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(sa)(sb)(sc)) (2)
Combining (1) and (2) with a lot of algebra,
we get
Area = x*y*z/(4*sqrt(w(wx)(wy)(wz)),
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 20070115 17:43:05 