1. Heron’s method
for finding
the area of a triangle
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2. 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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3. 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
15 cm
14 cm B A
13 cm
Heron showed that you only need the Pythagorean Theorem and some algebra
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4. Find the area of a triangle with side lengths 13 cm, 14 cm and 15 cm.
2 equations
2 unknowns, x and h
We can find h (and x ) and hence find the area of the triangle B A x
13 – x
13 cm
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5. Find the area of a triangle with side lengths 13 cm, 14 cm and 15 cm.2
15 cm
14 cm h B A x
13 – x
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6. Find the area of a triangle with side lengths 13 cm, 14 cm and 15 cm.2
15 cm
14 cm h B A x
13 – x
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7. Find the area of a triangle with side lengths 13 cm, 14 cm and 15 cm.2
15 cm
14 cm h B A x
13 – x
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8. Find the area of a triangle with side lengths 13 cm, 14 cm and 15 cm.2
15 cm
14 cm h B A x
13 – x
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9. Find the area of a triangle with side lengths 13 cm, 14 cm and 15 cm.2
Area =
x Base
x Height 1 2
168 13
Area =
x 13 x
15 cm
14 cm h
Area =
84 cm2 B A x
13 – x
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10. Take notes on another example
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11. Find the area of a triangle with side lengths 8 cm, 10 cm and 11 cm.2
11 cm
10 cm h B A x
8 – x
8 cm
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12. Find the area of a triangle with side lengths 8 cm, 10 cm and 11 cm.2
11 cm
10 cm h B A x
8 – x
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13. Find the area of a triangle with side lengths 8 cm, 10 cm and 11 cm.2
11 cm
10 cm h B A x
8 – x
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14. Find the area of a triangle with side lengths 8 cm, 10 cm and 11 cm.2
11 cm
10 cm h B A x
8 – x
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15. Find the area of a triangle with side lengths 8 cm, 10 cm and 11 cm.2
Area =
x Base
x Height 1 2 16
Area =
x 8 x
11 cm 2
10 cm h 4
Area =
≈ cm2 B A x
8 – x
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16. Heron’s Formula Η εξίσωση του Ήρωνα A = s (s – a ) (s – b ) (s – c )
where s = (a + b + c ) 1 2
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17. A = s (s – a ) (s – b ) (s – c ) where s = (a + b + c )
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 ) 1 2
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18. a triangle with sides 13 cm, 14 cm and 15 cm has area 84 cm2
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 ≈ cm2
Using Heron’s formula:
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19. a triangle with sides 13 cm, 14 cm and 15 cm has area 84 cm2
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 ≈ cm2
Using Heron’s formula:
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20. Heron’s Triangles
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21. 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] 13 5 12 17 4 25 6 30 8 7 60 84 3 5 15 24 37 41 20 29 9
180 12
210
210 40 35 21
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22. The answer is yes 84 cm2 Such triangles are known as Heron’s Triangles
The area of a right angled triangle whose 3 side lengths form a Pythagorean triple will always be an integer
Such triangles are known as Heron’s Triangles
They have integer side lengths and integer areas
These integer areas are always multiples of 6
They can be acute or obtuse
[it can be shown further that these areas are multiples of 6]
Can the area of a non right angled triangle whose 3 side lengths are integers, also be an integer?
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
14 cm
15 cm
84 cm2
13 cm
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23. Such triangles are known as Heron’s Triangles
They have integer side lengths and integer areas
These integer areas are always multiples of 6
They can be acute or obtuse
This is how you construct a Heron’s triangle:
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 40 25 24 84
384 7 32
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24. Such triangles are known as Heron’s Triangles
They have integer side lengths and integer areas
These integer areas are always multiples of 6
They can be acute or obtuse
We can always produce a second Heron triangle using the same 2 Pythagorean triangles:
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 40 25 24 25 7 7 32
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25. Another example of a pair of “Pythagorean triple” triangles added and subtracted to create a new pair of Heron’s triangles: 25 29 29 20 25 20 21 15 6 15 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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26. © T Madas