leg a, then the two controlling dimensions of the figure constructed on leg a might be xa and ya, some real multiples of a. The Similarity Condition is that these same multiples of b and c must also exist in the controlling dimensions of those figures, xb, yb, xc and yc. Thus, the areas of the three figures are kxaya, kxbyb and kxcyc. k is just another factor, it could be 1 or pi or any other real number. The result is that the Pythagorean Equation is just multiplied by kxy: (kxy)a² + (kxy)b² = (kxy)c²
True for similar Pentagrams.	The three pentagrams at the top of this page are Pythagorean Similar. Hence, the sum of the areas of the pentagrams drawn on the two legs is equal to the area of the pentagram drawn on the hypotenuse.

True for Circles & Semi-circles.	The sum of the areas of the circles drawn on the two legs is equal to the area of the circle drawn on the hypotenuse. (Can you write the appropriate equation?) Also, related to these Pythagorean circles, see Problem 4 below in the Problem Section.

What about Buffalos?	These buffalos are of the Buffalo Nickel fame. However, these three are also Pythagorean Buffalos in that they are Pythagorean Similar. Click below to see the Pythagorean relationship and a gallery of other familiar figures presented as Pythagorean Similar. Pythagorean Gallery
True for Boxes.	Suppose we extend the Pythagorean figure into the third dimension. Then we have ma² + mb² = mc² and that is neat! The sum of the volumes of the two smaller boxes equals the volume of the larger box, mc-squared.