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Three Altitudes (Posted on 2007-01-15) Difficulty: 3 of 5
If the lengths of the altitudes of a triangle are 4, 5, and 6, what is the area of the triangle?

Can you generalize?

See The Solution Submitted by Dennis    
Rating: 3.0000 (1 votes)

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Solution | Comment 1 of 3

Let a, b, and c be the lengths of the sides
of the triangle and h_a, h_b, and h_c the
corresponding altitudes. Then,
  Area = a*h_a/2 = b*h_b/2 = c*h_c/2      (1)
If s is the semiperimeter of the triangle,
then by Heron's formula we have
  Area = sqrt(s(s-a)(s-b)(s-c))           (2)
Combining (1) and (2) with a lot of algebra,
we get
  Area = x*y*z/(4*sqrt(w(w-x)(w-y)(w-z)), 
  where
    x = h_a*h_b
    y = h_b*h_c
    z = h_c*h_a
    w = (x+y+z)/2
For our problem,
    x = 4*5 = 20
    y = 5*6 = 30
    z = 6*4 = 24
    w = (20+30+24)/2 = 37
    and 
    Area = 3600/sqrt(57239)
         ~= 15.04723
 

  Posted by Bractals on 2007-01-15 17:43:05
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