brief history finding the shorter side a pythagorean puzzle pythagoras’ theorem using...

Brief History

Finding the shorter side

A Pythagorean Puzzle

Pythagoras’ Theorem

Using Pythagoras’ Theorem

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Further examples

Pythagoras was a Greek philosopher and religious leader.He was responsible for many important developments in maths,

astronomy, and

music.

Pythagoras (~560-480 B.C.)

His students formed a secret society called the Pythagoreans.

As well as studying maths, they were a political and religious organisation.

Members could be identified by a five pointed star they wore on their clothes.

The Secret Brotherhood

They had to follow some unusual rules. They were not allowed to wear wool, drink wine or pick up anything they had dropped! Eating beans was also strictly forbidden!

The Secret Brotherhood

A right angled triangle

A Pythagorean Puzzle

Ask for the worksheet and try this for yourself!

© R Glen 2001

Draw a square on each side.

A Pythagorean Puzzle

© R Glen 2001

xy

z

Measure the length of each side

A Pythagorean Puzzle

© R Glen 2001

Work out the area of each square.

A Pythagorean Puzzle

x

z

y

x²

y²

z²

© R Glen 2001

A Pythagorean Puzzle

x²

y²

z²

© R Glen 2001

A Pythagorean Puzzle

© R Glen 2001

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A Pythagorean Puzzle

© R Glen 2001

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2

A Pythagorean Puzzle

© R Glen 2001

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2

A Pythagorean Puzzle

© R Glen 2001

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2

3

A Pythagorean Puzzle

© R Glen 2001

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2

3

A Pythagorean Puzzle

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4 A Pythagorean Puzzle

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23

4

A Pythagorean Puzzle

© R Glen 2001

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45

A Pythagorean Puzzle

© R Glen 2001

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4

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What does this tell you about the areas of the three squares?

The red square and the yellow square together cover the green square exactly.The square on the longest side is equal in

area to the sum of the squares on the other two sides.

A Pythagorean Puzzle

© R Glen 2001

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23

4

5

Put the pieces back where they came from.

A Pythagorean Puzzle

© R Glen 2001

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45

A Pythagorean Puzzle

Put the pieces back where they came from.

© R Glen 2001

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A Pythagorean Puzzle

Put the pieces back where they came from.

© R Glen 2001

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2

3

4

5

A Pythagorean Puzzle

Put the pieces back where they came from.

© R Glen 2001

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23

4

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A Pythagorean Puzzle

Put the pieces back where they came from.

© R Glen 2001

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4

5

A Pythagorean Puzzle

Put the pieces back where they came from.

© R Glen 2001

This is called Pythagoras’ Theorem.

A Pythagorean Puzzle

x²

y²

z²

x²=y²+z²

© R Glen 2001

It only works with right-angled triangles.

hypotenuse

The longest side, which is always opposite the right-angle, has a special name:

This is the name of Pythagoras’ most famous discovery.

Pythagoras’ Theorem

x

z

y

x²=y²+z²

Pythagoras’ Theorem

x

y

x

x

y

y

z z

z

x

yz

Pythagoras’ Theorem

x²=y²+z²

1m

8m

Using Pythagoras’ Theorem

What is the length of the slope?

1m

8m

x

z=

y=

x²=y²+ z²

x²=1²+ 8²

x²=1 + 64

x²=65

?

Using Pythagoras’ Theorem

How do we find x?

We need to use the

square root button on the calculator.It looks like this √

Press

x²=65

√ , Enter 65 =

So x= √65 = 8.1 m (1 d.p.)

Using Pythagoras’ Theorem

Example 1

x

12cm

9cm

y

zx²=y²+ z²

x²=12²+ 9²

x²=144 + 81

x²= 225

x = √225= 15cm

x

6m4m

s

yz

x²=y²+ z²

s²=4²+ 6²

s²=16 + 36

s²= 52

s = √52

=7.2m (1 d.p.)

Example 2

What’s in the box?

24cm

7cm

25 cm

7m5m

8.6 m to 1 dp

Problem 1

Problem 2

7m

5m

hx

y

z

x²=y²+ z²

7²=h²+ 5²

49=h² + 25?

Finding the shorter side

49 = h² + 25

We need to get h² on its own.Remember, change side, change sign!

Finding the shorter side

+ 25

49 - 25 = h²

h²= 24

h = √24 = 4.9 m (1 d.p.)

169 = w² + 36

x

w

6m

13m

y

z

x²= y²+ z²

13²= w²+ 6²

169 – 36 = w²

w = √133 = 11.5m (1 d.p.)

w²= 133

Example 1

169 = w² + 36

Change side, change sign!

x

z x²= y²+ z²

11²= 9²+ PQ²

121 = 81 + PQ²

121 – 81 = PQ²

PQ = √40 = 6.3cm (1 d.p.)

PQ²= 40

y9cm

P

11cm

R

Q

Example 2

81

Change side, change sign!

What’s in the box 2?

9cm

A

11cm

C

B

6.3 cm to 1 dp

9m

4.5m

7.8m to 1 dp

Problem 1

Problem 2

x

y

z

x²=y²+ z²

r²=5²+ 7²

r²=25 + 49

r²= 74

r = √74

=8.6m (1 d.p.)

14m

5mr

r5m

7m

Example 1

½ of 14

?

x

y

z

23cm

38cm

p

38cm

23cm

x²= y²+ z²

38²= y²+ 23²1444 = y²+ 5291444 – 529 = y²

y = √915=30.2

y²= 915

So p =2 x 30.2 = 60.4cm

Example 2

+ 529

Change side, change sign!

What’s in the box 3?

20m

8mrr = 12.8m to 1 dp

30cm

42cm

pp = 58.8m to 1 dp

Problem 1

Problem 2