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/
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