Pythagoras Theorem

c is the length of the hypotenuse (side opposite the right angle). 

a and b are the lengths of the other two sides.

It does not matter whether a is the height and b the base or vice versa .

b

ac

This theorem is based on right-angled triangles only. a2 + b2 = c2

The next set of slides prove the theorem.

Before the proof you need to know the following:

1) The area of a triangle = ½ base x height

All the triangles below have the same area because the base and height are the same

height

base

2a

2b

2cbase

he

igh

t

1. We start with half the red square, which hasArea = ½ base x height

222 cba

b c

a

2. We move one vertex while maintaining the base & height, so that the area remains the same. This is called a SHEAR.

3. We rotate this triangle, which does not change its area.

base

height

4. We mark the base and height for this triangle.

(Area of green square) + (Area of red square) = Area of the blue square

The Pythagorean Theorem:The Pythagorean Theorem:

2a

2b

2c

base

height

5. We now do a shear on this triangle, keeping the same area.

Remember that this pink triangle is half the red square.

Half t

he re

d squar

e.

222 cba

(Area of green square) + (Area of red square) = Area of the blue square

The Pythagorean Theorem:The Pythagorean Theorem:

1. We start with half the red square, which hasArea = ½ base x height

2. We move one vertex while maintaining the base & height, so that the area remains the same. This is called a SHEAR.

3. We rotate this triangle, which does not change its area.

4. We mark the base and height for this triangle.

222 cba

(Area of green square) + (Area of red square) = Area of the blue square

The Pythagorean Theorem:The Pythagorean Theorem:

2a

2b

2c

6. The other half of the red square has the same area as this pink triangle, so if we copy and rotate it, we get this.

So, together these two pink triangles have the samearea as the red square.

7. We now take half of the green square and transform it the same way.

Half t

he re

d squar

e.

Half t

he re

d squar

e.

We end up with this triangle, which is half of the green square.

Hal

f th

e g

reen

sq

uar

e.

Hal

f th

e g

reen

sq

uar

e.

9. Together, they have they samearea as the green square.

So, we have shown that the red & green squarestogether have the same area as the blue square.

Shear

Rotate

Shear

8. The other half of the green square would give us this.

2c

a2 + b2 = c2 Replace letters

52 + 122 = c2 Calculate

25 + 144 = c2

169 = c2 Swap sides

c2 = 169

c = √169 Use the √ button

c = 13

Example 1

12

5c

a2 + b2 = c2 Replace letters

92 + b2 = 152 Calculate

81 + b2 = 225

Subtract 81 from both sides

b2 = 225 – 81 Calculate

b2 = 144

b = √144 Use the √ button

b = 12

Example 2

b

915

a=4

b=8

Find side XY

Look at red triangle WUY and find the hypotenuse

a2 + b2 = c2 Replace letters

42 + 82 = c2 Calculate

16 + 64 = c2

c2 = 80

c = √80 = 8.94

c

X

Y

U

W

8.94

a=4

c

Find side XY

Look at the green triangle XWY and find the hypotenuse

a2 + b2 = c2 Replace letters

42 + 8.942 = c2 Calculate

16 + 80 = c2

c2 = 96

c = √96 = 9.8

X

Y

U

W

b = 8.94

c=10

b=5 5

Find side WZW

ZXY

Look at green triangle WYX and find the height

a2 + b2 = c2 Replace letters

a2 + 52 = 102 Calculate

a2 + 25 = 100

a2 = 100 – 25

a2 = 75

a = √75 = 8.66

8.66a

5 5

Find side WZW

ZXY

Look at red triangle WYZ and find the hypotenuse

a2 + b2 = c2 Replace letters

8.662 + 102 = c2 Calculate

75 + 100 = c2

175 = c2

c2 = √ 175

c = √175 = 13.2

a=8.66

b=10

A 5 metre ladder is placed against a wall. The base is 1.5 metres from the wall.How high up the wall does the ladder reach?

Draw a triangle and put in the numbers… Using a2 + b2 = c2

a=1.5

c=5

1.52 + b2 = 52

2.25 + b2 = 25

b = 22.75 = 4.77m (to 2 d.p.)

b

b2 = 25 – 2.25 = 22.75