heron’s formula

30
By Rehan.V HERON’S FORMULA

Upload: rehanvipin2000

Post on 27-May-2015

1.184 views

Category:

Education


0 download

DESCRIPTION

a full 30 slide ppt with all the important facts

TRANSCRIPT

Page 1: Heron’s formula

By Rehan.V

HERON’S FORMULA

Page 2: Heron’s formula

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

CONTENTS.

Page 3: Heron’s formula

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.

Page 4: Heron’s formula

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.

Page 5: Heron’s formula

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 ?

Page 6: Heron’s formula

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.

Page 7: Heron’s formula

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.

Page 8: Heron’s formula

WHY HERON’S

FORMULA ??

Why is heron’s formula necessary?

Page 9: Heron’s formula

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…..

Page 10: Heron’s formula

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

Page 11: Heron’s formula

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.

Page 12: Heron’s formula

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.

Page 13: Heron’s formula

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

Page 14: Heron’s formula

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.

Page 15: Heron’s formula

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.

Page 16: Heron’s formula

= \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 2s\cdot 2(s-c)\cdot 2(s-b)}{4c^2}\\

= \frac{4s(s-a)(s-b)(s-c)}{c^2}

CONTINUATION

Page 17: Heron’s formula

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

Page 18: Heron’s formula

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

EXAMPLES :

Page 19: Heron’s formula

 

SOLUTION OF EXAMPLE 1)

Page 20: Heron’s formula

 

CONTINUE…

Page 21: Heron’s formula

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:

Page 22: Heron’s formula

• 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)

Page 23: Heron’s formula

( ) / 2 (13 14 15 ) / 2 21s a b c m m m m

CONTINUE..

21(21 13)(21 14)(21 15)

 

21*8*7*6= 

Page 24: Heron’s formula

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

CONTINUE …

Page 25: Heron’s formula

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

Page 26: Heron’s formula

HOW TO FIND THE AREA OF AN EQUILATERAL TRIANGLE

Page 27: Heron’s formula

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

SOLUTION

Page 28: Heron’s formula
Page 29: Heron’s formula

ANY QUESTIONS?

Page 30: Heron’s formula

The end

THE END .