Area of Triangles

Formulas:

K = 1/2 bc sin A

K = 1/2 ab sin C

K = 1/2 ac sin B

   PROOF:

  Draw a triangle on a rectangular coordinate system, and draw in its altitude.
  Label the base AC   with the origin being A,  and the upper vertex B. B has
    coordinates (x,y).

  The other three formulas can be obtained in a similar way.

Example:

Find the area of the triangle with A = 40o, b = 10 cm, and c = 14 cm.

K = 1/2 (10)(14)sin 40o

K = 44.995 sq cm
 

Hero's Formula:


          where  s = a + b + c
                                 2

Proof:

K = (1/2)bc sin A
K2 = (1/4)b2c2sin2A
     = (1/4)b2c2(1 - cos2A)
     = (1/4)b2c2(1 - cos A) (1 + cos A)
            Since a2 = b2 + c2 - 2bc cos A
                     a2 - b2 - c2 = - 2bc cos A


      = (1/16) (2bc -  b2 - c2+ a2 )(2bc +  b2 + c2 - a2 )
      = (1/16) [a2 -(b2 - 2bc + c2)] [(b2 + 2bc + c2) - a2 ]
      = (1/16) [a - (b - c)] [a + (b - c)] [b + c - a] [b + c + a]
      = (1/16) [a - b + c] [a + b - c] [b + c - a] [b + c + a]
      = (1/16) [a + b + c - 2b] [a + b + c - 2c] [a + b + c - 2a] [a + b + c]
                Let 2s = a + b + c
      = (1/16)(2s - 2b) (2s - 2c) (2s - 2a) (2s)
      = (1/16)[2(s - b)] [2(s - c)] [2(s - a)] (2s)
      = (s - b)(s - c)(s - a)s
      = s(s - a)(s - b)(s - c)

    where      s = a + b + c
                                                                        2


Problems


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