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Heron’s methodfor finding

the area of a triangle

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Heron, also Hero ( c 1st century AD)Greek mathematician and inventor.

He devised many machines such as a fire engine, a water organ, several coin operated devices and the earliest known form of a steam engine, shown opposite.

He proved that the angle of incidence in optics is equal to the angle of reflection.

In Mathematics he is credited with a formula which gives the area of a triangle if its three sides are known.

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A B

C

13 cm

14 c

m

15 cm

Find the area of a triangle with side lengths 13 cm, 14 cm and 15 cm.

Usually we solve such problem by:

application of the cosine rule to find one of the angles

use of the formula for the area of the triangle

12

sinA ab C=

Heron showed that you only need the Pythagorean Theorem and some algebra

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x 13 – xA B

C

13 cm

14 c

m

15 cm

Find the area of a triangle with side lengths 13 cm, 14 cm and 15 cm.

h

2x 2h+ 214=2(13 )x- 2h+ 215=

2 equations

2 unknowns, x and h

We can find h (and x ) and hence find the area of the triangle

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2x 2h+ 215= 213- 26x+

x 13 – xA B

C

14 c

m

15 cm

Find the area of a triangle with side lengths 13 cm, 14 cm and 15 cm.

h

2x 2h+ 214=2(13 )x- 2h+ 215=

1

2213 26x- 2x+ 2h+ 215=

2x 2h+ 215= 213- 26x+

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2x 2h+ 215= 213- 26x+

x 13 – xA B

C

14 c

m

15 cm

Find the area of a triangle with side lengths 13 cm, 14 cm and 15 cm.

h

2x 2h+ 214=1

2

215= 213- 26x+214215- 213+ 26x214 =

225- 169+ 26x196 =

140 26x=14026

x =7013

x =

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196 169169´= 4900

169-2h

2x 2h+ 215= 213- 26x+

x 13 – xA B

C

14 c

m

15 cm

Find the area of a triangle with side lengths 13 cm, 14 cm and 15 cm.

h

2x 2h+ 214=1

2

214= 2x-2h

196=2

27013

-2h

196= 4900169

-2h196 169

169´= 4900

169-2h

7013

x =

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196 169169´= 4900

169-2h

2x 2h+ 215= 213- 26x+

x 13 – xA B

C

14 c

m

15 cm

Find the area of a triangle with side lengths 13 cm, 14 cm and 15 cm.

h

2x 2h+ 214=1

2

7013

x =

33124 4900169

-=2h

28224169

=2h

28224169

=h16813

h =

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x 13 – xA B

C

14 c

m

15 cm

Find the area of a triangle with side lengths 13 cm, 14 cm and 15 cm.

h

7013

x = 16813

h =

Area = 12 x Base x Height

Area = 12 x 13 x 168

13

Area = 84 cm2

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Take notes on another example

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2h+2x 2h+ 211= 28- 16x+

8 – xxA B

C

8 cm

10 c

m

11 cm

Find the area of a triangle with side lengths 8 cm, 10 cm and 11 cm.

h

2x 2h+ 210=2(8 )x- 211=

1

228 16x- 2x+ 2h+ 211=2x 2h+ 211= 28- 16x+

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2x 2h+ 211= 28- 16x+

8 – xxA B

C

10 c

m

11 cm

Find the area of a triangle with side lengths 8 cm, 10 cm and 11 cm.

h

2x 2h+ 210=1

2

211= 28- 16x+210211- 28+ 16x210 =

121- 64+ 16x100 =

43 16x=4316

x =4316

x =

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100 256256´= 210= 1849

256-2h2h 2x-

2x 2h+ 211= 28- 16x+

8 – xxA B

C

10 c

m

11 cm

Find the area of a triangle with side lengths 8 cm, 10 cm and 11 cm.

h

2x 2h+ 210=1

2

4316

x =

100=2

24316

-2h

100= 1849256

-2h100 256

256´= 1849

256-2h

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100 256256´= 1849

256-2h

2x 2h+ 211= 28- 16x+

8 – xxA B

C

10 c

m

11 cm

Find the area of a triangle with side lengths 8 cm, 10 cm and 11 cm.

h

2x 2h+ 210=1

2

4316

x =

25600 1849256

-=2h

23751256

=2h

23751256

=h 3 263916

=3 2639

16h =

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8 – xxA B

C

10 c

m

11 cm

Find the area of a triangle with side lengths 8 cm, 10 cm and 11 cm.

h

4316

x = 3 263916

h =

Area = 12 x Base x Height

Area = 12 x 8 x 3 2639

16

Area = ≈ 38.5 cm2

2 3 2639

4

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Heron’s Formula

Η εξίσωση του Ήρωνα

A = s (s – a ) (s – b ) (s – c )

where s = (a + b + c )12

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The method we used to calculate the area of a triangle given its 3 sides, can be generalised to a triangle with sides a, b and c.

We can obtain a general result known as Heron’s formula.

It states that the area of triangle of side lengths a, b and c is given by:

[The algebra for deriving the general case, is a bit more involved and has been omitted]

A = s (s – a ) (s – b ) (s – c )

where s = (a + b + c )12

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We found earlier that:

• a triangle with sides 13 cm, 14 cm and 15 cm has area 84 cm2

• a triangle with sides 8 cm, 10 cm and 11 cm has area ≈ 38.53 cm2

Using Heron’s formula:12

( )s a b c= + +

( )( )( )A s s a s b s c= - - -

Û 12

(13 14 15)s = + + Û 21s =

21(21 13)(21 14)(21 15)A = - - -

21 8 7 6A = ´ ´ ´

7056A =

84A =

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Using Heron’s formula:12

( )s a b c= + +

( )( )( )29 29 29 292 2 2 2

8 10 11A = - - -

Û 12

(8 10 11)s = + + Û 292

s =

( )( )( )A s s a s b s c= - - -

29 13 9 72 2 2 2

A = ´ ´ ´

2375116

A = 38.53»

We found earlier that:

• a triangle with sides 13 cm, 14 cm and 15 cm has area 84 cm2

• a triangle with sides 8 cm, 10 cm and 11 cm has area ≈ 38.53 cm2

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Heron’s Triangles

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The area of a right angled triangle whose 3 side lengths form a Pythagorean triple will always be an integer

3

4 5

5

12

13

15

817

24

725

40

941

35

1237

212

0 29

6 30 60 84

180 210 210

[it can be shown further that these areas are multiples of 6]

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The area of a right angled triangle whose 3 side lengths form a Pythagorean triple will always be an integer[it can be shown further that these areas are multiples of 6]

Such triangles are known as Heron’s TrianglesThey have integer side lengths and integer areasThese integer areas are always multiples of 6They can be acute or obtuse

Can the area of a non right angled triangle whose 3 side lengths are integers, also be an integer?

13 cm

14 c

m

15 cm

The answer is yes

We saw earlier that the acute triangle with sides 13 cm, 14 cm and 15 cm has an area of 84 cm2

84 cm2

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Such triangles are known as Heron’s TrianglesThey have integer side lengths and integer areasThese integer areas are always multiples of 6They can be acute or obtuse

This is how you construct a Heron’s triangle:

7

25

84

32

40

384

Start with a “Pythagorean triple” triangle

Pick one of its perpendicular sides

Place next to it another “Pythagorean triple” triangle with a perpendicular side equal to the one chosen.

The new triangle will be a Heron triangle

In this example:

a 7,24,25 with a 24,32,40 (3,4,5) with respective areas 84 and 384

produces a 25,39,40 Heron triangle with an area of 468

24

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Such triangles are known as Heron’s TrianglesThey have integer side lengths and integer areasThese integer areas are always multiples of 6They can be acute or obtuse

We can always produce a second Heron triangle using the same 2 Pythagorean triangles:

7

25

32

40

24 2

5

subtracting:

a 7,24,25 from a 24,32,40 (3,4,5) with respective areas 84 and 384

produces a 25,32,40 Heron triangle with an area of 300

7

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21

20

29

15

25

6 15

20

29

25

Another example of a pair of “Pythagorean triple” triangles added and subtracted to create a new pair of Heron’s triangles:

21

20,21,29 Pythagorean with area 210

15,20,25 Pythagorean with area 150

25,29,39 Heron with area 360 by adding

6,25,29 Heron with area 60, by subtracting

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