Heron's formula provides a way to calculate the area of a triangle using only the lengths of its three sides. It was developed by the ancient Greek mathematician Heron of Alexandria around 60 AD. The formula was an important advancement as it allowed calculating the area of triangles without knowing the height. Over time, the formula has been expressed in different but equivalent ways and was independently discovered by Chinese mathematicians centuries later. Heron's formula remains useful for solving geometry problems and calculating areas whenever the triangle height is unknown.

1. By Rehan.V
HERON’S FORMULA

2. 1. Introduction.
2. About heron.
3. Evolution of it.
4. Proof of Heron’s formula.
5. 3 practical solutions.
CONTENTS.

3.  Heron's formula is named after Hero of Alexandria, a
Greek Engineer and Mathematician in 10 - 70 AD.
 You can use this formula to find the area of a
triangle using the 3 side lengths.
Therefore, you do not have to rely on the formula for
area that uses base and height.
 You only have to know the perimeter of the triangle
and the side’s measurement.
INTRODUCTION.

4.  This formula was founded by a Greek mathematician
called Heron in the year 60 A.D.
 It was invented to find the area of triangles.
 This was also mentioned in this book named Metrica.
 It was also mentioned in the book of the scientist
Archimedes but only in Heron’s book was it properly
understood by people.
 That’s why all the credit of the formula is given to
Heron.
INTRODUCTION.

5.  Heron of Alexandria (c. 10–70 AD) was an
ancient Greek mathematician and engineer who
was active in his native city of Alexandria, Roman
Egypt.
 He is considered the greatest experimenter of
antiquity and his work is representative of the
Hellenistic scientific tradition.
 Heron published a well recognized
description of a steam-power
device called an ’’aeolipile” .
WHO WAS HERON ?

6.  Heron was also known as the hero of Alexandria or
just hero.
 There were more than 18 scientists known as hero at
that time.
 Much of Hero's original writings and designs have
been lost, but some of his works were preserved in
Arab manuscripts.
 He is said to have been a follower of the Atomists.
Some of his ideas were derived from the works
of Ctesibius.
ABOUT HERON.

7.  He had at least 13 works on mathematics,
mechanics and physics.
 He devised many mechanical devices which actually
worked practically. One of them was his steam
engine.
 He made and wrote a method approximating square
and cube roots of numbers that are not perfect
squares or cubes.
 He was also one of the first to make a fountain
which shoots jets of water upwards.
MORE ABOUT HERON.

8. WHY HERON’S
FORMULA ??
Why is
heron’s
formula
necessary?

9. AREA OF SCALENE TRIANGLE
 In this triangle it is
impossible to find the
height which is
necessary to find the
are by the formula:
½ x (height) x(base)
 Hence we need the
Heron’s formula…..

10.  Heron’s formula was actually like this
I. Square root of (s (s-a)(s-b)(s-c) )
II. Where s stood for semi- perimeter.
III. a ,b and c stood for the sides.
IV. First you have to know the complete perimeter of
the triangle.
THE FORMULA

11.  Heron’s formula can be used to find out the area of a
triangle in case its height is unknown.
 The formula can be used for a scalene triangle in
which the height doesn’t definitely exist. The
formulas can also be used to find the area of
rhombus when only one of the diagonal and the
perimeter are only known .
 There are many places where the formula can be
used for example the area of flyovers.
THE USE OF THIS FORMULA.

12. The formula is credited to Heron (or Hero) of
Alexandria, and a proof can be found in his
book, Metrica, written c. A.D. 60. It has been
suggested that Archimedes knew the formula over
two centuries earlier, and since Metrica is a
collection of the mathematical knowledge
available in the ancient world, it is possible that the
formula predates the reference given in that work.
A formula equivalent to Heron's namely:
, where was
discovered by the Chinese independently of the
Greeks. It was published in Shushu
Jiuzhang ,written by Qin Jiushao and published
in A.D. 1247.

13.  The heron’s formula has not always been root of (s
(s-a)(s-b)(s-c) )
 It has been changed from a simple two step formula
to this current formula.
 The first formula that Heron worked through and
founding of the current formula are different.
 In the next slide this will be explained .
 The first derivation of the formula requires a lot of
higher studies so I will only explain the basic.
EVOLUTION OF THE FORMULA

14.  The Heron’s formula has many other methods in
which it can be written in these are a few methods –
 A = ¼ square root of (a+b+c)(-a+b+c)(a-b+c)(a+b-c).
 A = ¼ square root of 2(a^2*b^2+a^2*c^2+c^2*b^2)-
(a^4+b^4+c^4)
It was derived from the Chinese in the form of –
 A=1/2 square root of a^2*c^2-(a^2+c^2-b^2/2)
 Since there was no trigonometry at that time it was
very hard to prove the formula back in heron’s time.
THE CHANGING OF THE FORMULA.

15.  We can only prove Heron’s formula through advanced
techniques so most of us wouldn’t understand it but we
can also try it with Pythogaros theorem.
 By the Pythagorean theorem we have b^2=h^2+d^2 and
a^2=h^2+(c-d)^2 according to the figure at the right.
Subtracting these yields a^2-b^2=c^2-2cd. Thus
 d=frac{-a^2+b^2+c^2}{2c}.
 Then we get for the height of the triangle that
 h^2 & = b^2-d^2=left(frac{2bc}{2c}right)^2-
left(frac{-a^2+b^2+c^2}{2c}right)^2
THE PROOF OF THIS FORMULA.

16.  = frac{(2bc-a^2+b^2+c^2)(2bc+a^2-b^2-
c^2)}{4c^2}
 = frac{((b+c)^2-a^2)(a^2-(b-c)^2)}{4c^2}
 = frac{(b+c-a)(b+c+a)(a+b-c)(a-b+c)}{4c^2}
 = frac{2(s-a)cdot 2scdot 2(s-c)cdot 2(s-
b)}{4c^2}
 = frac{4s(s-a)(s-b)(s-c)}{c^2}
CONTINUATION

17.  We now apply this result to the formula for the area
A of the triangle that involves a height, in this case
height h from side c:
 A & = {ch}{2}
 & = sqrt{frac{c^2}{4}cdot frac{4s(s-a)(s-b)(s-
c)}{c^2}}
 & = sqrt{s(s-a)(s-b)(s-c)}
CONTINUATION

18. • 1) Find the area of a triangle having sides :
AB = 4 cm
BC = 3 cm
CD = 5 cm
EXAMPLES :

19. 
SOLUTION OF EXAMPLE 1)

20. 
CONTINUE…

21. 2) Rahul has a garden, which is triangular in shape.
The sides of the garden are 13 m, 14 m, and 15 m
respectively. He wants to spread fertilizer in the
garden and the total cost required for doing it is Rs 10
per m2. He is wondering how much money will be
required to spread the fertilizer in the garden
EXAMPLE 2:

22. • Given a = 13 m , b = 14 m and c = 15 m
So , we will find
the area of the
triangle by
using Heron’s
formula.
SOLUTION OF EXAMPLE 2)

23. CONTINUE..
21(21 13)(21 14)(21 15)   21*8*7*6=

24.  Given the rate = Rs 10 per sq. m .
 Now :
Total cost = Rs. 10 * 84 = Rs 840/-
CONTINUE …

25.  What equilateral triangle would have the same area as a
triangle with sides 6, 8 and 10?
 This is not the same as the normal questions and will require
to know the area of the area of the equilateral triangle.
CONCEPT BASED QUESTION

26. HOW TO FIND THE AREA OF AN EQUILATERAL
TRIANGLE

27.  First of all we will find the area of the triangle having sides : a
= 6 units , b = 8 units and c = 10 units
SOLUTION

29. ANY
QUESTIONS
?

30. The end
THE END .