The *Pythagorean* (or *Pythagoras’*) *Theorem* is the statement that for a right angle triangle the sum of (the areas of) the two small squares formed on the sides of the triangle equals (the area of) the big one.

In algebraic terms, **a² + b² = c²** where **c** is the hypotenuse while **a** and **b** are the other two sides of the triangle.

The theorem is of fundamental importance in Euclidean Geometry where it serves as a basis for the definition of distance between two points. (See your text books in the chapter “Coordinate Geometry”).

There are many proofs of this theorem, some graphical in nature and others

using algebra. I like this one.

If you prefer an even easier picture to remember but a bit more algebra you could use **Heron’s formula** to prove Pythagoras’s theorem. It is the last formula on page 9 of the Formulae and Tables booklet and gives the area of a triangle as √(s(s-a)(s-b)(s-c) where s is half the length of the perimeter and a,b and c are the lengths of the sides.

Take an isosceles triangle made of two congruent right angled triangles as shown.

Its area = ab. Its perimeter length = 2b+2c, so s = b+c). Applying Heron’s formula gives

ab = √(b+c)(b+c-2b)(b+c-c)(b+c-c)

= √(b+c)(c-b)(b^{2})

=√(c^{2}-b^{2})(b^{2})

a^{2}b^{2 }= (c^{2}-b^{2})(b^{2})

a^{2 }= (c^{2}-b^{2})

c^{2 }= a^{2}+ b^{2}

The Theorem is *reversible* which means that its *converse* is also true. The converse states that a triangle whose sides satisfy a² + b² = c² is necessarily right angled.

In Euclidean Geometry this theorem serves as a basis for the definition of distance between two points. This in turn has many practical applications.

Combining Pythagoras’ theorem with the concept of a coordinate plane leads to the formula for the length of a line segment in co-ordinate geometry, the equation of a circle (because a radius is a line segment and we describe a circle by giving the position of its centre point and the radius around it), the modulus of a complex number, the modulus or norm (F&T page 17) of a vector or the value of anything that can be represented by the length of a line segment on a plane or graph with units marked off on two axes at right angles to each other. Modulus is just another name for the absolute value, that is the value of its length without worrying about whether it is positive or negative in direction.

**Length of line segment and anything that can be represented by a line segment**

The relationship of sides |x_{2}−x_{1}| and |y_{2}−y_{1}| to side d is the same as that of sides a and b to side c. We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. To find the length c, take the square root of both sides of the Pythagorean Theorem.

c^{2 }=a^{2}+b^{2 }→ c=√(a^{2}+b^{2})

It follows that the distance formula is given as

d^{2}=(x_{2}−x_{1})^{2}+(y_{2}−y_{1})^{2 }→ d=√(x_{2}−x_{1})^{2}+(y_{2}−y_{1})^{2}

We do not have to use the absolute value symbols in this definition because any number squared is positive.

For any point P a distance r from the origin, x^{2} + y^{2} = r^{2}

.

Taking the special case where the radius of the circle is one unit long (the **unit circle**) and applying Pythagoras’ theorem gives us the trigonometrical identity sin^{2} +cos^{2} = 1 and provides a way to move between Cartesian (xy) coordinates and polar coordinates (using angles referenced to the positive x-axis). We simply replace the x coordinate with cos(Ѳ) and the y coordinate with sin(Ѳ), where Ѳ is the angle made with the positive x-axis. If the radius is made say 20 units long then x = 20cos(Ѳ) and the y coordinate = 20sin(Ѳ).

In the above both the x and the y axes are real number lines. But if we make the y axis an imaginary number line marked off in units of *i,* where *i* is the square root of -1, to deal with complex numbers, then we can see that a complex number represented by x+y*i* on an Argand diagram can be represented by r[cos(Ѳ)+ i sin(Ѳ)] in trigonometrical form. In this case we call the angle (Ѳ) the amplitude or argument*, *and r = √( x^{2}+y^{2}) is the absolute value or modulus of x+y*i. *

# Equation of a Circle

A circle can be describe by giving the coordinates of its centre point and the length of its radius. The length of the radius can be given by Pythagoras’ theorem. If the circle is centred at the origin, (0,0), then the equation of the circle is given by x^{2} + y^{2} = r^{2 }as described above.

If we slide the circle away from the origin so that its centre is located at a point (h,k) we can apply the same reasoning to come up with the equation of a circle as given on page 19 of the F&T booklet. That is (x-h)^{2} + (y-k)^{2} = r^{2}. This can be understood by examining the diagram below. This is the same as the equation for the length of a line segment given on page 18 of the F&T booklet. It just looks a bit different because they have used (x_{2}-x_{1}) on one page and (x-h) on the other, and in one give the formula in terms of the distance between the end points of the line segment and in the other in terms of the square of the length of the radius.

But r = √( x^{2}+y^{2}) is the same as x^{2}+y^{2} = r^{2}.

In other words the formula for a circle is just the distance formula applied to the radius of the circle, and both sides of the equation are squared just to get rid of the root sign.

The other version of the equation of a circle on page 19 of the Formulae and Tables booklet

x^{2} + y^{2} + 2gx + 2fy + c = 0

is obtained directly from (x-h)^{2} + (y-k)^{2} = r^{2 } by renaming the centre coordinates (h,k) as (-g,-f), and expanding the left hand side to get

x^{2} + g^{2} + 2gx + y^{2 }+ f ^{2} + 2fy = r^{2}.

Then bringing the r^{2 }to the left hand side

x^{2} + g^{2} + 2gx + y^{2 }+ f ^{2} + 2fy – r^{2}.= 0

and gathering all the constant number terms into a single value c to give

x^{2} + y^{2} + 2gx + 2fy + c = 0.

So c = g^{2 }+ f ^{2 }– r^{2 . } Or r^{2. } = g^{2 }+ f ^{2}-c

*resulting in the given formula for the radius r = *√(g^{2 }+ f ^{2}-c)

**The Cosine Law**

We can also use Pythagoras’ theorem in the form of the distance formula to derive the **cosine law** given on page 16 of the F&T booklet quite easily. The law of cosines generalizes the Pythagorean theorem, which holds only for right triangles.

This theorem states that in any triangle with angles A,B and C and corresponding opposite sides a, b, and c,

a^{2} = b^{2 }+ c^{2 }– 2bc cos A

If A is 90^{0} then cos A is zero and this simplifies to a^{2} = b^{2 }+ c^{2 }for a right angle triangle.

First draw a scalene triangle and name the vertices A,B and C. The capital letters represent the angles and the small letters represent the side lengths that are opposite these angles.

Split triangle ABC into 2 right angled triangles by drawing a vertical line from vertex B down to the line AC. You can call this new line h (AD). You now have two right angled triangles (ABD and BCD).

Introduce a new variable x, which will be the length of the line AD in the left right angled triangle. Therefore the length of CD in the other right angled triangle will be b – x.

Next, use Pythagoras Theorem in both triangles to write down two formulas that give h²:

h² = c² – x² **(using triangle ABD)**

h² = a² – (b – x)² **(using triangle BCD)**

Combining these two formulae you get:

c² – x² = a² – (b-x)²

Expand and simplify the double bracket on the right hand side of the formula (be careful with your negatives):

c² – x² = a² – [(b-x)(b-x)]

c² – x² = a² – [b² – 2bx +x²]

c² – x² = a² – b² + 2bx – x²

Rearranging and cancelling out the x² on both sides you get:

a² = b² + c² – 2bx

All you need to do now is get rid of the letter x. This is done by using basic trigonometry in the triangle ABD. Since CosѲ = adjacent/hypotenuse, then x can be expressed as:

x = cCosA

Finally substitute x = cCosA into a² = b² + c² – 2bx:

a² = b² + c² – 2bcCosA

And this is your cosine rule proved.

The Theorem also leads to the trigonometric identity **Sin**^{2}**A + Sin**^{2}**B = 1**

SinA = a/c SinB =b/c

Sin^{2}A + Sin^{2}B = a^{2}/c^{2} + b^{2}/c^{2}

= (a^{2} + b^{2) }/c^{2}

= c^{2) }/c^{2}

= 1

And also since sin90° = 1, this also gives **Sin**^{2}**A + Sin**^{2}**B + Sin**^{2}**C= 2**

Wherever all three sides of a right triangle are integers, their lengths form a *Pythagorean triple* (or *Pythagorean numbers*). There is a general formula for obtaining all such numbers.

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Below this line only for those that are particularly interested and have lots of time to spare. If just want to pass exams ignore beyond this point.

Click here for a collection of over a 100 approaches to proving the theorem. Many of the proofs are accompanied by interactive Java illustrations. Fortunately you do not need to know all these. This is only shown to drive home that there can often be more than one right way to answer a question. For exams you only need to know the ones in your school text books.