# 03 04 Section 8 4 Area of Triangle by HC120831142657

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```									   Section 8.4
Area of a Triangle
1
Area  absin C
2
1
Area  6  8 sin 30
2
 12
Note: s is the “semi-perimeter.”
457
s       8
2
A  8(8  4)(8  5)(8  7)

 8(4)(3)(1)
4 6
Proof of Heron’s Formula
Note: Since Heron of Alexandria didn’t have the advantage of trigonometry,
this is NOT the proof he gave.
a           B
C
c
b
A

1
Area    absin C
2
1
Let   C
2
Then C  2

And sin C  sin 2  2 sin  cos
1         C    C
Area  ab 2 sin cos
2         2    2
Proof of Heron’s Formula
Note: Since Heron of Alexandria didn’t have the advantage of trigonometry,
this is NOT the proof he gave.
a 2  b2  c2
a           B                                     1
C                                             1  cosC               2ab
So             
c                                2                  2
b
a 2  2ab  b 2  c 2      ( a  b) 2  c 2
A                                                    
4ab                     4ab
C     C                   (a  b  c)(a  b  c)
Area  ab sin      cos                   
2     2                            4ab
abc     
 ab
1  cos C 1  cos C             But a  b  c  2       c   2s  c 
2         2                                    2      
abc
a2  b2  c2             And a  b  c  2        2s
But cos C                                                2 
2ab
1  cosC 2( s  c)2s                ( s  c) s
So                                  
2        4ab                        ab
Proof of Heron’s Formula
Note: Since Heron of Alexandria didn’t have the advantage of trigonometry,
this is NOT the proof he gave.
1  cosC ( s  c) s
C
a           B                                  
2        ab
c
b
In a very similar manner, we can prove that
A
1  cosC ( s  a)(s  b)
C     C                            
Area  ab sin      cos                          2           ab
2     2
1  cos C 1  cos C
1  cos C 1  cos C               So Area  ab
 ab                                                     2         2
2         2
a2  b2  c2                               ( s  a )( s  b) ( s  c) s
But cos C                                       ab
2ab                                           ab             ab

 s(s  a)(s  b)(s  c)

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