Area of Triangles When Altitudes Are Given

Date: 12/16/2002 at 02:39:51
From: Ridhi Kashyap
Subject: Area of Triangles when altitudes are given

Dear Dr. Math,

I was wondering how to find the area of a triangle when the lengths of 
all three altitudes are given. 

An example: Find the area of a triangle with altitudes of lengths 20 
cm, 28 cm. and 35 cm. 

Please help; the question keeps haunting me...
Thank you, 
Ridhi

Date: 12/16/2002 at 09:52:38
From: Doctor Floor
Subject: Re: Area of Triangles when altitudes are given

Hi, Ridhi,

Thanks for your question.

Let us denote the area of the triangle ABC by D, the sidelengths by 
a, b, c, and the corresponding altitudes by ha, hb, hc, respectively. 
Then we know that

  a*ha = b*hb = c*hc = 2D

and thus

  a/2 = D/ha
  b/2 = D/hb
  c/2 = D/hc.

We can use this to rewrite Heron's formula

 D = sqrt(s(s-a)(s-b)(s-c))

(where s = (a+b+c)/2) into

 D = sqrt[(D/ha + D/hb + D/hc)*(-D/ha + D/hb + D/hc)*
          (D/ha - D/hb + D/hc)*(D/ha + D/hb - D/hc) ]

   = D^2*sqrt[(1/ha + 1/hb + 1/hc)*(-1/ha + 1/hb + 1/hc)*
              (1/ha - 1/hb + 1/hc)*(1/ha + 1/hb - 1/hc) ]

so that we finally find the formula

 1/D = sqrt[(1/ha + 1/hb + 1/hc)*(-1/ha + 1/hb + 1/hc)*
            (1/ha - 1/hb + 1/hc)*(1/ha + 1/hb - 1/hc) ]

or

  D = 1/sqrt[(1/ha + 1/hb + 1/hc)*(-1/ha + 1/hb + 1/hc)*
             (1/ha - 1/hb + 1/hc)*(1/ha + 1/hb - 1/hc) ].

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/