If the lengths of the altitudes of a triangle are 4, 5, and 6, what is the area of the triangle?

Can you generalize?

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