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### Proof of Hero's formula

```
Date: 09/08/97 at 13:25:23
From: Russ Boyd
Subject: Hero's formula

Could you tell me where to find a proof of Hero's formula or help on
how to derive it?  I am a teacher and my class is working on it.
```

```
Date: 09/08/97 at 16:05:14
From: Doctor Wilkinson
Subject: Re: Hero's formula

The book "Journey through Genius" has a proof very much like Hero's.
You can derive it by brute force from the law of cosines.

-Doctor Wilkinson,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 09/08/97 at 16:35:37
From: Doctor Anthony
Subject: Re: Hero's formula

We use the formula  area = (1/2)bc.sin(A)

And sin(A) = 2 sin(A/2).cos(A/2)    so area = bc.sin(A/2).cos(A/2)

Next we find expressions for cos(A/2) and sin(A/2) in terms of
s, a, b, c, where a, b, c are the sides of the triangle and s is
the semi-perimeter.

So  s = (a+b+c)/2

b^2 + c^2 - a^2
cos(A) =   2cos^2(A/2) - 1    =   ----------------
2bc

b^2 + c^2 - a^2
2cos^2(A/2) = 1 + ------------------
2bc

(b+c)^2 - a^2      (b+c-a)(b+c+a)
=  --------------   = --------------
2bc                2bc

(2s-2a)2s
=   -----------
2bc

s(s-a)
cos^2(A/2) = ---------
bc

b^2 + c^2 - a^2
Next  cos(A) = 1 - 2 sin^2(A/2) = ----------------
2bc

b^2 + c^2 - a^2
2sin^2(A/2) = 1 - ----------------
2bc

a^2 - (b-c)^2      (a-b+c)(a+b-c)
=  --------------  =  --------------
2bc                  2bc

(2s-2b)(2s-2c)
=  --------------
2bc

(s-b)(s-c)
sin^2(A/2)  =  ------------
bc

We can now return to the formula for the area of the triangle:

(1/2)bc.sin(A) = bc.sin(A/2).cos(A/2)

bc.sqrt[s(s-a)(s-b(s-c)]
= ------------------------
bc

= sqrt[s(s-a)(s-b)(s-c)]

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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