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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