Pythagoras’ theorem
mc-TY-pythagoras-2009-1
Pythagoras’ theorem is well-known from schooldays. In this unit we revise the theorem and use
it to solve problems involving right-angled triangles. We will also meet a less-familiar form of the
theorem.
In order to master the techniques explained here it is vital that you undertake plenty of practice
exercises so that they become second nature.
After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
• state Pythagoras’ theorem
• use Pythagoras’ theorem to solve problems involving right-angled triangles.
Contents
1. Introduction 2
2. The theorem of Pythagoras a2 + b2 = c2 2
3. A further application of the theorem 5
4. Applications in cartesian geometry 6
5. A final result: 7
www.mathcentre.ac.uk 1 c mathcentre 2009
, 1. Introduction
The Theorem of Pythagoras is a well-known theorem. It is also a very old one, not only does it
bear the name of Pythagoras, an ancient Greek, but it was also known to the ancient Babylonians
and to the ancient Egyptians. Most school students learn of it as a2 + b2 = c2 . The actual
statement of the theorem is more to do with areas. So, let’s have a look at the statement of
the theorem.
2. The Theorem of Pythagoras
The theorem makes reference to a right-angled triangle such as that shown in Figure 1. The
side opposite the right-angle is the longest side and is called the hypotenuse.
hyp
ote
nus
e
Figure 1. A right-angled triangle with hypotenuse shown.
What the theorem says is that the area of the square on the hypotenuse is equal to the sum of the
areas of the squares on the two shorter sides. Figure 2 shows squares drawn on the hypotenuse
and on the two shorter sides. The theorem tells us that area A + area B = area C.
C
A c
a
b
B
Figure 2. A right-angled triangle with squares drawn on each side.
An excellent demonstration of this is available on the accompanying video. If we denote the
lengths of the sides of the triangle as a, b and c, as shown, then area A = a2 , area B = b2 and
area C = c2 . So, using Pythagoras’ theorem
area A + area B = area C
a2 + b2 = c2
This is the traditional result.
www.mathcentre.ac.uk 2 c mathcentre 2009
mc-TY-pythagoras-2009-1
Pythagoras’ theorem is well-known from schooldays. In this unit we revise the theorem and use
it to solve problems involving right-angled triangles. We will also meet a less-familiar form of the
theorem.
In order to master the techniques explained here it is vital that you undertake plenty of practice
exercises so that they become second nature.
After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
• state Pythagoras’ theorem
• use Pythagoras’ theorem to solve problems involving right-angled triangles.
Contents
1. Introduction 2
2. The theorem of Pythagoras a2 + b2 = c2 2
3. A further application of the theorem 5
4. Applications in cartesian geometry 6
5. A final result: 7
www.mathcentre.ac.uk 1 c mathcentre 2009
, 1. Introduction
The Theorem of Pythagoras is a well-known theorem. It is also a very old one, not only does it
bear the name of Pythagoras, an ancient Greek, but it was also known to the ancient Babylonians
and to the ancient Egyptians. Most school students learn of it as a2 + b2 = c2 . The actual
statement of the theorem is more to do with areas. So, let’s have a look at the statement of
the theorem.
2. The Theorem of Pythagoras
The theorem makes reference to a right-angled triangle such as that shown in Figure 1. The
side opposite the right-angle is the longest side and is called the hypotenuse.
hyp
ote
nus
e
Figure 1. A right-angled triangle with hypotenuse shown.
What the theorem says is that the area of the square on the hypotenuse is equal to the sum of the
areas of the squares on the two shorter sides. Figure 2 shows squares drawn on the hypotenuse
and on the two shorter sides. The theorem tells us that area A + area B = area C.
C
A c
a
b
B
Figure 2. A right-angled triangle with squares drawn on each side.
An excellent demonstration of this is available on the accompanying video. If we denote the
lengths of the sides of the triangle as a, b and c, as shown, then area A = a2 , area B = b2 and
area C = c2 . So, using Pythagoras’ theorem
area A + area B = area C
a2 + b2 = c2
This is the traditional result.
www.mathcentre.ac.uk 2 c mathcentre 2009
