Cut the knot: learn to enjoy mathematics
A math books store at a unique math study site. Shopping at the store helps maintain the site. Thank you.
Math & English enrichment at SchoolPlus-Online
HoodaMath: games and movies
Sites for teachers
Sites for parents
Terms of use
Awards
Interactive Activities

CTK Exchange
CTK Wiki Math
CTK Insights - a blog
Math Help

III Millennium Olympiad

Games & Puzzles
What Is What
Arithmetic/Algebra
Geometry
Probability
Outline Mathematics
Make an Identity
Book Reviews
Stories for Young
Eye Opener
Analog Gadgets
Inventor's Paradox
Did you know?...
Proofs
Math as Language
Things Impossible
Visual Illusions
My Logo
Math Poll
Cut The Knot!
MSET99 Talk
Other Math sites
Front Page
Movie shortcuts
Personal info
Privacy Policy

Guest book
News sites

Recommend this site

Games to relax

Sites for teachers
Sites for parents

Education & Parenting

Manifesto: what CTK is about Buying a book is a commitment to learning Table of content Things you can find on CTK Chronology of updates Email to Cut The Knot Recommend this page

A Proof of the Pythagorean Theorem From Heron's Formula

Let the sides of a triangle have lengths a,b and c. Introduce the semiperimeter p = (a + b + c)/2 and the area S. Then Heron's formula asserts that

  S2 = p(p - a)(p - b)(p - c)

W.Dunham analyzes the original Heron's proof in his Journey through Genius.

For the right triangle with hypotenuse c, we have S = ab/2. We'll modify the right hand side of the formula by noting that

  p - a = (- a + b + c)/2, p - b = (a - b + c)/2, p - c = (a + b - c)/2

It takes a little algebra to show that

 
16S2= (a + b + c)(- a + b + c)(a - b + c)(a + b - c)
 = 2a2b2 + 2a2c2 + 2b2c2 - (a4 + b4 + c4)

For the right triangle, 16S2 = 4a2b2. So we have

  4a2b2= 2a2b2 + 2a2c2 + 2b2c2 - (a4 + b4 + c4)

Taking all terms to the left side and grouping them yields

  (a4 + 2a2b2 + b4) - 2a2c2 - 2b2c2 + c4 = 0

With a little more effort

  (a2 + b2)2 - 2c2(a2 + b2) + c4 = 0

And finally
  [(a2 + b2) - c2]2 = 0

Remark

For a quadrilateral with sides a, b, c and d inscribed in a circle there exists a generalization of Heron's formula discovered by Brahmagupta. In this case, the semiperimeter is defined as p = (a + b + c + d)/2. Then the following formula holds

  S2 = (p - a)(p - b)(p - c)(p - d)

Since any triangle is inscribable in a circle, we may let one side, say d, shrink to 0. This leads to Heron's formula.

References

  1. W. Dunham, Journey through Genius, Penguin Books, 1991

Copyright © 1996-2009 Alexander Bogomolny

33058406Page copy protected against web site content infringement by Copyscape


Search:
Keywords:

Google
Web CTK