heron's formula

HERON’S FORMULA

By – Dhruv JhaRounak Jha

IX - F

Contents

Heron’s FormulaAbout Heron Needs of

Heron’s Formula

Area of Triangles and some

other shapes

Questions

Born / Died When Heron lived is not well established. Most accounts place his life around 0 AD.

Heron, (also known as Hero) was a Greek mathematician. Some authorities place his birthday early 150 BCE in Ptolemaic, Egypt. While other scholars have dated his birth to be 250 CE in late Roman Empire. Nothing is really known of Hero's life, but what we do know comes from clues in the 14 known books by him.

About Heron

As a student, Hero spent most of his time in the Library at the University of Alexandria. He loved to be in the library, because of the series of gardens vast collection of books. Hero was strongly influenced by the writings of Ctesibius of Alexandria. He may have been a student of Ctesibius. When older he taught at the University of Alexandria, and taught mathematics, mechanics, and physical science. He wrote many books and he used them as texts for his students, and manuals for technicians, and were written in Greek, Latin and Egyptian.

About Heron

Some of Heron's books Baroulkos Berlopoeica (in Greek and Roman Artillery, Technical Treatises, 1971) Catoptrica (in Latin) Chieroballistra (in Greek and Roman Artillery; Technical Treatises,

1971) Dioptra (partical English translation, 1963) Eutocuis Geometrica Mechanica (3 volumes, in Arabic) Metrica (3 volumes) Peri Automatopoitikes (Automata, 1971) Peri Metron (also called Mensurae) Pneumatica (2 volumes: The Pneumatica of Hero Of Alexandria, 1851) Stereometrica

About Heron

One of his books, Metrica, was lost until the end of the 19th century. Scholars knew of its existence only through one of his other books, Eutocuis. In 1894, historian Paul Tannery discovered a fragment of the book in Paris. Then, in 1896, R. Schone found a complete copy in Constantinople. This book is the most famous book that Hero wrote. It consists of 3 volumes, and shows ways to calculate area and volume, and their divisions.

About Heron

He was an accomplished inventor and mechanical engineer. Among his inventions were a reaction steam turbine, a vending machine, and a wind-powered organ.

About Heron

i)area=1/2 (b x h) Let there be a scalene triangle, the lengths of its

sides are known but the height is not known. To find its area none of above listed formulae is applicable. In fact, we require the height corresponding to a base. But we do not have any clue for the same. Heron , an encyclopedic writer in Applied Mathematics gave a formula for finding area of triangle in terms of lengths of its three sides as discussed in next slide.

Herons Formula

o We can calculate the area of a triangle if we know the lengths of all three sides, using a formula that has been known for nearly 2000 years.

o It is called "Heron's Formula" after Hero of Alexandria

o The formula can be written as :o S=

How to Calculate Area Of Triangle Using Herons Formula?

Herons formula can be used to measure the area of a triangle whose sides are given. It helps you to find the area of a triangle where the height is not given. This includes scalene, isosceles and equilateral triangles

It can be used in our daily life in the following ways.-(i) When we buy a piece of land we can find its area by

using herons formula(ii) Imagine someone gave you a triangular figure(it

can be 2d or 3d) now u can find its area by just applying herons formula.

Need of Heron's Formula

Example : If a farmer wishes to find the area of his field which is in shape of quadrilateral. He needs to divide the quadrilateral in triangular parts and he uses the Heron’s formula for the area of a triangular part.

Application Of Herons Formula

Area of a right angled triangle

AREA OF AN EQUILATERAL TRIANGLE

CONTINUED

Continued

Area of an isosceles triangle

Continued

Formulas of some shapes which are related to triangles

RECTANGLE If l and b denote respectively the length and breadth of a rectangle, then

(i) Perimeter = 2(l + b)

(ii) Area = l x b

If a denote the length of each side of a square , then

(i) Perimeter = 4a

(ii) Area = a² = (side)²

(iii) Area = ½ (Diagonal)²

SQUARE

(i) Perimeter = 2(AB + BC) = 2(l + b )

(ii) Area = Base x Height

Parallelogram

If d1 and d2 are the lengths of the diagonals of a rhombus of side a, then

(i) Perimeter = 4a = 4(side)

(ii) Area = ½ (d1 x d2)

Rhombus

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

Examples :

Given Sides : AB = a = 4 cm AC = b = 3 cm BC = c = 5 cm

Solution of Example 1

Continued

Continued

Q2 Find the area of a quadrilateral ABCD in which AB = 3 cm, B = 90°CD = 4 cm, DA = 5 cm, and AC = 5 cm.

5 cm

5 cm

4 cm

4 cm

3 cm

D

A B

C

Area of ∆ ABC = ½ × AB × BC = ( ½ × 3 × 4) cm² = 6 cm² For Δ ACD: Let a = 5 cm, b = 4 cm and c

= 5 cm. then, s = ½ × (a + b + c) s = ½ ( 5 + 4 + 5 ) cm s = ½ × 14 cm s = 7 cm

Solution

Now, s – a = ( 7 – 5 ) = 2 cm s – b = ( 7 – 4 ) = 3 cm s – c = ( 7 – 5 ) = 2 cm Area of ∆ ACD = √s(s — a)(s —b)(s—c) = √ 7 (2) (3) (2) = √ 2 × 2 × 3 × 7 = 2 × √3 ×√7 = 2 × √21 = 2 × 4.58 = 9.167 /- Area of quadrilateral ABCD = Area of

∆ABC + Area of ∆ ACD = 6 + (9.16) cm² = 15.2 cm² (approx.)